Tagged Questions

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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A binomial inequality with factorial fractions

Prove that $$\left(1+\frac{1}{n}\right)^n<\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!}$$ for $n>1 , n \in \mathbb{N}$.
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Limit of alternating sum with binomial coefficient

I need to find a limit, or approximation for $\sum\limits_{k=1}^{n} (-1)^k {n \choose k} \log(a+bk)$ for, say, an $a,b\in (0,10)$. It is not so important what values $a$ and $b$ have. It would be ...
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when is $\frac{1}{n}\binom{n}{r}$ an integer

So I am considering for which values of n is $a_n =\frac{1}{n}\binom{n}{r}$ an integer for all $1\leq r \leq n-1$. The first thing I did was to check the Pascal Triangle. So I guess n has to be ...
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Number of Elements in a Conjugacy Class of $S_N$ (Derivation)

Consider the conjugacy classes of the symmetric group $S_N$. Each conjugacy class consists of permutations that have the same cycle structure. We see that the number of possible cycle structures is ...
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Formula for binomial coefficients

Does someone know, if the subsequent formula holds for $m \ge n \ge i \ge 1$ and if yes, can give a reference. $$\sum_{k=i}^{m-n+i}\binom{k}{i}\binom{m-k}{n-i} = \binom{m+1}{n+1}$$ Thank you very ...
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Binomial Sum: Values

I need this as lemma. Regard the sums: $$S_k:=\sum_{n=0}^N\binom{N}{n}(-1)^{N-n}n^k\quad(k\in\mathbb{N}_0)$$ Then it holds: $$S_k\stackrel{k<N}{=}0\quad S_k\stackrel{k=N}{=}N!$$ How can I check ...
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Simplify triangular sum of triangular numbers: $\sum_{i=1}^{n}(\frac12i(i+1))$

I'd like to simplify this expression, which sums up the first $n$ triangular numbers: $$\sum_{i=1}^{n}(\frac12i(i+1))$$ which is equal to: $$\sum_{i=0}^{n}((n-i)(i+1))$$ Is it even possible without ...
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Prove equality between binomial coefficients.

Using the Binomial theorem, prove that: $$\binom{m+n}{k}=\sum_{j=0}^k \binom{n}{j}\binom{m}{k-j},\; 0\leq k\leq m+n$$
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Prove inequality with binomial coefficient: $6 + \frac{4^n}{2 \sqrt{n}} \le \binom{2n}n$

I have to prove inequality, where $n \in N$ $$6 + \frac{4^n}{2 \sqrt{n}} \le {2n \choose n}$$ I have checked and it is true when $n\ge4$, however I have no idea how I should start. Can anyone give a ...
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$\sum_{i=0}^m \binom{m-i}{j}\binom{n+i}{k} =\binom{m + n + 1}{j+k+1}$ Combinatorial proof

Is there a simple combinatorial proof for the following identity? $$\sum_{0\leq i \leq m} \binom{m-i}{j}\binom{n+i}{k} =\binom{m + n + 1}{j+k+1}$$ where $m,j \geq 0$, $k \geq n \geq 0$.
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Kolmogorov-Zurbenko filter - Calculation of coefficients

I'm currently researching the Kolmogorov-Zurbenko filter and trying to implement it myself as a way to smooth one-dimensional signal strength values. The basic filter per se is pretty easy to ...
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Number of ways to distribute indistinguishable balls into distinguishable boxes of given size

I need to find a formula for the total number of ways to distribute $N$ indistinguishable balls into $k$ distinguishable boxes of size $S\leq N$ (the cases with empty boxes are allowed). So I mean ...
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What is the best way of proving inequalities including binomial coefficients in general? Lets look at this example: $$\frac{n^k}{k^k}\leq\binom{n}{k}\leq\frac{n^k}{k!}$$ What I started with is: $$\... 2answers 114 views Proof by induction \sum_{k=1}^{n} k \binom{n}{k} = n\cdot 2^{n-1} for each natural number n [duplicate] Prove by induction that \sum_{k=1}^{n} k \binom{n}{k} = n\cdot 2^{n-1} for each natural number n. 1answer 104 views Binomial Expression [duplicate] Please give me feedback on my answer to this question. Question: For all  n\geq1:\binom{2n}{0}+\binom{2n}{2}+\binom{2n}{4}+\cdots+\binom{2n}{2k}+\cdots+\binom{2n}{2n}  is equal to  \binom{2n}{1}+\... 3answers 94 views Binomial coefficient problem I still haven't quite realized how to solve binomial coefficient problems like this, can someone show me an elaborated way of solving this? I need to write this expression in a more simplified way: \... 1answer 191 views Have you seen this formula for factorial? Let p always be a prime. n! = \prod_{p\leq n}p^{\lfloor \frac{n}{p}\rfloor}. Then \binom{n}{r} = \prod_{p\leq r}p^{\lfloor n/p \rfloor -\lfloor (n-r)/p \rfloor - \lfloor r/p \rfloor} \times ...... 1answer 66 views Equating coefficients of binomial expansion modulo p In this answer: http://math.stackexchange.com/a/652909 Ted equates mod p the coefficients of$$\sum_{n=0}^{pa} \binom{pa}{n} x^n$$and$$\sum_{i=0}^{a} \binom{a}{i} x^{pi}$$to get that$$\binom{pa}...
I'm asked to use the fact that $\begin{pmatrix}n\\r\end{pmatrix}+\begin{pmatrix}n\\r-1\end{pmatrix}=\begin{pmatrix}n+1\\r\end{pmatrix}$ to show, by induction, that (a+b)^n=\begin{pmatrix}n\\0\end{...
Let both $a$,$b$ and $\theta$ be real numbers not equal to a negative integer. Let $m$ be a positive integer. I have shown that the following equality holds: \begin{eqnarray} &&S^{a}_{b,\...