It is also a matrix and also an array; all scalars are also vectors, and all scalars are also matrix, and all scalars are also array You can put this solution on YOUR website! See the picture below. Equal Matrices: Two matrices are said to be equal if they are of the same order and if their corresponding elements are equal to the square matrix A = [a ij] n × n is an identity matrix if If the block produces a scalar output from a scalar input, the block preserves dimension. If you multiply any number to a diagonal matrix, only the diagonal entries will change. 8) Unit or Identity Matrix. Nonetheless, it's still a diagonal matrix since all the other entries in the matrix are . A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. Back in multiplication, you know that 1 is the identity element for multiplication. The following rules indicate how the blocks in the Communications Toolbox process scalar, vector, and matrix signals. The column (or row) vectors of a unitary matrix are orthonormal, i.e. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. However, there is sometimes a meaningful way of treating a $1\times 1$ matrix as though it were a scalar, hence in many contexts it is useful to treat such matrices as being "functionally equivalent" to scalars. While off diagonal elements are zero. Long Answer Short: A $1\times 1$ matrix is not a scalar–it is an element of a matrix algebra. A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. [] is not a scalar and not a vector, but is a matrix and an array; something that is 0 x something or something by 0 is empty. Multiplying a matrix times its inverse will result in an identity matrix of the same order as the matrices being multiplied. Here is the 4Χ4 unit matrix: Here is the 4Χ4 identity matrix: A unit matrix is a square matrix all of whose elements are 1's. Yes it is. All the other entries will still be . The same goes for a matrix multiplied by an identity matrix, the result is always the same original non-identity (non-unit) matrix, and thus, as explained before, the identity matrix gets the nickname of "unit matrix". In this post, we are going to discuss these points. 2. Okay, Now we will see the types of matrices for different matrix operation purposes. For an example: Matrices A, B and C are shown below. References [1] Blyth, T.S. In the next article the basic operations of matrix-vector and matrix-matrix multiplication will be outlined. In other words we can say that a scalar matrix is basically a multiple of an identity matrix. #1. It is never a scalar, but could be a vector if it is 0 x 1 or 1 x 0. An identity matrix is a square matrix whose upper left to lower right diagonal elements are 1's and all the other elements are 0's. and Robertson, E.F. (2002) Basic Linear Algebra, 2nd Ed., Springer [2] Strang, G. (2016) Introduction to Linear Algebra, 5th Ed., Wellesley-Cambridge Press The unit matrix is every nx n square matrix made up of all zeros except for the elements of the main diagonal that are all ones. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. Basis. This topic is collectively known as matrix algebra. The scalar matrix is basically a square matrix, whose all off-diagonal elements are zero and all on-diagonal elements are equal. If a square matrix has all elements 0 and each diagonal elements are non-zero, it is called identity matrix and denoted by I. 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# Binomial theorem

Series binomial theorem series contents page contents binomial theorem notation n as a nonnegative integer proof of the binomial theorem proof when n and k are. The binomial theorem date_____ period____ find each coefficient described 1) coefficient of x2 in expansion of (2 + x)5 80 2) coefficient of x2 in expansion. A polynomial with two terms is called a binomial we have already learned to multiply binomials and to raise binomials to powers, but raising a binomial to a high. Posts about binomial theorem written by ujjwal gulecha. Yes, pascal's triangle and the binomial theorem isnâ€™t particularly exciting but it can, at least, be enjoyable we dare you to prove us wrong.

Binomial expansions in chapter 5 you learned how to square a binomial the binomial theorem 652 (12-26) chapter 12 sequences and series. Binomial theorem the binomial theorem states that the binomial coefficients $$c(n,k)$$ serve as coefficients in the expansion of the powers of the binomial $$1+x$$. Binomial theorem : akshay mishra xi a , k v 2 , gwalior in elementary algebra, the binomial theorem describes the algebraic expansion of powers of a binomial. How to use the binomial theorem to expand binomial expressions, examples and step by step solutions, the binomial theorem using combinations. Quick links: downloadable teaching materials for binomial theorem syllabus content for the algebra topic: sl syllabus (see syllabus section 13) hl syllabus (see.

Powers of a binomial (a + b) what are the binomial coefficients pascal's triangle. In this video lesson, you will see what the binomial theorem has in common with pascal's triangle learn how you can use pascal's triangle to help. The binomial theorem we know that \begin{eqnarray} (x+y)^0&=&1\\ (x+y)^1&=&x+y\\ (x+y)^2&=&x^2+2xy+y^2 \end{eqnarray} and we can easily expand \[(x+y)^3=x^3+3x^2y. Expanding a binomial expression that has been raised to some large power could be troublesome one way to solve it is to use the binomial theorem.

Explains how to use the binomial theorem, and displays the theorem's relationship to pascal's triangle. Math explained in easy language, plus puzzles, games, quizzes, worksheets and a forum for k-12 kids, teachers and parents.

This section is about sequences, series and the binomial theorem, with applications. Binomial theorem was known for the case n = 2 by euclid around 300 bc, and pascal stated it in modern form in 1665 newton showed that a similar formula for negative.

## Binomial theorem

In this lesson, students will learn the binomial theorem and get practice using the theorem to expand binomial expressions the theorem is broken. Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. Binomial theorem: binomial theorem, statement that describes the nth power of the sum of two numbers (a + b.

When a binomial is raised to whole number powers, the coefficients of the terms in the expansion form a pattern. Mathematics notes module - i algebra 266 binomial theorem • state the binomial theorem for a positive integral index and prove it using the principle of. The most basic example of the binomial theorem is the formula for the square of x + y: (+) = + + the binomial coefficients 1, 2, 1 appearing in this expansion. Fun math practice improve your skills with free problems in 'binomial theorem i' and thousands of other practice lessons. 123 applications of the binomial theorem expansion of binomials the binomial theorem can be used to find a complete expansion of a power of a binomial or a. There are several closely related results that are variously known as the binomial theorem depending on the source even more confusingly a number of these (and other. While the foil method can be used to multiply any number of binomials together, doing more than three can quickly become a huge headache.

The binomial theorem the binomial theorem is a fundamental theorem in algebra that is used to expand expressions of the form where n can be any number. Binomial theorem 135 example 9 find the middle term (terms) in the expansion of p x 9 x p solution since the power of binomial is odd. Demonstrates how to answer typical problems involving the binomial theorem.

Binomial theorem
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# Smallest set such that all arithmetic progression will always contain at least a number in a set

Let $$S= \left\{ 1,2,3,...,100 \right\}$$ be a set of positive integers from $$1$$ to $$100$$. Let $$P$$ be a subset of $$S$$ such that any arithmetic progression of length 10 consisting of numbers in $$S$$ will contain at least a number in $$P$$. What is the smallest possible number of elements in $$P$$ ?

Denote $$|P|$$ as the number of elements in $$P$$. We shall find the smallest possible value of $$|P|$$.

For $$|P|=16$$, we have the answer by @RobertIsrael below.

However, for $$|P|<16$$, I can neither find such set $$P$$ nor prove that $$|P|$$ cannot be less than $$16$$. So my question is:

Is it true that $$|P| \geq 16$$? How can I prove it? If not, what is the minimum amount of elements in $$P$$ ?

Also, I am wondering that:

If we replace 10 with an even number $$n$$,and $$100$$ with $$n^2$$, can we find the minimum of $$|P|$$ ?

Any answers or comments will be appreciated. If this question should be closed, please let me know. If this forum cannot answer my question, I will delete this question immediately.

• it is not too unusual that questions here get answered, say, after a year, and not immediately. Jun 27 '19 at 6:24
• @DimaPasechnik Thanks. I just afraid that my question will be forgotten and cannot be answered. Jun 27 '19 at 7:02
• good questions don't get forgotten. they pop up in searches, etc. Jun 27 '19 at 8:21
• This can be considered as a set-covering problem. Although set covering is NP-complete, I suspect this one is within the reach of current technology. Jun 27 '19 at 12:37
• For the last question (replacing 10 with $n$), have you computed the optimal number for $n\le 9$ and checked the OEIS? Jun 27 '19 at 14:44

Considering the complement of $$P$$ in $$[1,100]$$, you are asking how large can a subset of $$[1,100]$$ be given that it does not contain any $$10$$-term arithmetic progression. The more general question

How large can a subset of $$[1,N]$$ be given that it does not contain any $$k$$-term arithmetic progression?

is one of the central problems in combinatorial number theory. There is no chance to give a precise answer, as an "explicit" function of $$N$$ and $$k$$, and it quite likely that this is impossible already in your special situation where $$N=n^2$$ and $$k=n$$.

Here is an argument showing that if $$P\subset[1,n^2]$$ meets every $$n$$-term progression contained in $$[1,n^2]$$, then $$|P|>n+n^{0.5+o(1)}$$. (See also the paragraph at the very end for the estimate $$|P|\ge 12$$ in your special case where $$P\subset[1,100]$$ and we want to block all $$10$$-term progressions.) It would be interesting to improve these estimates or at least to decide whether $$|P|>Cn$$ holds true with an absolute constant $$C>1$$.

Write $$K:=|P|$$, $$\Delta:=K-n$$, and $$P=\{p_1,\dotsc,p_K\}$$ where $$1\le p_1<\dotsb. Notice that $$p_1\le n$$ and $$p_K\ge n^2-(n-1)$$, whence $$p_K-p_1\ge(n-1)^2$$.

