Coefficients involved in the Binomial Theorem. $\binom{n}{k}$ counts the subsets of size $k$ of a set of size $n$.

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Expected Value of a Binomial distribution?

If $\mathrm P(X=k)=\binom nkp^k(1-p)^{n-k}$ for a binomial distribution, then from the definition of the expected value $$\mathrm E(X) = \sum^n_{k=0}k\mathrm P(X=k)=\sum^n_{k=0}k\binom ...
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Proving $\sum_{k=0}^n{2k\choose k}{2n-2k\choose n-k}=4^n$ [duplicate]

Some background. I was asked to find an arithmetic function $f$ such that $f*f=\mathbf 1$ where $\mathbf 1$ is the constant function 1 and $*$ denotes Dirichlet convolution. I was able to prove that ...
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2answers
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Prove $(1-x)^{2k+1} \sum\limits_{n\ge 0}\binom{n+k-1}{k}\binom{n+k}{k} x^n = {\sum\limits_{j\ge 0} \binom{k-1}{j-1}\binom{k+1}{j} x^j} $

I stumbled upon the identity $$(1-x)^{2k+1} \sum\limits_{n\ge 0}\binom{n+k-1}{k}\binom{n+k}{k} x^n = {\sum\limits_{j\ge 0} \binom{k-1}{j-1}\binom{k+1}{j} x^j}. $$ The right-hand side is a polynomial. ...
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1answer
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Binomial Congruence

How can we show that $\dbinom{pm}{pn}\equiv\dbinom{m}{n}\pmod {p^3}$ for positive integers m and n and p a prime greater than 5? I can do it for mod p^2 but Im stuck here.
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1answer
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Are these two binomial sums known? Proven generalization to the Hockey Stick patterns in Pascal's Triangle

English translation. You can see the original - deprecated - in Portuguese here Hi, I arrived at a generalization for the Hockey Stick Patterns, from our beloved Pascal's Triangle. This ...
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2answers
310 views

A conjecture including binomial coefficients

Question: $$\sum_{k=1}^{n}k\binom{2n}{n+k}=\frac n2\binom{2n}{n}$$ is true for every $n\in \mathbb N$? If this is true, then how can we prove this? When I was playing with numbers, I conjectured ...
12
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382 views

How prove this $\sum_{k=0}^{n} \frac{\binom{2n-k}{n}}{2^{2n-k}}=1 $

Show that $$\sum_{k=0}^{n}\dfrac{\binom{2n-k}{n}}{2^{2n-k}}=1$$ I think this problem can be solved with nice methods, such as algebraic ones. Or can I use probability methods? Thank you
12
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7answers
535 views

Prove that $2^n < \binom{2n}{n} < 2^{2n}$

Prove that $2^n < \binom{2n}{n} < 2^{2n}$. This is proven easily enough by splitting it up into two parts and then proving each part by induction. First part: $2^n < \binom{2n}{n}$. The ...
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4answers
866 views

Combinatorial proof that binomial coefficients are given by alternating sums of squares?

A student recently asked whether there was a combinatorial proof of the following identity: $\begin{equation*} \sum^n_{k=1}(-1)^{n-k}k^2 = {n+1 \choose 2}. \end{equation*}$ I was in a rush and ...
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Inequality $\binom{2n}{n}\leq 4^n$

I would like to prove the following inequality, for $n=0,1,2,...$, $$ \binom{2n}{n}\leq 4^n.$$ I already proved it by induction, and I'm looking for another proof.
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1answer
163 views

Interpretation of a combinatorial identity

I am trying to find an combinatorial interpretation for the following combinatorial identity involving iterated binomial coefficients, which appeared in the November 1980 edition of The American Math ...
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0answers
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Solution of $\large\binom{x}{n}+\binom{y}{n}=\binom{z}{n}$ with $n\geq 3$

I found this question in an old problem set. There's no hint or solution mentioned. For $n \geq 3$, prove or disprove the existence of $(x,y,z) \in \mathbb N^3, ...
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1answer
220 views

Summation of weighted squares of binomial coefficients

Show that $$\sum_{k=0}^n \left[ \frac{n-2k}{n} {n\choose k}\right]^2=\frac{2}{n}{2n-2 \choose n-1}.$$
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Combinatorial proof for $\binom{n}{a}\binom{a}{k}\binom{n-a}{b-k} = \binom{n}{b}\binom{b}{k}\binom{n-b}{a-k}$?

