# 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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### Proving $\sum\limits_{k=0}^{n} (-1)^{k} \binom{n}{k} = 0$

Show that $$\sum_{k=0}^{n} (-1)^{k} \binom{n}{k} = 0$$ So for odd $n$ we have an even number of terms. So $\binom{n}{k} = \binom{n}{n-k}$ which have opposite signs. Thus the sum is $0$. For even $n$ ...
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### Sum with binomial coefficients: $\sum_{k=1}^m \frac{1}{k}{m \choose k}$

I got this sum, in some work related to another question: $$S_m=\sum_{k=1}^m \frac{1}{k}{m \choose k}$$ Are there any known results about this (bounds, asymptotics)?
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### Prove that $\dfrac{(n^2)!}{(n!)^n}$ is an integer for every $n \in \mathbb{N}$

Prove that $\dfrac{(n^2)!}{(n!)^n}$ is an integer for every $n \in \mathbb{N}$ I know that there are tools in Number theory to proves the required but I want to use the tool that says that if you ...
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### Proving that $\sum_{i=0}^{n}\binom{n}{i}i^{n-i}(n-i)^{i}\le\frac{1}{2}n^n$

How can we prove that $$\displaystyle\sum_{i=0}^{n}\binom{n}{i}i^{n-i}(n-i)^{i}\le\dfrac{1}{2}n^n$$ where $\displaystyle\binom{n}{i}=\dfrac{n!}{i!(n-i)!}$. This inequality is very interesting. I ...
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### Combinatorial interpretation of Binomial Inversion

It is known that if $f_n = \sum\limits_{i=0}^{n} g_i \binom{n}{i}$ for all $0 \le n \le m$, then $g_n = \sum_{i=0}^{n} (-1)^{i+n} f_i \binom{n}{i}$ for $0 \le n \le m$. This sort of inversion is ...
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### On computing: $\gcd \left({2n \choose 1}, {2n \choose 3},\cdots, {2n \choose 2n-1}\right)$

I would like to calculate $$d=\gcd \left({2n \choose 1}, {2n \choose 3},\cdots, {2n \choose 2n-1}\right)$$ We have: $$\sum_{k=0}^{n-1}{2n \choose 2k+1}=2^{2n-1}$$ $$d=2^k, 0\leq k\leq2n-1$$ ...
### Proving $\sum_{k=0}^{2m}(-1)^k{\binom{2m}{k}}^3=(-1)^m\binom{2m}{m}\binom{3m}{m}$ (Dixon's identity)
I found the following formula in a book without any proof: $$\sum_{k=0}^{2m}(-1)^k{\binom{2m}{k}}^3=(-1)^m\binom{2m}{m}\binom{3m}{m}.$$ I don't know how to prove this at all. Could you show me how ...