For any $$d\in[1,n]$$, the set $$P$$ contains an element from every residue class modulo $$d$$, and it follows that there are at most $$K-d$$ pairs of consecutive elements of $$P$$ with the difference equal to $$d$$; also, if $$d>n$$, then there are no such pairs at all. Let $$a$$ and $$r$$ be defined by \begin{align*} K-1 &= \Delta+(\Delta+1)+\dotsb+(\Delta+(a-1))+r \\ &= a\Delta+\frac{a(a-1)}2 + r,\quad 0\le r<\Delta+a. \tag{1} \end{align*} Since there are totally $$K-1$$ pairs of consecutive elements of $$P$$, of them at most $$\Delta$$ pairs at distance $$n$$, at most $$\Delta+1$$ pairs at distance $$n-1$$, etc, we conclude that \begin{align*} p_K-p_1 &\le n\Delta+(n-1)(\Delta+1)+\dotsb+(n-(a-1))(\Delta+(a-1))+(n-a)r \\ &= \Delta na+(n-\Delta)\cdot\frac{a(a-1)}2-\frac{a(a-1)(2a-1)}{6}+(n-a)r. \end{align*} Recalling the estimate $$p_K-p_1\ge(n-1)^2$$, and using ($$1$$), we get \begin{align*} (n-1)^2 &\le \Delta na+(n-\Delta)\cdot\frac{a(a-1)}2-\frac{a(a-1)(2a-1)}{6}+(n-a)r \\ &= n\Big(a\Delta+\frac{a(a-1)}2 + r\Big) - \Delta\cdot\frac{a(a-1)}2 - \frac{a(a-1)(2a-1)}{6} - ar \\ &= n(K-1) - \Delta\cdot\frac{a(a-1)}2 - \frac{a(a-1)(2a-1)}{6} - ar. \tag{2} \end{align*}

We now assume, aiming at a contradiction, that $$\Delta with an absolute constant $$0. From (1) we get then $$K-1 \ge \Delta a + \frac{a(a-1)}2 \ge \frac12\,a^2 - 1$$ implying $$a\le\sqrt{2K}$$; hence, $$\Delta a=O(n^{0.5+c})$$ and $$r=a+\Delta=O(n^{0.5})$$. As a result, $$\frac12\,a^2 = K-1+\frac12\,a-\Delta a - r > K - O(n^{0.5+c}),$$ leading to $$a>(1-o(1))\sqrt{2K}$$.

With these estimates in mind, from (2) we obtain $$n^2 + O(n) \le nK - \frac12\,\Delta a^2 - \frac13\,a^3;$$ that is, $$\Delta n \ge \frac12\,\Delta a^2 + \frac13\,a^3 + O(n).$$ Consequently, $$n^{1+c} \ge \Delta n \ge \frac13\,a^3 + O(n) \ge (1-o(1))(2K)^{1.5} + O(n) > n^{1.5} + O(n),$$ a contradiction.

As an illustration of this approach, let's show that one needs at least $$12$$ elements to block every $$10$$-term progression in $$[1,100]$$. Suppose for a contradiction that $$P\subset[1,100]$$ is an $$11$$-element set blocking all such progressions. There are $$|P|-1=10$$ pairs of consecutive elements of $$P$$. Of these ten pairs, there is at most one pair with distance $$10$$ between its two elements, at most two pairs with distance $$9$$, at most three pairs with distance $$8$$, and at most four pairs with distance $$7$$. Therefore the largest element of $$P$$ exceeds the smallest one by at most $$1\cdot 10+2\cdot 9 + 3\cdot 8 + 4\cdot 7=80$$. It follows that either the smallest element of $$P$$ is at least $$11$$, or its largest element is at most $$90$$; but then $$P$$ does not block at least one of the progressions $$[1,10]$$ and $$[91,100]$$, a contradiction.

Using a tabu search procedure, I have found a solution for $$|P|=17$$, namely $${1, 11, 18, 25, 31, 32, 33, 36, 44, 51, 58, 65, 69, 70, 77, 84, 91}$$. I don't know if this is optimal.

EDIT: Found a solution for $$|P|=16$$, namely $$10, 15, 22, 29, 36, 43, 53, 55, 56, 57, 58, 68, 73, 74, 84, 91$$

• I'm working on $|P|=16$. So far I've found a $P$ with $|P|=16$, namely $\{9, 18, 28, 29, 31, 40, 42, 51, 53, 56, 65, 69, 70, 77, 84, 91\}$, that intersects all but one of these arithmetic progressions, the exception being $({36, 43, 50, 57, 64, 71, 78, 85, 92, 99})$. Jun 27 '19 at 16:50
• I'm using a tabu search over sets of a given size to maximize the number of a.p.'s that intersect the set. Possible moves consist of replacing a member of the set with a nonmember. Jun 27 '19 at 17:00
• Thank you. Your answer is correct. How long did it take to find those numbers? Can you find the boundary of $|P|$? Jun 28 '19 at 8:00
• So is 16 optimal?
– EGME
Jun 28 '19 at 20:43
• My brute-force confirms that there no 15. Jul 2 '19 at 5:19


# Mutlivariable Calculus: Surface Area

This was a question a students had asked me earlier today regarding surface area.

Find the surface area of the hemisphere $x^2+y^2+z^2 = 4$ bounded below by $z=1$.

I decided to approach this problem using spherical coordinates and got the following

\begin{eqnarray} \int_{0}^{2\pi}\int_{0}^{\pi/3}4\sin\phi d\phi d\theta & = & 4\int_{0}^{2\pi}d\theta\int_{0}^{\pi/3}\sin\phi d\phi\\ & = & 8\pi\cos\phi|_{\pi/3}^{0}\\ & = & 8\pi(1-\frac{1}{2}) = 4\pi \end{eqnarray}

I also solved this problem using single variable calculus as follows. I can represent the sphere as a circle of $h^2+z^2 =4$. Thus we have the following:

\begin{eqnarray} SA = \int_a^b2\pi f(z)ds & = & 2\pi\int_1^2\sqrt{4-z^2}\sqrt{1+\frac{z^2}{4-z^2}}dz\\ & = & 2\pi\int_1^2\sqrt{4-z^2}\sqrt{\frac{4}{4-z^2}} dz\\ & = & 2\pi\int_1^22dz=4\pi z|_1^2=4\pi \end{eqnarray}

As you see, I got the same answer for both approaches.

The student, and a few others, comes back later during the day and tells me the answer I got was incorrect. He does not tell me what the professor got, he just told me it was wrong. I asked myself "why?" Is there something I missed?

Thanks in advance for any feedback.

• Maybe there is something wrong with the text of the problem. The surface whose area has been calculated is not a hemisphere but a spherical cap. – Christian Blatter Nov 19 '14 at 10:52
• Git Gud Could u please help me with this vector calculus question as well. I really need help in this. Thanks – ys wong Nov 23 '14 at 7:38

I don't see what you did wrong here...

To me, to solve these types of problems you have to think geometrically--there isn't going to be some way to do it just from a knowledge of multivariable calculus.

First, how to find the total surface area of the sphere--that will help. You need to break the sphere up into circles stacked on top of each other, then find the $dA$:

$$dA = 2\pi r_{\phi} h$$

$h$ is easy to find, it's just $rd\phi$, $r_\phi$ is the radius at the given azimuth: $r_\phi = r\sin(\phi)$ which gives:

$$dA = 2\pi r^2\sin(\phi)d\phi\\ A = \int dA = 2\pi r^2\left.\int_{0}^{\pi}\sin(\phi)d\phi = -2\pi r^2\cos(\phi)\right|_0^\pi = 4\pi r^2$$

So the correct integral should be:L

$$A = 2\pi r^2\left.\int_{0}^{\phi_0} \sin(\phi)d\phi = -2\pi r^2 \cos(\phi)\right|_{0}^{\phi_0} = 2\pi r^2\left(1 - \cos(\phi_0)\right)$$

In this case, $\phi_0$ satisfies that $z = r\cos(\phi) = 2\cos(\phi_0) = 1 \rightarrow \cos(\phi_0) = \frac{1}{2}$ and thus:

$$A = 2\pi 2^2\left(1 - \frac{1}{2}\right) = 4\pi$$


Tag Archives: factoring

Infinite Ways to an Infinite Geometric Sum

One of my students, K, and I were reviewing Taylor Series last Friday when she asked for a reminder why an infinite geometric series summed to $\displaystyle \frac{g}{1-r}$ for first term g and common ratio r when $\left| r \right| < 1$.  I was glad she was dissatisfied with blind use of a formula and dove into a familiar (to me) derivation.  In the end, she shook me free from my routine just as she made sure she didn’t fall into her own.

STANDARD INFINITE GEOMETRIC SUM DERIVATION

My standard explanation starts with a generic infinite geometric series.

$S = g+g\cdot r+g\cdot r^2+g\cdot r^3+...$  (1)

We can reason this series converges iff $\left| r \right| <1$ (see Footnote 1 for an explanation).  Assume this is true for (1).  Notice the terms on the right keep multiplying by r.

The annoying part of summing any infinite series is the ellipsis (…).  Any finite number of terms always has a finite sum, but that simply written, but vague ellipsis is logically difficult.  In the geometric series case, we might be able to handle the ellipsis by aligning terms in a similar series.  You can accomplish this by continuing the pattern on the right:  multiplying both sides by r

$r\cdot S = r\cdot \left( g+g\cdot r+g\cdot r^2+... \right)$

$r\cdot S = g\cdot r+g\cdot r^2+g\cdot r^3+...$  (2)

This seems to make make the right side of (2) identical to the right side of (1) except for the leading g term of (1), but the ellipsis requires some careful treatment. Footnote 2 explains how the ellipses of (1) and (2) are identical.  After that is established, subtracting (2) from (1), factoring, and rearranging some terms leads to the infinite geometric sum formula.

$(1)-(2) = S-S\cdot r = S\cdot (1-r)=g$

$\displaystyle S=\frac{g}{1-r}$

STUDENT PREFERENCES

I despise giving any formula to any of my classes without at least exploring its genesis.  I also allow my students to use any legitimate mathematics to solve problems so long as reasoning is justified.

