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If $p$ is a prime and $m,n\in\mathbb{N}$, prove that $\binom{pm}{pn}\equiv\binom{m}{n}$ mod $p$.
Hint: Compare the binomial expansions of $(1+x)^{pm}$ and of $(1+x^m)^p$ in $\mathbb{F}_p[x]$.
If $R$ is a commutative ring, then the set of all polynomials with coefficients in $R$ is denoted by $R[x]...
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Why do two ways of expanding the same formal polynomial lead to matching coefficients?
There are a few proofs in which the technique is to expand the product of some formal polynomials in $\mathbb{R}[x_1,x_2,\ldots,x_k]$ in more than one distinct way and then we can match up the ...
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Vandermonde identity in a ring
Let $R$ be a commutative $\mathbb{Q}$-algebra. For $r \in R$ and $n \in \mathbb{N}$ we can define the binomial coefficient $\binom{r}{n}$ as usual by $\binom{r}{0}=1$ and $\binom{r}{n+1}=\frac{r-n}{n+...
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