I have to prove the following using a combinatorial proof: $\binom{n}{a}\binom{a}{k}\binom{n-a}{b-k} = \binom{n}{b}\binom{b}{k}\binom{n-b}{a-k}$ Ok, so here is what I have worked out so far: We ...
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1answer
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Hard binomial sum [closed]

How to prove this relation? $$\sum_{i=0}^{n}\frac{2^{-2i}\binom{2i}{i}}{n+i+2}=\frac{2^{4n+2}-\binom{2n+1}{n}^2}{(2n+3)2^{2n+1}\binom{2n+1}{n}}$$ That seems difficult!
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3answers
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Evaluate a sum with binomial coefficients

$$\text{Find} \ \ \sum_{k=0}^{n} (-1)^k k \binom{n}{k}^2$$ I expanded the binomial coefficients within the sum and got $$\binom{n}{0}^2 + \binom{n}{1}^2 + \binom{n}{2}^2 + \dots + \binom{n}{n}^2$$ ...
13
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1answer
215 views

Fitting a closed curve on the roots of ${x \choose k}-c$

Let $${n \choose k} = \frac1{(n+1) \operatorname{B}(n-k+1, k+1)}.$$ be the generalized binomial coefficient, and here $\operatorname{B}$ is the Beta function. Let $f_{k,c}(x)$ be the following ...
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Asymptotics of the sum of squares of binomial coefficients

We are trying to estimate the cardinality $K(n,p)$ of so-called Kuratowski monoid with $p$ positive and $n$ negative linearly ordered idempotent generators. In particular, we are interesting in the ...
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5answers
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Proving identities like $\sum_{k=1}^nk{n\choose k}^2=n{2n-1\choose n}$ combinatorially

I have to give a combinatorial proof of $$\sum_{k=1}^nk{n\choose k}^2=n{2n-1\choose n}.$$ I find it difficult to solve such problems. I'm not a brilliant person and never will be so I need to have ...
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Binomial Sum Related to Fibonacci: $\sum\binom{n-i}j\binom{n-j}i=F_{2n+1}$

How would I prove $$ \sum\limits_{\vphantom{\large A}i\,,\,j\ \geq\ 0}{n-i \choose j} {n-j \choose i} =F_{2n+1} $$ where $n$ is a nonnegative integer and $\{F_n\}_{n\ge 0}$ is a sequence of ...
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Let $a_k=\frac1{\binom{n}k}$, $b_k=2^{k-n}$. Compute $\sum_{k=1}^n\frac{a_k-b_k}k$

Let $a_k=\frac1{\binom{n}k}$, $b_k=2^{k-n}$. Compute $$\sum_{k=1}^n\frac{a_k-b_k}k$$ By computing some partial sums, the answers are 0. It seems an inductive argument is possible.
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An extrasensory perception strategy :-)

Inspired by classical Joseph Banks Rhine experiments demonstrating an extrasensory perception (see, for instance, the beginning of the respective chapter of Jeffrey Mishlove book “The Roots of ...
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3answers
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Prove that $\prod_{k=1}^{\infty} \big\{(1+\frac1{k})^{k+\frac1{2}}\big/e\big\} = \dfrac{e}{\sqrt{2\pi}}$

This result, $$\prod_{k=1}^{\infty} \big\{\big(1+\frac1{k}\big)^{k+\frac1{2}}\big/e\big\} = \dfrac{e}{\sqrt{2\pi}}$$ is in a paper by Hirschhorn in the current issue of the Fibonacci Quarterly (vol. ...
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Proof of the identity $\sum_{k=0}^{\min[p,q]}{p\choose k}{q\choose k}{n+k\choose p+q}={n\choose p}{n\choose q}$

Prove the identity: $$\sum_{k=0}^{\min[p,q]}{p\choose k}{q\choose k}{n+k\choose p+q}={n\choose p}{n\choose q}.$$
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Combinatorial Proof of $\binom{\binom{n}{2}}{2} = 3 \binom{n}{3}+ 3 \binom{n}{4}$ for $n \geq 4$