In my experiences, about half of my students opt for a formulaic approach to infinite geometric sums while an equal number prefer the quick “multiply-by-r-and-subtract” approach used to derive the summation formula.  For many, apparently, the dynamic manipulation is more meaningful than a static rule.  It’s very cool to watch student preferences at play.

K’s VARIATION

K understood the proof, and then asked a question I hadn’t thought to ask.  Why did we have to multiply by r?  Could multiplication by $r^2$ also determine the summation formula?

I had three nearly simultaneous thoughts followed quickly by a fourth.  First, why hadn’t I ever thought to ask that?  Second, geometric series for $\left| r \right|<1$ are absolutely convergent, so K’s suggestion should work.  Third, while the formula would initially look different, absolute convergence guaranteed that whatever the “$r^2$ formula” looked like, it had to be algebraically equivalent to the standard form.  While I considered those conscious questions, my math subconscious quickly saw the easy resolution to K’s question and the equivalence from Thought #3.

Multiplying (1) by $r^2$ gives

$r^2 \cdot S = g\cdot r^2 + g\cdot r^3 + ...$ (3)

and the ellipses of (1) and (3) partner perfectly (Footnote 2), so K subtracted, factored, and simplified to get the inevitable result.

$(1)-(3) = S-S\cdot r^2 = g+g\cdot r$

$S\cdot \left( 1-r^2 \right) = g\cdot (1+r)$

$\displaystyle S=\frac{g\cdot (1+r)}{1-r^2} = \frac{g\cdot (1+r)}{(1+r)(1-r)} = \frac{g}{1-r}$

That was cool, but this success meant that there were surely many more options.

EXTENDING

Why stop at multiplying by r or $r^2$?  Why not multiply both sides of (1) by a generic $r^N$ for any natural number N?   That would give

$r^N \cdot S = g\cdot r^N + g\cdot r^{N+1} + ...$ (4)

where the ellipses of (1) and (4) are again identical by the method of Footnote 2.  Subtracting (4) from (1) gives

$(1)-(4) = S-S\cdot r^N = g+g\cdot r + g\cdot r^2+...+ g\cdot r^{N-1}$

$S\cdot \left( 1-r^N \right) = g\cdot \left( 1+r+r^2+...+r^{N-1} \right)$  (5)

There are two ways to proceed from (5).  You could recognize the right side as a finite geometric sum with first term 1 and ratio r.  Substituting that formula and dividing by $\left( 1-r^N \right)$ would give the general result.

Alternatively, I could see students exploring $\left( 1-r^N \right)$, and discovering by hand or by CAS that $(1-r)$ is always a factor.  I got the following TI-Nspire CAS result in about 10-15 seconds, clearly suggesting that

$1-r^N = (1-r)\left( 1+r+r^2+...+r^{N-1} \right)$.  (6)

Math induction or a careful polynomial expansion of (6) would prove the pattern suggested by the CAS.  From there, dividing both sides of (5) by $\left( 1-r^N \right)$ gives the generic result.

$\displaystyle S = \frac{g\cdot \left( 1+r+r^2+...+r^{N-1} \right)}{\left( 1-r^N \right)}$

$\displaystyle S = \frac{g\cdot \left( 1+r+r^2+...+r^{N-1} \right) }{(1-r) \cdot \left( 1+r+r^2+...+r^{N-1} \right)} = \frac{g}{1-r}$

In the end, K helped me see there wasn’t just my stock approach to an infinite geometric sum, but really an infinite number of parallel ways.  Nice.

FOOTNOTES

1) RESTRICTING r:  Obviously an infinite geometric series diverges for $\left| r \right| >1$ because that would make $g\cdot r^n \rightarrow \infty$ as $n\rightarrow \infty$, and adding an infinitely large term (positive or negative) to any sum ruins any chance of finding a sum.

For $r=1$, the sum converges iff $g=0$ (a rather boring series). If $g \ne 0$ , you get a sum of an infinite number of some nonzero quantity, and that is always infinite, no matter how small or large the nonzero quantity.

The last case, $r=-1$, is more subtle.  For $g \ne 0$, this terms of this series alternate between positive and negative g, making the partial sums of the series add to either g or 0, depending on whether you have summed an even or an odd number of terms.  Since the partial sums alternate, the overall sum is divergent.  Remember that series sums and limits are functions; without a single numeric output at a particular point, the function value at that point is considered to be non-existent.

2) NOT ALL INFINITIES ARE THE SAME:  There are two ways to show two groups are the same size.  The obvious way is to count the elements in each group and find out there is the same number of elements in each, but this works only if you have a finite group size.  Alternatively, you could a) match every element in group 1 with a unique element from group 2, and b) match every element in group 2 with a unique element from group 1.  It is important to do both steps here to show that there are no left-over, unpaired elements in either group.

So do the ellipses in (1) and (2) represent the same sets?  As the ellipses represent sets with an infinite number of elements, the first comparison technique is irrelevant.  For the second approach using pairing, we need to compare individual elements.  For every element in the ellipsis of (1), obviously there is an “partner” in (2) as the multiplication of (1) by r visually shifts all of the terms of the series right one position, creating the necessary matches.

Students often are troubled by the second matching as it appears the ellipsis in (2) contains an “extra term” from the right shift.  But, for every specific term you identify in (2), its identical twin exists in (1).  In the weirdness of infinity, that “extra term” appears to have been absorbed without changing the “size” of the infinity.

Since there is a 1:1 mapping of all elements in the ellipses of (1) and (2), you can conclude they are identical, and their difference is zero.

Probability, Polynomials, and Sicherman Dice

Three years ago, I encountered a question on the TI-Nspire Google group asking if there was a way to use CAS to solve probability problems.  The ideas I pitched in my initial response and follow-up a year later (after first using it with students in a statistics class) have been thoroughly re-confirmed in my first year teaching AP Statistics.  I’ll quickly re-share them below before extending the concept with ideas I picked up a couple weeks ago from Steve Phelps’ session on Probability, Polynomials, and CAS at the 64th annual OCTM conference earlier this month in Cleveland, OH.

BINOMIALS:  FROM POLYNOMIALS TO SAMPLE SPACES

Once you understand them, binomial probability distributions aren’t that difficult, but the initial conjoining of combinatorics and probability makes this a perennially difficult topic for many students.  The standard formula for the probability of determining the chances of K successes in N attempts of a binomial situation where p is the probability of a single success in a single attempt is no less daunting:

$\displaystyle \left( \begin{matrix} N \\ K \end{matrix} \right) p^K (1-p)^{N-K} = \frac{N!}{K! (N-K)!} p^K (1-p)^{N-K}$

But that is almost exactly the same result one gets by raising binomials to whole number powers, so why not use a CAS to expand a polynomial and at least compute the $\displaystyle \left( \begin{matrix} N \\ K \end{matrix} \right)$ portion of the probability?  One added advantage of using a CAS is that you could use full event names instead of abbreviations, making it even easier to identify the meaning of each event.

The TI-Nspire output above shows the entire sample space resulting from flipping a coin 6 times.  Each term is an event.  Within each term, the exponent of each variable notes the number of times that variable occurs and the coefficient is the number of times that combination occurs.  The overall exponent in the expand command is the number of trials.  For example, the middle term– $20\cdot heads^3 \cdot tails^3$ –says that there are 20 ways you could get 3 heads and 3 tails when tossing a coin 6 times. The last term is just $tails^6$, and its implied coefficient is 1, meaning there is just one way to flip 6 tails in 6 tosses.

The expand command makes more sense than memorized algorithms and provides context to students until they gain a deeper understanding of what’s actually going on.

FROM POLYNOMIALS TO PROBABILITY

Still using the expand command, if each variable is preceded by its probability, the CAS result combines the entire sample space AND the corresponding probability distribution function.  For example, when rolling a fair die four times, the distribution for 1s vs. not 1s (2, 3, 4, 5, or 6) is given by

The highlighted term says there is a 38.58% chance that there will be exactly one 1 and any three other numbers (2, 3, 4, 5, or 6) in four rolls of a fair 6-sided die.  The probabilities of the other four events in the sample space are also shown.  Within the TI-Nspire (CAS or non-CAS), one could use a command to give all of these probabilities simultaneously (below), but then one has to remember whether the non-contextualized probabilities are for increasing or decreasing values of which binomial outcome.

Particularly early on in their explorations of binomial probabilities, students I’ve taught have shown a very clear preference for the polynomial approach, even when allowed to choose any approach that makes sense to them.

TAKING POLYNOMIALS FROM ONE DIE TO MANY

Given these earlier thoughts, I was naturally drawn to Steve Phelps “Probability, Polynomials, and CAS” session at the November 2014 OCTM annual meeting in Cleveland, OH.  Among the ideas he shared was using polynomials to create the distribution function for the sum of two fair 6-sided dice.  My immediate thought was to apply my earlier ideas.  As noted in my initial post, the expansion approach above is not limited to binomial situations.  My first reflexive CAS command in Steve’s session before he share anything was this.

By writing the outcomes in words, the CAS interprets them as variables.  I got the entire sample space, but didn’t learn gain anything beyond a long polynomial.  The first output– $five^2$ –with its implied coefficient says there is 1 way to get 2 fives.  The second term– $2\cdot five \cdot four$ –says there are 2 ways to get 1 five and 1 four.  Nice that the technology gives me all the terms so quickly, but it doesn’t help me get a distribution function of the sum.  I got the distributions of the specific outcomes, but the way I defined the variables didn’t permit sum of their actual numerical values.  Time to listen to the speaker.

