Mathematics Stack Exchange
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Questions tagged [binomial-coefficients]

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

The binomial coefficient $\dbinom{n}{k}$ can be defined in several equivalent ways for $n$ and $k$ non-negative integers:

  1. The number of subsets of size $k$ of a set of size $n$.
  2. Element $k$ of row $n$ in Pascal's triangle (counting the first element or row as $0$).
  3. The coefficient of $x^k$ in $(1+x)^n$.
  4. $\dfrac{n!}{k!(n-k)!}$

The binomial theorem says that $$(x+y)^n=\sum_{k=0}^n\binom{n}{k}x^{n-k}y^k$$ using the convention that $0^0=1$.

Binomial coefficients can be extended for arbitrary complex $\alpha$ through the formula: $$\binom{\alpha}{k}=\frac{\alpha(\alpha-1)(\alpha-2)\dots(\alpha-k+1)}{k(k-1)(k-2)\dots1}$$

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Identity for convolution of central binomial coefficients: $\sum\limits_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=2^{2n}$

It's not difficult to show that $$(1-z^2)^{-1/2}=\sum_{n=0}^\infty \binom{2n}{n}2^{-2n}z^{2n}$$ On the other hand, we have $(1-z^2)^{-1}=\sum z^{2n}$. Squaring the first power series and comparing terms gives us $$\sum_{k=0}^n…
Skatche
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Combinatorial proof of summation of $\sum\limits_{k = 0}^n {n \choose k}^2= {2n \choose n}$

I was hoping to find a more "mathematical" proof, instead of proving logically $\displaystyle \sum_{k = 0}^n {n \choose k}^2= {2n \choose n}$. I already know the logical Proof: $${n \choose k}^2 = {n \choose k}{ n \choose n-k}$$ Hence summation can…
Lance C
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Proof that a Combination is an integer

From its definition a combination $\binom{n}{k}$, is the number of distinct subsets of size $k$ from a set of $n$ elements. This is clearly an integer, however I was curious as to why the expression $$\frac{n!}{k!(n-k)!}$$ always evaluates to an…
Akusete
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Combinatorial proof that $\sum \limits_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} (-1)^k = 2^n \binom{n}{n/2}$ when $n$ is even

In my answer here I prove, using generating functions, a statement equivalent to $$\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} (-1)^k = 2^n \binom{n}{n/2}$$ when $n$ is even. (Clearly the sum is $0$ when $n$ is odd.) The nice expression on the…
Mike Spivey
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Proof of the hockey stick/Zhu Shijie identity $\sum\limits_{t=0}^n \binom tk = \binom{n+1}{k+1}$

After reading this question, the most popular answer use the identity $$\sum_{t=0}^n \binom{t}{k} = \binom{n+1}{k+1}.$$ What's the name of this identity? Is it the identity of the Pascal's triangle modified. How can we prove it? I tried by…
hlapointe
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A comprehensive list of binomial identities?

Is there a comprehensive resource listing binomial identities? I am more interested in combinatorial proofs of such identities, but even a list without proofs will do.
user813
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$n$th derivative of $e^{1/x}$

I am trying to find the $n$'th derivative of $f(x)=e^{1/x}$. When looking at the first few derivatives I noticed a pattern and eventually found the following formula $$\frac{\mathrm d^n}{\mathrm dx^n}f(x)=(-1)^n e^{1/x} \cdot \sum _{k=0}^{n-1} k!…
Listing
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How to reverse the $n$ choose $k$ formula?

If I want to find how many possible ways there are to choose k out of n elements I know you can use the simple formula below: $$ \binom{n}{k} = \frac{n! }{ k!(n-k)! } .$$ What if I want to go the other way around though? That is, I know I want to…
Daniel Scocco
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Solutions to $\binom{n}{5} = 2 \binom{m}{5}$

In Finite Mathematics by Lial et al. (10th ed.), problem 8.3.34 says: On National Public Radio, the Weekend Edition program posed the following probability problem: Given a certain number of balls, of which some are blue, pick 5 at random. The…
Nick Matteo
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How do I count the subsets of a set whose number of elements is divisible by 3? 4?

Let $S$ be a set of size $n$. There is an easy way to count the number of subsets with an even number of elements. Algebraically, it comes from the fact that $\displaystyle \sum_{k=0}^{n} {n \choose k} = (1 + 1)^n$ while $\displaystyle…
Qiaochu Yuan
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You have to estimate $\binom{63}{19}$ in $2$ minutes to save your life.

This is from the lecture notes in this course of discrete mathematics I am following. The professor is writing about how fast binomial coefficients grow. So, suppose you had 2 minutes to save your life and had to estimate, up to a factor of $100$,…
capablanca79
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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 $$\mathrm E(X) = \sum^n_{k=0}k\mathrm P(X=k)=\sum^n_{k=0}k\binom nkp^k(1-p)^{n-k}$$ but the expected value of a Binomal…
user31280
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How often can a number occur in Pascals Triangle?

Each number occurs at least twice as $a={a\choose1}={a\choose a-1}$. If a number occurs somewhere else in the triangle (most likely twice, if it's not of the form ${2a\choose a}$) then that number occurs $4$ times. After that, it becomes…
vrugtehagel
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Evaluating the series $\sum\limits_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}} $

Wolfram MathWorld states that $$ \sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}} = \frac{ \pi \sqrt{3}}{18} \Bigg[ \psi_{1} \left(\frac{1}{3} \right) - \psi_{1} \left(\frac{2}{3} \right) \Bigg]- \frac{4}{3} \zeta(3) \, , $$ where $\psi_{1}(x)$ is…
Random Variable
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Beautiful identity: $\sum_{k=m}^n (-1)^{k-m} \binom{k}{m} \binom{n}{k} = \delta_{mn}$

Let $m,n\ge 0$ be two integers. Prove that $$\sum_{k=m}^n (-1)^{k-m} \binom{k}{m} \binom{n}{k} = \delta_{mn}$$ where $\delta_{mn}$ stands for the Kronecker's delta (defined by $\delta_{mn} = \begin{cases} 1, & \text{if } m=n; \\ 0, & \text{if }…
fosco
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