# Questions tagged [binomial-coefficients]

For questions involving the 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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### number theory, binomial coefficients divisibility

Let $p$ be a prime number greater than $3$ and $n$ a positive integer. Suppose $\nu _p(n) = r$. Prove that $\dbinom{np}{p} - n$ is divisible by $p^{3+r}$ The problem is here: https://poti.impa.br/...
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### The upper bound of $\sum_{k=0}^{m} \binom{n}{k}x^k(1-x)^{n-k}$

As the title mentioned, I want to calculte $$\sum_{k=0}^{m} \binom{n}{k}x^k(1-x)^{n-k}.$$ Note that if $m=n$, the result is simply 1. However, when $m<n$, this seems to ...
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### $\sum_{i=1}^n(-1)^{n-i}(i-x)^{n-1}{n-1\choose i-1}$ doesn't seem to depend on $x$ [closed]

Experimenting with a computer, the sum $$\sum_{i=1}^n(-1)^{n-i}(i-x)^{n-1}{n-1\choose i-1}$$ doesn't seem to depend on $x$. Is that right? Why?
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### How to calculate an upper bound for $\sum_{k=0}^n \binom{n}{k} \frac{(-x)^k}{\sqrt{k+a}}$

As the title mentioned, I want to get a closed-form result of $$\sum_{k=0}^n \binom{n}{k} \frac{(-x)^k}{\sqrt{k+a}},$$ where $x\in[0,1]$ is a real number, and $a$ is a ...
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### Combinatoric equation involving three unknowns

Find $n\in\mathbb{N^*}$ and $p, q \in\mathbb{Q}$, such that: $$(pn+q)\binom{nk+1}{k}+(qn+p)\binom{nk+2}{k-1}+(pn+q)\binom{nk+1}{k-1}=\binom{nk+2}{k+1}$$ for $\forall k\in\mathbb{N^*}$ Here's my ...
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### Prove $\lim_{n\to\infty}\binom{n}{k}\left(\frac{\mu}{n}\right)^{k}\left(1-\frac{\mu}{n}\right)^{n-k}=\frac{\mu^k}{e^\mu\cdot{k!}}$

The problem is to prove the following equality: $$\lim_{n\to\infty}\binom{n}{k}\left(\frac{\mu}{n}\right)^{k}\left(1-\frac{\mu}{n}\right)^{n-k}=\frac{\mu^k}{e^\mu\cdot{k!}}$$ This is what I have ...
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### Calculating $\sum\limits_{k=0}^n\binom{n}{k}/\left(2^k+2^{n-k}\right)$

I am trying to find a closed form for $\sum\limits_{k=0}^n\frac{\binom{n}{k}}{2^k+2^{n-k}}$. I saw on quora that integration can be used to rewrite portions of such equations, and so I attempted this. ...
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### A Combinatoric Identity Involving Binomial Coefficients: $\sum^{n}_{r=1} {n \choose r} (-1)^r \frac{2^r-1}{2r}= \sum^{n}_{r=1}\frac{(-1)^r}{2r}$
Recently I came across this combinatoric identity: $$$$\sum^{n}_{r=1} {n \choose r} (-1)^r \frac{2^r-1}{2r}$$ = \sum^{n}_{r=1}\frac{(-1)^r}{2r}$$ I have verified that ...