He suggested using a common variable, X, for all faces with the value of each face expressed as an exponent.  That is, a standard 6-sided die would be represented by $X^1+X^2+ X^3+X^4+X^5+X^6$ where the six different exponents represent the numbers on the six faces of a typical 6-sided die.  Rolling two such dice simultaneously is handled as I did earlier with the binomial cases.

NOTE:  Exponents are handled in TWO different ways here.  1) Within a single polynomial, an exponent is an event value, and 2) Outside a polynomial, an exponent indicates the number of times that polynomial is applied within the specific event.  Coefficients have the same meaning as before.

Because the variables are now the same, when specific terms are multiplied, their exponents (face values) will be added–exactly what I wanted to happen.  That means the sum of the faces when you roll two dice is determined by the following.

Notice that the output is a single polynomial.  Therefore, the exponents are the values of individual cases.  For a couple examples, there are 3 ways to get a sum of 10 $\left( 3 \cdot x^{10} \right)$, 2 ways to get a sum of 3 $\left( 2 \cdot x^3 \right)$, etc.  The most commonly occurring outcome is the term with the largest coefficient.  For rolling two standard fair 6-sided dice, a sum of 7 is the most common outcome, occurring 6 times $\left( 6 \cdot x^7 \right)$.  That certainly simplifies the typical 6×6 tables used to compute the sums and probabilities resulting from rolling two dice.

While not the point of Steve’s talk, I immediately saw that technology had just opened the door to problems that had been computationally inaccessible in the past.  For example, what is the most common sum when rolling 5 dice and what is the probability of that sum?  On my CAS, I entered this.

In the middle of the expanded polynomial are two terms with the largest coefficients, $780 \cdot x^{18}$ and $780 \cdot x^{19}$, meaning a sums of 17 and 18 are the most common, equally likely outcomes when rolling 5 dice.  As there are $6^5=7776$ possible outcomes when rolling a die 5 times, the probability of each of these is $\frac{780}{7776} \approx 0.1003$, or about 10.03% chance each for a sum of 17 or 18.  This can be verified by inserting the probabilities as coefficients before each term before CAS expanding.

With thought, this shouldn’t be surprising as the expected mean value of rolling a 6-sided die many times is 3.5, and $5 \cdot 3.5 = 17.5$, so the integers on either side of 17.5 (17 & 18) should be the most common.  Technology confirms intuition.

ROLLING DIFFERENT DICE SIMULTANEOUSLY

What is the distribution of sums when rolling a 4-sided and a 6-sided die together?  No problem.  Just multiply two different polynomials, one representative of each die.

The output shows that sums of 5, 6, and 7 would be the most common, each occurring four times with probability $\frac{1}{6}$ and together accounting for half of all outcomes of rolling these two dice together.

A BEAUTIFUL EXTENSION–SICHERMAN DICE

My most unexpected gain from Steve’s talk happened when he asked if we could get the same distribution of sums as “normal” 6-sided dice, but from two different 6-sided dice.  The only restriction he gave was that all of the faces of the new dice had to have positive values.  This can be approached by realizing that the distribution of sums of the two normal dice can be found by multiplying two representative polynomials to get

$x^{12}+2x^{11}+3x^{10}+4x^9+5x^8+6x^7+5x^6+4x^5+3x^4+2x^3+x^2$.

Restating the question in the terms of this post, are there two other polynomials that could be multiplied to give the same product?  That is, does this polynomial factor into other polynomials that could multiply to the same product?  A CAS factor command gives

Any rearrangement of these eight (four distinct) sub-polynomials would create the same distribution as the sum of two dice, but what would the the separate sub-products mean in terms of the dice?  As a first example, what if the first two expressions were used for one die (line 1 below) and the two squared trinomials comprised a second die (line 2)?

Line 1 actually describes a 4-sided die with one face of 4, two faces with 3s, and one face of 2.  Line 2 describes a 9-sided die (whatever that is) with one face of 8, two faces of 6, three faces of 4, two faces of 2, and one face with a 0 ( $1=1 \cdot x^0$).  This means rolling a 4-sided and a 9-sided die as described would give exactly the same sum distribution.  Cool, but not what I wanted.  Now what?

Factorization gave four distinct sub-polynomials, each with multitude 2.  One die could contain 0, 1, or 2 of each of these with the remaining factors on the other die.  That means there are $3^4=81$ different possible dice combinations.  I could continue with a trail-and-error approach, but I wanted to be more efficient and elegant.

What follows is the result of thinking about the problem for a while.  Like most math solutions to interesting problems, ultimate solutions are typically much cleaner and more elegant than the thoughts that went into them.  Problem solving is a messy–but very rewarding–business.

SOLUTION

Here are my insights over time:

1) I realized that the $x^2$ term would raise the power (face values) of the desired dice, but would not change the coefficients (number of faces).  Because Steve asked for dice with all positive face values.  That meant each desired die had to have at least one x to prevent non-positive face values.

2) My first attempt didn’t create 6-sided dice.  The sums of the coefficients of the sub-polynomials determined the number of sides.  That sum could also be found by substituting $x=1$ into the sub-polynomial.  I want 6-sided dice, so the final coefficients must add to 6.  The coefficients of the factored polynomials of any die individually must add to 2, 3, or 6 and have a product of 6.  The coefficients of $(x+1)$ add to 2, $\left( x^2+x+1 \right)$ add to 3, and $\left( x^2-x+1 \right)$ add to 1.  The only way to get a polynomial coefficient sum of 6 (and thereby create 6-sided dice) is for each die to have one $(x+1)$ factor and one $\left( x^2+x+1 \right)$ factor.

3) That leaves the two $\left( x^2-x+1 \right)$ factors.  They could split between the two dice or both could be on one die, leaving none on the other.  We’ve already determined that each die already had to have one each of the x, $(x+1)$, and $\left( x^2+x+1 \right)$ factors.  To also split the $\left( x^2-x+1 \right)$ factors would result in the original dice:  Two normal 6-sided dice.  If I want different dice, I have to load both of these factors on one die.

That means there is ONLY ONE POSSIBLE alternative for two 6-sided dice that have the same sum distribution as two normal 6-sided dice.

One die would have single faces of 8, 6, 5, 4, 3, and 1.  The other die would have one 4, two 3s, two 2s, and one 1.  And this is exactly the result of the famous(?) Sicherman Dice.

If a 0 face value was allowed, shift one factor of x from one polynomial to the other.  This can be done two ways.

The first possibility has dice with faces {9, 7, 6, 5, 4, 2} and {3, 2, 2, 1, 1, 0}, and the second has faces {7, 5, 4, 3, 2, 0} and {5, 4, 4, 3, 3, 2}, giving the only other two non-negative solutions to the Sicherman Dice.

Both of these are nothing more than adding one to all faces of one die and subtracting one from from all faces of the other.  While not necessary to use polynomials to compute these, they are equivalent to multiplying the polynomial of one die by x and the other by $\frac{1}{x}$ as many times as desired. That means there are an infinite number of 6-sided dice with the same sum distribution as normal 6-sided dice if you allow the sides to have negative faces.  One of these is

corresponding to a pair of Sicherman Dice with faces {6, 4, 3, 2, 1, -1} and {1,5,5,4,4,3}.

CONCLUSION:

There are other very interesting properties of Sicherman Dice, but this is already a very long post.  In the end, there are tremendous connections between probability and polynomials that are accessible to students at the secondary level and beyond.  And CAS keeps the focus on student learning and away from the manipulations that aren’t even the point in these explorations.

Enjoy.

Powers of 2

Yesterday, James Tanton posted a fun little problem on Twitter:

So, 2 is one more than $1=2^0$, and 8 is one less than 9=2^3\$, and Dr. Tanton wants to know if there are any other powers of two that are within one unit of a perfect square.

While this problem may not have any “real-life relevance”, it demonstrates what I describe as the power and creativity of mathematics.  Among the infinite number of powers of two, how can someone know for certain if any others are or are not within one unit of a perfect square?  No one will ever be able to see every number in the list of powers of two, but variables and mathematics give you the tools to deal with all possibilities at once.

For this problem, let D and N be positive integers.  Translated into mathematical language, Dr. Tanton’s problem is equivalent to asking if there are values of D and N for which $2^D=N^2 \pm 1$.  With a single equation in two unknowns, this is where observation and creativity come into play.  I suspect there may be more than one way to approach this, but my solution follows.  Don’t read any further if you want to solve this for yourself.

Because D and N are positive integers, the left side of $2^D=N^2 \pm 1$,  is always even.   That means $N^2$, and therefore N must be odd.

Because N is odd, I know $N=2k+1$ for some whole number k.  Rewriting our equation gives $2^D=(2k+1)^2 \pm 1$, and the right side equals either $4k^2+4k$ or $4k^2+4k+2$.

Factoring the first expression gives $2^D=4k^2+4K=4k(k+1)$.   Notice that this is the product of two consecutive integers, k and $k+1$, and therefore one of these factors (even though I don’t know which one) must be an odd number.  The only odd number that is a factor of a power of two is 1, so either $k=1$ or $k+1=1 \rightarrow k=0$.  Now, $k=1 \longrightarrow N=3 \longrightarrow D=3$ and $k=0 \longrightarrow N=1 \longrightarrow D=0$, the two solutions Dr. Tanton gave.  No other possibilities are possible from this expression, no matter how far down the list of powers of two you want to go.

But what about the other expression?  Factoring again gives $2^D=4k^2+4k+2=2 \cdot \left( 2k^2+2k+1 \right)$.  The expression in parentheses must be odd because its first two terms are both multiplied by 2 (making them even) and then one is added (making the overall sum odd).  Again, 1 is the only odd factor of a power of two, and this happens in this case only when $k=0 \longrightarrow N=1 \longrightarrow D=0$, repeating a solution from above.