For $n \geq 4$, show that $\binom{\binom{n}{2}}{2} = 3 \binom{n}{3}+ 3 \binom{n}{4}$. LHS: So we have a set of $\binom{n}{2}$ elements, and we are choosing a $2$ element subset. RHS: We are ...
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Choice Problem: choose 5 days in a month, consecutive days are forbidden

I'm "walking" through the book "A walk through combinatorics" and stumbled on an example I don't understand. Example 3.19. A medical student has to work in a hospital for five days in January. ...
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The maximum of $\binom{n}{x+1}-\binom{n}{x}$

The following question comes from an American Olympiad problem. The reason why I am posting it here is that, although it seems really easy, it allows for some different and really interesting ...
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Truncated alternating binomial sum

It is easily checked that $\displaystyle\sum_{i\ =\ 0}^{n}\left(\, -1\,\right)^{i} \binom{n}{i} = 0$, for example by appealing to the binomial theorem. I'm trying to figure out what happens with the ...
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Formula for $\sum_{k=0}^n k^d {n \choose 2k}$

If $d \geq 1$ is an integer, is there a general formula for $$\sum_{k=0}^n k^d {n \choose 2k}\,?$$ We know that $\sum_{k=0}^n k {n \choose 2k} = \frac{n2^n}{8}$ and $\sum_{k=0}^n k^2 {n \choose 2k} = ...
6
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Binomial probability with summation

Show that $$\sum_{k=0}^{m} \frac{m!(n-k)!}{n!(m-k)!} = \frac{n+1}{n-m+1}$$ Attempt: It becomes: $$\sum_{k=0}^{m } \frac{\binom{m}{k}}{\binom{n}{k}}$$ Telescoping, pairing, binomial theorem don't ...
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Negative Exponents in Binomial Theorem

I'm looking at extensions of the binomial formula to negative powers. I've figured out how to do $n \choose k$ when $n < 0 $ and $k \geq 0$: $${n \choose k} = (-1)^k {-n + k - 1 \choose k}$$ So ...
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Sums of binomial coefficients

Does anyone know something about the following sums? $$ S_m(n)=\sum\limits_{k=o}^n(-1)^k{mn\choose mk} $$ Notice that $S_m(n)=0$ for odd $n$, so we only consider $S_m(2n)$. It holds that $S_0(2n)=1$, ...
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4answers
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Closed form for $\sum_{k=0}^{n} \binom{n}{k}\frac{(-1)^k}{(k+1)^2}$

How can I calculate the following sum involving binomial terms: $$\sum_{k=0}^{n} \binom{n}{k}\frac{(-1)^k}{(k+1)^2}$$ Where the value of n can get very big (thus calculating the binomial ...
4
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5answers
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Spivak's Calculus - Exercise 4.a of 2nd chapter

4 . (a) Prove that $$\sum_{k=0}^l \binom{n}{k} \binom{m}{l-k} = \binom{n+m}{l}.$$ Hint: Apply the binomial theorem to $(1+x)^n(1+x)^m$. I'm having a hard time trying to solve the problem ...
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How do I prove that there infinitely many rows of Pascal's triangle with only odd numbers?

This is exercise number $59$ from Chapter $2$ of Hugh Gordon's Discrete Probability. Show that there are infinitely many rows of Pascal's Triangle that consist entirely of odd numbers. ...
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Sum of product of binomial coefficient

Is the following true? $$\sum_{x_1+x_2+...+x_n=n}\ \ \, \prod_{i=1}^{n}{k_i\choose x_i}={\sum_{i=1}^{n}k_i \choose n} .$$ I tried to use the multinomial theorem, but it doesn't seem applicable.
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Obtaining binomial coefficients without “counting subsets” argument

I want to obtain the formula for binomial coefficients in the following way: Elementary ring theory shows that $(X+1)^n\in\mathbb Z[X]$ is a degree $n$ polynomial, for all $n\geq0$, so we can write ...
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4answers
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Orthogonality for Binomial Coefficients

Could somebody explain to me where these two formulas come from as applications of the binomial theorem? $$\sum_{k=0}^n {n \choose k}(-1)^kk^r=0$$ for non-negative integers $r\lt n$. And ...
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A peculiar binomial coefficient identity