Because no other algebraic solutions are possible, the two solutions Dr. Tanton gave in the problem statement are the only two times in the entire universe of perfect squares and powers of two where elements of those two lists are within a single unit of each other.

Math is sweet.


# Distinguishable objects into distinguishable boxes

How many ways are there to distribute $15$ distinguishable objects into $5$ distinguishable boxes so that the boxes have one, two, three, four, and five objects in them respectively?

$\begin{gather} &\_\_\_ &\_\_\_ &\_\_\_ &\_\_\_ &\_\_\_ \\ &1 &2 &3 &4 &5 \end{gather}$

The lines represent the $5$ distinguishable boxes and the numbers below represent how many distinguishable objects each box must hold. I'm thinking I have $C\left(15,1\right)$ options for the first box then $C\left(14,2\right)$ for the second box, all the way to $C\left(5,5\right)$ for the fifth box. I multiply all those combinations together because of the product rule and I have no idea if that's the right answer.

• Good clear correct analysis. I would say that for each option for the first box there are $\dots$. Surely it is not true that you have no idea whether this is the right answer! – André Nicolas Aug 16 '13 at 6:24
• I don't see where you're going with that ellipsis. What do you mean that it's not true? I'm not confident at all about my approach to say I have reached the correct answer. – Kasper-34 Aug 16 '13 at 6:30
• I just meant it should be made clearer why we multiply. Note that if it is not specified which boxes contain $1,2,\dots,5$ then we need to multiply your answer by $5!$. – André Nicolas Aug 16 '13 at 6:33
• Well I'm multiplying because of the product rule. I think? I would multiply by $5!$ if I wasn't restricted, because I could put them in any order such as $5,3,1,2,4$? – Kasper-34 Aug 16 '13 at 6:41
• Okay, in that case I agree with you. The use of the word "respectively" makes me think they must be in the order $1,2,3,4,5$ and only that order. – Kasper-34 Aug 16 '13 at 7:10

Ways to put the labels $\{1,2,3,4,5\}$ on the boxes according as how many objects they contain: 5!. Then, as you correctly presumed,

$\binom{15}{1}$ ways to select an object for the one-object box;

$\binom{14}{2}$ ways to select two objects for the two-object box;

$\binom{12}{3}$ ways to select three objects for the three-object box;

$\binom{9}{4}$ ways to select four objects for the four-object box;

$\binom{5}{5}=1$ way to put the remaining five objects into the five-object box.

I think the answer is $$5!\binom{15}{1}\binom{14}{2}\binom{12}{3}\binom{9}{4}.$$

If the labels of the boxes are fixed and cannot be reassigned (i.e., according as how many objects they contain), then the term $5!$ should be suppressed.

• I would say in this particular case since it used the word "respectively" there is only one way to order the boxes, which means we can leave out the multiplication of $5!$. – Kasper-34 Aug 16 '13 at 7:17


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# At the rate of f floors per m minutes, how many floors does an elevato

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21 Nov 2015, 07:34
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At the rate of f floors per m minutes, how many floors does an elevator travel in s seconds?

(A) $$\frac{fs}{60m}$$

(B) $$\frac{ms}{60f}$$

(C) $$\frac{fm}{s}$$

(D) $$\frac{fs}{m}$$

(E) $$\frac{60s}{fm}$$

Explanation: You’re given a rate and a time, and you’re looking for distance. This is clearly a job for the rate formula. Since the rate is in terms of minutes and the time is in seconds, you’ll need to convert one or the other; it’s probably easier to convert s seconds to minutes than the rate to floors per second. Since 1 minute equals 60 seconds, s seconds equals $$\frac{s}{60}$$ minutes. Now we can plug our rate and time into the rate formula: $$r=\frac{d}{t}$$
$$\frac{f}{m}=d/\frac{s}{60}$$

Now, cross-multiply:

$$dm = \frac{fs}{60}$$

$$d=\frac{fs}{60m}$$, choice (A).
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At the rate of f floors per m minutes, how many floors does an elevato  [#permalink]

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21 Nov 2015, 14:46
gmatser1 wrote:
At the rate of f floors per m minutes, how many floors does an elevator travel in s seconds?

(A) $$\frac{fs}{60m}$$

(B) $$\frac{ms}{60f}$$

(C) $$\frac{fm}{s}$$

(D) $$\frac{fs}{m}$$

(E) $$\frac{60s}{fm}$$

Looks like a good candidate for the INPUT-OUTPUT approach.

Let's INPUT some values for f, m and s.
Let's say that f = 8 floors, m = 2 minutes, and s = 30 seconds

That is, the elevator travels at a rate of 8 floors per 2 minutes.
How many floors does an elevator travel in 30 seconds?

Well, 8 floors in 2 minutes translates to 4 floors in 1 minute, and 2 floors in 30 seconds.

So, when f = 8, m = 2, and s = 30, the answer to the question (OUTPUT) is 2 floors

Now, let's plug f = 8, m = 2, and s = 30 into each answer choice and see which one yields an OUTPUT of 2

(A) $$\frac{(8)(30)}{60(2)}$$ = 2 GREAT!

(B) $$\frac{(2)(30)}{60(8)}$$ = 1/8 ELIMINATE

(C) $$\frac{(8)(2)}{(30)}$$ = 8/15 ELIMINATE

(D) $$\frac{(8)(30)}{(2)}$$ = 120 ELIMINATE

(E) $$\frac{60(30)}{(8)(2)}$$ = some big number ELIMINATE

For more information on this question type and this approach, we have some free videos:
- Variables in the Answer Choices - http://www.gmatprepnow.com/module/gmat- ... /video/933
- Tips for the Algebraic Approach - http://www.gmatprepnow.com/module/gmat- ... /video/934
- Tips for the Input-Output Approach - http://www.gmatprepnow.com/module/gmat- ... /video/935

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Brent
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Re: At the rate of f floors per m minutes, how many floors does an elevato  [#permalink]

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21 Nov 2015, 15:19
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f/m=floors per minute
f/60m=floors per one second
fs/60m=floors per s seconds
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Re: At the rate of f floors per m minutes, how many floors does an elevato  [#permalink]

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29 Sep 2017, 10:31
gmatser1 wrote:
At the rate of f floors per m minutes, how many floors does an elevator travel in s seconds?

(A) $$\frac{fs}{60m}$$

(B) $$\frac{ms}{60f}$$

(C) $$\frac{fm}{s}$$

(D) $$\frac{fs}{m}$$

(E) $$\frac{60s}{fm}$$

We have a rate of (f floors)/(m minutes) and need to determine how many floors an elevator travels in s seconds = s/60 minutes, and thus:

f/m x s/60 = fs/60m

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Re: At the rate of f floors per m minutes, how many floors does an elevato  [#permalink]

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14 Jan 2018, 23:19
Let's let f = 60 in m = 1 minutes as it will make the calculation easy!
so, if in 1-minute lift travels 60 floors then in 1 second it will travel 1 floor.
Plugging the values as f=60,s=1,m=1 the result should be 1

Jkay
Re: At the rate of f floors per m minutes, how many floors does an elevato &nbs [#permalink] 14 Jan 2018, 23:19
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Evaluate: $\frac{1}{(1+1)!} + \frac{2}{(2+1)!}+...+\frac{n}{(n+1)!}$ using combinatorics.

Evaluate $\frac{1}{(1+1)!} + \frac{2}{(2+1)!}+...+\frac{n}{(n+1)!}$. This is from a combinatorics textbook so I'd like a combinatorial proof. I find doing this kind of problem difficult especially when you have to sum - I don't know how to construct a sensible analogy using the addition principle.

Similar question that appears just before this question in the text: Combinatorics problem involving series summation

• I'm not sure whether a combinatoric proof can sum those fractions, but induction shows the sum is $1-1/(n+1)!$.
– J.G.
Jul 7 '18 at 5:36
• @J.G. I linked a similar problem maybe that will help... (didn't help me) Jul 7 '18 at 5:42

Consider a uniformly at random selected permutation of $\{1,2,\dots,n,n+1\}$.

The probability that $2$ appears before $1$ is $\frac{1}{(1+1)!}$

Given that this does not occur, then $1$ and $2$ appear in the correct order. The probability then that $3$ appears before at least one of $2$ or $1$ as well as $1$ and $2$ appearing in the correct order is $\frac{2}{(2+1)!}$.

Given that this does not occur, then $1,2,3$ all appear in the correct order. The probability then that $4$ appears before at least on of $3,2,1$ and $1,2,3$ all appearing in the correct order is $\frac{3}{(3+1)!}$...

...

Given that this does not occur, then $1,2,3,\dots,n$ all appear in the correct order. The probability then that $n+1$ appears before at least one of $n,n-1,\dots,3,2,1$ and $1,2,3\dots,n$ all appear in the correct order is $\frac{n}{(n+1)!}$

Given that this does not occur, then $1,2,3,\dots,n,n+1$ all appear in the correct order. This occurs with probability $\frac{1}{(n+1)!}$

Note that these are all mutually exclusive and exhaustive events, so they add up to equal $1$. Note further that the sum you are interested in is the sum of all of the events except the last one. We have then

$$\frac{1}{(1+1)!}+\frac{2}{(2+1)!}+\dots+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}$$

Rephrased, by multiplying the expression by $(n+1)!$, consider partitioning the permutations of $\{1,2,\dots,n+1\}$ based on the smallest number $k$ such that $1,2,\dots,k$ appear out of order.

That is to say, if $k$ is the smallest number such that $1,2,\dots,k$ appear out of order then $1,2,\dots,k-1$ must appear in order while $k$ does not appear after $1,2,\dots,k-1$. To count how many permutations satisfy this condition first pick the spaces that $1,2,\dots,k-1,k$ occupy simultaneously and then pick which of those positions $k$ occupies noting that it cannot be the last. $1,2,\dots,k-1$ appear in their normal order in the remaining selected positions. Then all other elements are distributed among the other spaces. This occurs in

$$\binom{n+1}{k}(k-1)(n+1-k)!=\frac{(n+1)!}{k!(n+1-k)!}(k-1)(n+1-k)!=\frac{k-1}{k!}(n+1)!$$

which you should recognize as following the sequence $0,\frac{1}{2!},\frac{2}{3!},\frac{3}{4!},\frac{4}{5!},\dots$ with the additional factor of $(n+1)!$ which we introduced earlier, otherwise mimicking the desired sum.

By including also the additional case of the identity permutation, the above forms a partition of the permutations of $\{1,2,\dots,n+1\}$. It follows that their respective totals add up to $(n+1)!$.

By removing the identity permutation as well as dividing by the factor of $(n+1)!$ that we introduced, this yields the desired identity

$$\frac{1}{(1+1)!}+\frac{2}{(2+1)!}+\dots+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}$$

• Hello can you clarify what you mean by the last part: "based on the smallest number k such that 1,2,…,k appear out of order." Jul 7 '18 at 7:12
• @helios321 Added more details. If you are just having difficulty understanding the phrasing I used, perhaps an example will help. $\color{red}{1}5\color{blue}{4}\color{red}{2}6\color{red}{3}$ is an example of a permutation where $\color{blue}{4}$ is the smallest number which occurs out of order since $\color{red}{123}$ appear in the correct order. By "appearing in the correct order" that is not to say they are adjacent, but merely that $1$ appears before $2$, that $2$ appears before $3$, etc... Jul 7 '18 at 16:19
• Thanks I figured it out now. Interesting it seems the expression multiplied by $(n+1)!$ is the same as in this question math.stackexchange.com/questions/2334537/…, both equal $(n+1)!-1$, but the other is counted by fixing the position. Jul 8 '18 at 0:23

Solution

Notice that$$\frac{k}{(k+1)!}=\frac{(k+1)-1}{(k+1)!}=\frac{1}{k!}-\frac{1}{(k+1)!}.$$

Hence, $$\sum_{k=1}^n \frac{k}{(k+1)!}=\left(\frac{1}{1!}-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\cdots+\left(\frac{1}{n!}-\frac{1}{(n+1)!}\right)=1-\frac{1}{(n+1)!}.$$

Using generating functions, which are widely used in combinatorics: $$a_n=\sum\limits_{k=1}^{n}\frac{k}{(k+1)!}$$ which is the same as $$a_n=a_{n-1}+\frac{n}{(n+1)!}$$ with generating function $$f(x)=\sum\limits_{n}\color{red}{a_n}x^n =a_0+\sum\limits_{n=1}\left(a_{n-1}+\frac{n}{(n+1)!}\right)x^n=\\ a_0+x\sum\limits_{n=1}a_{n-1}x^{n-1}+\sum\limits_{n=1}\frac{n}{(n+1)!}x^n=\\ a_0+xf(x)+\sum\limits_{n=1}\frac{n+1}{(n+1)!}x^n-\sum\limits_{n=1}\frac{1}{(n+1)!}x^n=\\ a_0+xf(x)+\left(\sum\limits_{n=1}\frac{1}{(n+1)!}x^{n+1}\right)'-\frac{1}{x}\sum\limits_{n=1}\frac{1}{(n+1)!}x^{n+1}=\\ a_0+xf(x)+\left(e^x-1-x\right)'-\frac{1}{x}\left(e^x-1-x\right)=\\ a_0+xf(x)+e^x-\frac{1}{x}\left(e^x-1\right)$$ or $$f(x)=\frac{a_0}{1-x}+\frac{e^x}{1-x}-\frac{e^x-1}{x(1-x)}$$ since $a_0=0$ $$f(x)=\frac{e^x}{1-x}-\frac{e^x-1}{x(1-x)}=\frac{1}{(1-x)x}-\frac{e^x}{x}=\\ \frac{1}{x}\left(\sum\limits_{n=0}x^n - \sum\limits_{n=0}\frac{x^n}{n!}\right)=\sum\limits_{n=1}\color{red}{\left(1-\frac{1}{(n+1)!}\right)}x^{n}$$ as a result $$a_n=1-\frac{1}{(n+1)!}, n\geq1$$

• I don't know what the down-voter was up to, but by all means this is a combinatorial proof! Jul 7 '18 at 12:31
• I'm not the downvoter, but usually your kind of proof is considered to be an algebraic one, whereas a combinatoric proof takes a set with known cardinality and constructs a bijection with another set showing the wanted formula or binomial identity. Jul 8 '18 at 14:02
• @MarkusScheuer or you assume that for a set of size $n$ the number of favourable cases is $a_n$ and then try to derive the recurrence for $n+1$ from $n$. And after, solve the recurrence with generating functions. A technique very often used in enumerative combinatorics. Jul 8 '18 at 14:17
• No doubt, this technique is great and I also appreciate it and use it often. The point is, this type of proof is usually not denoted as combinatorial proof. We find for instance in R. P. Stanleys Enumerative Combinatorics in the introductory chapter How to Count: A proof that shows that a certain set $S$ has a certain number $m$ of elements by constructing an explicit bijection between $S$ and some other set that is known to have $m$ elements is called a combinatorial proof or bijective proof. - As we can see the key term denoting a combinatorial proof is bijection. Jul 8 '18 at 14:43


Since a right kite can be divided into two right triangles, the following metric formulas easily follow from well known properties of right triangles. Okay, so that sounds kind of complicated. The triangle ABD is isosceles. It looks like the kites you see flying up in the sky. A second identifying property of the diagonals of kites is that one of the diagonals bisects, or halves, the other diagonal. In Euclidean geometry, a kite is a quadrilateral whose four sides can be grouped into two pairs of equal-length sides that are adjacent to each other. So it doesn't always look like the kite you fly. A Square is a Kite? 3. • diagonals which alwaysmeet at right angles. Kite properties include (1) two pairs of consecutive, congruent sides, (2) congruent non-vertex angles and (3) perpendicular diagonals. Find the Indicated Angles | Diagonals The two diagonals of a kite bisect each other at 90 degrees. 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Charlene puts together two isosceles triangles so that they share a base, creating a kite. The sketch below shows how to construct a kite. See, a kite shape looks like a diamond whose middle has been shifted upwards a bit. It can be viewed as a pair of congruent triangles with a common base. In this section, we will discuss kite and its theorems. E-learning is the future today. Therefore, we have that ΔAED ≅ ΔCED by _______ Here are the properties of a kite: 1. Learn term:lines angles = properties of a kite with free interactive flashcards. In this section, we will discuss kite and its theorems. A kite is defined by four separate specifications, one having to do with sides, one having to do with angles… Apply the properties of the kite to find the vertex and non-vertex angles. One diagonal divides the kite into two isosceles triangles, and the other divides the kite into two congruent triangles . What are the Properties of a Kite? The main diagonal of a kite bisects the other diagonal. 3. In a kite, the measures of the angles are 3x °, 75°, 90°, and 120°.Find the value of x.What are the measures of the angles that are congruent? 3. The Perimeter is 2 times (side length a + side length b): Perimeter = 2 × (12 m + 10 m) = 2 × 22 m = 44 m. When all sides have equal length the Kite will also be a Rhombus. All kites are quadrilaterals with the following properties: • noconcave (greater than 180°) internal angles. The angles The problem. Apply the properties of the kite to find the vertex and non-vertex angles. Convex: All its interior angles measure less than 180°. By definition, a kite is a polygon with four total sides (quadrilateral). Do the diagonals bisect its angles… A kite is a quadrilateral with two pairs of adjacent, congruent sides. Properties of Kites. A kite is a quadrilateral with two pairs of adjacent, congruent sides. The main diagonal of a kite bisects the other diagonal. Sometimes one of those diagonals could be outside the shape; then you have a dart. Use this interactive to investigate the properties of a kite. 1. A kite is the combination of two isosceles triangles. The smaller diagonal of a kite … Section 7.5 Properties of Trapezoids and Kites 441 7.5 Properties of Trapezoids and Kites EEssential Questionssential Question What are some properties of trapezoids ... Measure the angles of the kite. Find the Indicated Angles | Vertex and Non-Vertex Angles. 2. And then we could say statement-- I'm taking up a lot of space now-- statement 11, we could say measure of angle DEC plus measure of angle DEC is equal to 180 degrees. One diagonal is the perpendicular bisector of the other. Solve for x | Find the Indicated Angles in a Kite. When all the angles are also 90° the Kite will be a Square. The two non-vertex angles are always congruent. Covid-19 has led the world to go through a phenomenal transition . Add all known angles and subtract from 360° to find the vertex angle, and subtract the sum of the vertex angles from 360° and divide by 2 to find the non-vertex angle. 2. It looks like the kites you see flying up in the sky. In every kite, the diagonals intersect at 90 °. These sides are called as distinct consecutive pairs of equal length. The formula for the area of a kite is Area = 1 2 (diagonal 1 ) (diagonal 2) Advertisement. Kite. A Kite is a flat shape with straight sides. Metric formulas. • noparallel sides. Parallel, Perpendicular and Intersecting Lines. Use the appropriate properties and solve for x. Kite properties. Explanation: . Other important polygon properties to be familiar with include trapezoid properties , parallelogram properties , rhombus properties , and rectangle and square properties . 4. Types of Kite. Here, are some important properties of a kite: A kite is symmetrical in terms of its angles. ... Properties of triangle. 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Fibonacci equality, proving it someway

$F_{2n} = F_n(F_n+2F_{n-1})$

$F_n$ is a nth Fibonacci number. I tried by induction but i didn't get anywhere

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An answer just by induction on $n$ of the equality $F_{2n}=F_n(F_{n-1}+F_{n+1})$ is as follows:

For $n=2$ we have $3=F_4=(1+3)\cdot 1=(F_1+F_3)F_2$.