While inventing exercises for a discrete math text I'm writing I came up with this $$ \binom{\binom{n}{2}}{2}=3\binom{n+1}{4} $$ It's an easy result to prove, but it got me wondering Is this pure ...
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Does this really converge to 1/e? (Massaging a sum)

Short version: can we prove that $$\sum_{k=0}^n (-1)^k \binom{n}{k}^2 \frac{k!}{n^{2k}} \to \frac1e$$ as $n \to \infty$? Long version: First, consider $$a_n = \sum_{k=0}^n \frac{(-1)^k}{k!}$$ It is ...
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Asymptotics of $\sum_{k=0}^{n} {\binom n k}^a$

I need to estimate the asymptotics of $$\sum_{k=0}^{n} {\binom n k}^a, \quad a>2, \quad a \in \mathbb{N}$$ In particular, I'm pretty much interested in $a=4$ case, but if the general solution ...
7
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Combinatorial Identity $(n-r) \binom{n+r-1}{r} \binom{n}{r} = n \binom{n+r-1}{2r} \binom{2r}{r}$

Show that $(n-r) \binom{n+r-1}{r} \binom{n}{r} = n \binom{n+r-1}{2r} \binom{2r}{r}$. In the LHS $\binom{n+r-1}{r}$ counts the number of ways of selecting $r$ objects from a set of size $n$ where ...
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1answer
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How to transform the product to sum?

I just wonder that how to prove that $$ \prod_{m=1}^{n}\Big(x-2\cos\frac{m\pi}{n+1}\Big)=\sum_{k=0}^{[n/2]}(-1)^{k}\binom{n-k}{k}x^{n-2k}. $$ Similarly, how to transform the product $$ ...
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Wanted: Insight into formula involving binomial coefficients

There is the formula $$\sum_i (-1)^i\binom{a}{k+i}\binom{l+i}{b} = (-1)^{a+k} \binom{l-k}{b-a}.$$ Only finitely many summands are non-zero (those for $i\in\{b-l,\ldots,a-k\}$), so the sum is finite. ...
6
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1answer
847 views

Relation between different ways of accessing bernoulli numbers with matrices

First Variant: Bernoulli numbers can easily be expressed by linear algebra equations. For example just using the recursion formula $$\sum_{k=0}^{n-1}{n\choose k}B_k=0$$ which is equation (34) from ...
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How prove this $\sum_{k=0}^n \binom{n}{k} \binom{(p-1)n}{k} \binom{pn+k}{k} = \binom{pn}{n}^2 $

I think the following equality is true ($p\in \mathbb{N},p\ge 2$): $$\sum_{k=0}^n \binom{n}{k} \binom{(p-1)n}{k} \binom{pn+k}{k} = \binom{pn}{n}^2 $$ when $p=2$, then ...
5
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2answers
318 views

Proof of inequality $\sum\limits_{k=0}^{n}\binom n k\frac{5^k}{5^k+1}\ge\frac{2^n\cdot 5^n}{3^n+5^n}$

Show that $$\sum_{k=0}^{n}\binom n k\frac{5^k}{5^k+1}\ge\frac{2^n\cdot 5^n}{3^n+5^n}$$ where $$\binom n k=\frac{n!}{k!(n-k)!}$$
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votes
3answers
586 views

A Binomial Coefficient Sum

In my work on $f$-vectors in polytopes, I ran across an interesting sum which has resisted all attempts of algebraic simplification. Does the following binomial coefficient sum simplify? \begin{align} ...
4
votes
2answers
298 views

maybe this sum have approximation $\sum_{k=0}^{n}\binom{n}{k}^3\approx\frac{2}{\pi\sqrt{3}n}\cdot 8^n,n\to\infty$

prove or disprove this $$\sum_{k=0}^{n}\binom{n}{k}^3\approx\dfrac{2}{\pi\sqrt{3}n}\cdot 8^n,n\to\infty?$$ this problem is from when Find this limit ...
4
votes
3answers
374 views

Number of binary strings with $n$ ones and $m$ zeros

$f(n,m)$ is the number of binary strings with up to $n$ ones and up to $m$ zeros. Prove that the number of possible strings is: $${n+m+2 \choose n+1} -1$$ I got to the point that: $$\sum_{a=0}^n ...