To go from $n$ to $n+1$:

\begin{align} F_{2n+2}&=3F_{2n}-F_{2n-2}\\ &=3(F_{n-1}+F_{n+1})F_n-(F_{n-2}+F_n)F_{n-1}\\ &=F_{n-1}(3F_n-F_n-F_{n-2})+3F_{n+1}F_n\\ &=F_{n-1}(F_n+F_{n-1})+3F_{n+1}F_n\\ &=F_{n-1}F_{n+1}+3F_{n+1}F_n\\ &=F_{n+1}(3F_n+F_{n-1})\\ &=F_{n+1}(2F_n+F_{n+1})\\ &=F_{n+1}(F_n+F_{n+2}) \end{align}

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A combinatorial interpretation:

$F_n$ is the number of ways to tile a row of $(n-1)$ squares with $1\times 1$ blocks and $1\times 2$ blocks.

The left hand side is the number of ways to tile a $1\times (2n-1)$ block with $1\times 1$ and $1\times 2$ blocks. Consider the middle square (the $(n-1)$th square.)

Case 1: It is used in a $1\times 1$ block. Then, there are $F_{n}$ ways to tile each of the $1 \times (n-1)$ blocks on each side of the middle, so $F_n^2$ total.

Case 2: It is used in a $1\times 2$ block. This block contains the $(n-1)$th square and either the $n$th or the $(n-2)$th square. In either case, there are $F_{n-1}$ ways to tile the shorter side and $F_n$ ways to tile the longer side.

We thus have $F_{2n} = F_n^2 + 2F_nF_{n-1},$ as desired.

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I didn't have mentioned about your way of doing it. I can't fully understand your solution. The man said it's solutionable by induction –  matiit Oct 14 '12 at 15:27
Not useful to the OP, perhaps, but still very nice. –  Brian M. Scott Oct 14 '12 at 15:46

This is very related to the answer at Showing that an equation holds true with a Fibonacci sequence: $F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$

That question has the identity:

$F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$

which can be modeled to your identity by letting $n=m$

$F_{2n} = F_{n-1}F_n + F_n F_{n+1}$

$F_{2n} = F_n(F_{n-1} + F_{n+1})$

Setting the expression inside the parenthesis to:

$F_{n-1} + F_{n+1} = F_{n-1}+ F_{n-1} + F_{n} = F_n + 2F_{n-1}$

We get

$F_{2n} = F_n(F_n+2F_{n-1})$

Which is your identity. So work backwards to that identity and use the proof at the linked question to prove your relation.

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Here’s a purely computational proof. Let $$A=\pmatrix{F_2&F_1\\F_1&F_0}=\pmatrix{1&1\\1&0}\;;$$ a straightforward induction shows that $$A^n=\pmatrix{F_{n+1}&F_n\\F_n&F_{n-1}}$$ for all $n\ge 1$. Then

\begin{align*} \pmatrix{F_{m+n+1}&F_{m+n}\\F_{m+n}&F_{m+n-1}}&=A^{m+n}\\ &=A^mA^n\\\\ &=\pmatrix{F_{m+1}&F_m\\F_m&F_{m-1}}\pmatrix{F_{n+1}&F_n\\F_n&F_{n-1}}\\\\ &=\pmatrix{F_{m+1}F_{n+1}+F_mF_n&F_{m+1}F_n+F_mF_{n-1}\\F_mF_{n+1}+F_{m-1}F_n&F_mF_n+F_{m-1}F_{n-1}}\;, \end{align*}

so $F_{m+n}=F_{m+1}F_n+F_mF_{n-1}$. Take $m=n$, and this becomes

$$F_{2n}=F_{n+1}F_n+F_nF_{n-1}=F_n\left(F_{n+1}+F_{n-1}\right)=F_n\left(F_n+2F_{n-1}\right)\;.$$

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Exact duplicate of Zchpyvr's 21-minute prior answer (you've simply inlined the lemma that he links to). –  Bill Dubuque Oct 14 '12 at 19:06
@Bill: I’m not terribly surprised. However, I didn’t see his answer until after I posted, and when I did see it, I was busy and didn’t feel like chasing down the link. And quite frankly, I don’t consider it an exact duplicate for that very reason. –  Brian M. Scott Oct 14 '12 at 21:52
It is most certainly an exact duplicate. As I said, you've simply inlined the link in the other answer. Lacking anything new, it should be deleted for the sake of the readers. –  Bill Dubuque Oct 14 '12 at 21:58
@Bill: You have a rather inexact definition of exact. The fact that the information is right on the page is a difference, in convenience if nothing else. And deleting it does not serve the readers; slightly the reverse, if anything, for that same reason. The only person whom it might possibly ill serve is Zchpyvr, and I’ve upvoted his answer. And that’s the end of it as far as I’m concerned. –  Brian M. Scott Oct 14 '12 at 22:04
Posting duplicate answers potentially wastes many reader's time, since they may read two or more answers when it would have sufficed to read one. Not to mention that the abstraction in Zchpyvr's answer gained by calling the lemma by name (vs. value) only serves to make the answer more comprehensible. By your argument, every textbook should inline the proof of all lemma's so that they are "right on the page". That is, of course, absurd. –  Bill Dubuque Oct 14 '12 at 22:14

Another direct proof, using the fact that

$$F_n=\frac{\phi^n-(1-\phi)^n}{\sqrt5}\tag1$$

where

$$\phi=\frac{1+\sqrt5}{2},\qquad1-\phi=\frac{1-\sqrt5}{2}=-\frac{1}{\phi}.$$

From $(1)$ we have

$$s_n\equiv\frac{F_{2n}}{F_n}=\frac{\phi^{2n}-(1-\phi)^{2n}}{\phi^n-(1-\phi)^n}=\phi^n+(1-\phi)^n$$

then

\begin{align} s_n-F_n&=\phi^n+(1-\phi)^n-\frac{\phi^n-(1-\phi)^n}{\sqrt5}=\\ &=\frac{1}{\sqrt5}\left[\sqrt5\phi^n+\sqrt5(1-\phi)^n-\phi^n+(1-\phi)^n\right]=\\ &=\frac{1}{\sqrt5}\left[-(1-\sqrt5)\phi^n+(1+\sqrt5)(1-\phi)^n\right]=\\ &=\frac{1}{\sqrt5}\left[-2(1-\phi)\phi^n+2\phi(1-\phi)^n\right]=\\ &=-\frac{2\phi(1-\phi)}{\sqrt5}\left[\phi^{n-1}-(1-\phi)^{n-1}\right]=2F_{n-1} \end{align}

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This can be done also using the fact that

$$F_n = \frac{1}{\sqrt{5}}\left(\sigma^n-\bar{\sigma}^n\right),$$

from which it is easy to get

$$F_{2n} = F_n L_n,$$

(where $L_n$ is the $n$-th Lucas number) and we have only to prove

$$L_n= F_n+2F_{n-1},$$

that is true since $\{L_n\}_{n\in\mathbb{N}}$ and $\{F_n+2F_{n-1}\}_{n\in\mathbb{N}}$ are sequences with the same characteristic polynomial ($x^2-x-1$) and the same starting values $L_1=F_1+2F_0=1$, $L_2=F_2+2F_1=3$.

-


# Solutions to Congruence Modulo 50

1. Jan 12, 2013

### knowLittle

1. The problem statement, all variables and given/known data
Find all solutions to the equation $35x\equiv 10mod50$

3. The attempt at a solution

gcd( 35,50)= 5
So, there is a solution to this, since 5| 10. Also, there is a theorem that guarantees the existence of exactly 5 solutions.

Now, dividing $35x\equiv 10mod50$ over 5 gives:
$7x\equiv 2mod10$
Now, what multiple of 7 gives us $\equiv 2mod10$
{ 2, 12, 22, 32, 42,...} Here, we found 42 that is a multiple of 7 and satisfies $\equiv 2mod10$
We can write $7x\equiv 42mod10$ . Now, I divided the expression by 7 and got $x\equiv 6mod10$

Now, there is another theorem that tells me this
6+(50/5)t, t=0, 1, ..., 4
I get: 6, 16, 26, 36, 46
So, the solutions are
$x\equiv 6mod50$ , $x\equiv 16 mod50$, $x\equiv 26 mod50$, $x\equiv 36mod 50$, $x\equiv 46mod 50$
These are the 5 solutions of $35x\equiv 10mod50$

I found other solutions online:
So x = 6, 16, 22, 28, 34, 40, 46 modulo 50 are the
solutions to the congruence 35x ≡ 10 mod 50.

Am I incorrect?

Thank you.

2. Jan 12, 2013

### kru_

Your method is correct. You can check your own solutions to verify that they are correct.

35*6 = 210 which is congruent to 10 mod 50.
35*16 = 560 which is also congruent to 10 mod 50.
Similarly for the other solutions that you found.

You can verify that 22, 28, and 40 are not solutions.
35*22 = 770 which is 20 mod 50.
35*28 = 980, is 30 mod 50.
etc.

3. Jan 12, 2013

### knowLittle

I have read somewhere that division is not defined in modular arithmetic. Can someone tell me how this affect my solution?

@kru: This is puzzling, since I found those other solutions at a .edu site.

4. Jan 12, 2013

### HallsofIvy

Division is not necessarily defined in modular arithmetic because there may be "0 divisors". For example, in modulo 6, 2(3)= 6= 0 (mod 6). If we had an equation of the form 3x= 1 (mod 6) we can immediately check that 3(1)= 3, 3(2)= 6, 3(3)= 9= 3, 3(4)= 12= 0, and 3(5)= 15= 3 mod 6. There is NO x such that 3x= 1 and so we could not, for example, "divide by 3" to get "1/3" as an answer. If we are working "modulo" a prime number, that doesn't happen and we can define "division".

The way I would do "35x= 10 mod 50" is this. This is the same as saying 35x= 50n+ 10 for some integer n- a linear "diophantine equation". The first thing we can do divide through by 5 to 7x= 10n+ 2 or 7x- 10n= 2. Now 7 divides into 10 once with remainder 3: 3= 10- 7. 3 divides into 7 twice with remainder 1: 1= 7- 2(3). We can replace that "3" with 10- 7 from the first equation: 1= 7- 2(10- 7)= 3(7)- 2(10)= 1 (The "Euclidean Divison Algorithm"). Multiply through by 2 to get 6(7)- 4(10)= 2.

So one solution to 7x- 10n= 2 is x= 6, n= 4. It is possible to write out the "general solution" but since 6 itself is between 0 and 10, x= 6 satisfies 7(6)= 2 (mod 10) and so 35(6)= 210= 10 (mod 50).

5. Jan 12, 2013

### knowLittle

According to Wikipedia, Diophantine equations are written as follows:
ax + by = c
The Diphantine equation that you are really writing is this
35x-50n=10?

I understand everything, until you change the equation 1=7 -2(10-7)= 3(7)-2(10)=1. I understand that 21-20=1, but why changing from 7- 2(10-7) to 3(7)-2(10)?
Also, I am acquainted with Euclid's GCD algorithm:
Euclid(a,b)
if b==0
return a
else return Euclid (b, a mod b)

Is there a way to use it without having to trace it?

Is this all solutions for 35x $\equiv$ 10 mod 50? Also, is it correct that there has to be exactly 5 solutions, since the gcd of 35, 50 is 5?
Is my solution correct?


# Math Help - Complex numbers-finding real number pairs

1. ## Complex numbers-finding real number pairs

Hello,
I am having trouble with this question:

"Find all possible real number pairs p, q such that 3+5i/1+pi =q+4i"

Im sure it's easy but I think I am overlooking something. I multiplied both sides by the conjugate of 1+pi....ie.(1-pi) but I think it's wrong.

Any help would be appreciated.

Regards,
Neverquit

2. Multiply both sides by 1+pi

Then you get...

3 + 5i = (1 + pi)(q + 4i) = q + pqi + 4i - 4p

=> 3 + 5i = (q - 4p) + 4i + pqi

=> 3 + i = (q - 4p) + pqi

So you want to solve...

q - 4p = 3

pq = 1

3. ## found 1 solution but not the other

I found the solutions 0.25, -1 after re-arranging pq = 1 to q = 1/p to get 1/p - 4p = 3 to get quadratic 4p^2+3p-1=0.

There is also apparently another solution of 4, -1 which I can’t find.

How do I find it?

4. This is a tricky question.
$\frac{{3 + 5i}}{{1 + pi}} = \frac{{\left( {3 + 5i} \right)\left( {1 - pi} \right)}}{{1 + p^2 }} = \frac{{3 + 5p}}{{1 + p^2 }} + \frac{{\left( {5 - p} \right)i}}{{1 + p^2 }}$
Now set real part equal to real part and imaginary equal to imaginary.
$\frac{3+5p}{1+p^2}=q~\&~\frac{5-3p}{1+p^2}=4$
From that we get $p=\frac{1}{4}~\&~p=-1$.

5. Originally Posted by Neverquit
I found the solutions 0.25, -1 after re-arranging pq = 1 to q = 1/p to get 1/p - 4p = 3 to get quadratic 4p^2+3p-1=0.

There is also apparently another solution of 4, -1 which I can’t find.

How do I find it?
I think you are misinterpreting your own answer. When you solve the system of equations, you will find that

$p=1/4 ~~\mbox{ or }~~ p=-1$

These are two separate solutions, not a single solution. You need to find the value of $q$ that pairs with each of these solutions for $p$. So, you need to plug each value of $p$ back into the system of equations and find the corresponding values of $q$. Since $pq=1$, it's a pretty straightforward calculation:

$p=1/4 \implies q=4$

$p=-1 \implies q=-1$

Therefore, the solutions are:

First solution: $p=1/4 \mbox{ and } q=4$

Second solution: $p=-1 \mbox{ and } q=-1$

You might also write this as:

$(p,q) = (1/4,4) \mbox{ or } (p,q) = (-1,-1)$

But you definitely would not say that the solutions are $(1/4,-1)$ and $(4,-1)$.

6. Originally Posted by Neverquit
Hello,
I am having trouble with this question:

"Find all possible real number pairs p, q such that 3+5i/1+pi =q+4i"

Im sure it's easy but I think I am overlooking something. I multiplied both sides by the conjugate of 1+pi....ie.(1-pi) but I think it's wrong.

Any help would be appreciated.

Regards,
Neverquit
$\frac{3+5i}{1+pi} =q+4i$

Multiply top and bottom of LHS by $1-pi$ to give:

$\frac{(3+5p)+(5-3p)i}{1+p^2}=q+4i$

Equate real and imaginary parts to get:

$3+5p=(1+p^2)q$

and:

$5-3p=4(1+p^2)$

Now the problem is to find all solutions to this pair of equations.

CB

7. Having Plato and CP post in this thread has made me question whether my answer is wrong. Is it?

Solving the equations I arrived at...

q - 4p = 3

pq = 1

gives

q=1/p,

So subbing that into the first equation will give you a polynomial (if you multiply both sides by p) which gives you p = 1/4 and -1.

Hence q = 4 and -1.

Solutions are (1/4,4) and (-1, -1)

Are these the only solutions? Why must you multiply the top and bottom by the conjugate?

Having Plato and CP post in this thread has made me question whether my answer is wrong. Is it?
Your solution was correct. Multiplying a complex fraction by the conjugate of the denominator is probably just a habit for them. But it's not necessary at all in this particular problem.

9. ## Using the conjugate

Originally Posted by Plato
This is a tricky question.
$\frac{{3 + 5i}}{{1 + pi}} = \frac{{\left( {3 + 5i} \right)\left( {1 - pi} \right)}}{{1 + p^2 }} = \frac{{3 + 5p}}{{1 + p^2 }} + \frac{{\left( {5 - p} \right)i}}{{1 + p^2 }}$
Now set real part equal to real part and imaginary equal to imaginary.
$\frac{3+5p}{1+p^2}=q~\&~\frac{5-3p}{1+p^2}=4$
From that we get $p=\frac{1}{4}~\&~p=-1$.
I think the solution that Plato gives using the congugate is what the text books author had in mind as the question is shortly after conjugates of complex numbers is explained.

Deastar, your solution still gives the same answer in the text book so it must be correct.

...........

10. ## generalisation

Originally Posted by Neverquit
Hello,
I am having trouble with this question:

"Find all possible real number pairs p, q such that 3+5i/1+pi =q+4i"

Im sure it's easy but I think I am overlooking something. I multiplied both sides by the conjugate of 1+pi....ie.(1-pi) but I think it's wrong.

Any help would be appreciated.

Regards,
Neverquit
the basic idea behind questions of your type is equaslising real and imaginary parts.


# What is the probability that Jan and Jon are chosen?

Jan , Jon and $10$ other children are in a classroom. The principal of the school walks in and choose $3$ children at random. What is the probability that Jan and Jon are chosen?

My approach:

Including Jon and Jan, number of ways of selection is $1\cdot1\cdot{10\choose 1} = 10$.
Total way of selection is ${12\choose 3} = 220$

So, probability is $\frac{10}{220}= \frac{1}{22}.$

But when I solving as follows:

\begin{align*} P(\text{selection of Jon and Jan}) &= 1- P(\text{not selection of Jon and Jan}) \\ &= 1- \frac{{10\choose 3}}{{12\choose 3}}\\ &= \frac{5}{11}. \end{align*}

Which approach is correct and why alternative one is wrong?

Note: My previous post had some mistakes, so I deleted that

I will call $A$, $B$ the events selecting Jon, Jan respectively. You did not take the complement correctly \begin{align*} P(A\cap B) &= 1-\color{red}{P(\bar A \cup\bar B)}\tag 1 \\ &= 1-[P(\bar A)+P(\bar B)-P(\bar A\bar B)]\tag 2 \\ &=1-\left[\frac{\binom{11}{3}}{\binom{12}{3}}+\frac{\binom{11}{3}}{\binom{12}{3}}-\frac{\binom{10}{3}}{\binom{12}{3}}\right]\\ &=\frac{1}{22} \end{align*} where $(1)$ is true by DeMorgan's law, and $(2)$ is true by inclusion-exclusion. As you can see the two methods give the same value.

Your first approach gives the probability that Jon AND Jan are selected, the second gives the probability that Jon OR Jan are selected.

When you computed the probability of not selecting Jon AND Jan, you didn't include the situations where Jon was selected but not Jan, and vice versa.

Lets say the even Jon is selected is A, Jan is selected B.

Then $P(A \wedge B)=1-P(Not (A \wedge B))= 1- P(Not A \vee Not B)$

$P(Not A \vee Not B)= P(Not A)+ P(Not B)- P(Not A \wedge Not B)$

Here $\wedge$ means and, and $\vee$ means or. Think of a Venn diagram, if we look at the union of two circles, the total area is equal to the sum of the circles minus the intersection, because we counted that part twice.

$P(Not A)=1-P(A)=1-(1*\binom{11}{2})/220)=3/4$

or

$P(Not A)=\binom{11}{3}/220=3/4$

$P(Not B)=3/4$

$P(Not A \wedge Not B)= \binom{10}{3}/220=6/11$

Now $3/4+3/4-6/11=21/22$ So $1-21/22=1/22$