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Let $\alpha = (\alpha_{1},\dots, \alpha_{n}) \text{ and } \beta = (\beta_{1},\dots, \beta_{n}) \in \mathbb{N}^{n}$ be two vectors of positive integers, and assume that $\beta_{i} \le \alpha_{i}$ for all $1\le i \le n$. As usual, let $\alpha ! = (\alpha_{1} !) \cdots (\alpha_{n}!)$ and $$\binom{\alpha}{\beta} = \frac{\alpha!}{\beta! (\alpha-\beta)!}.$$ We set $\vert \alpha \vert = \alpha_{1} + \dots + \alpha_{n} $. My question: Is it true that $$ \binom{\alpha}{\beta} \le \binom{\vert \alpha \vert }{\vert \beta \vert} $$

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The inequality is true, and in fact it's rather simple to proof it. Observe that $\binom{\alpha}{\beta} = \prod_{i=1}^{n} \binom{\alpha_{i}}{\beta_{i}}$. The binomial coefficient $\binom{\alpha_{i}}{\beta_{i}}$ is the coefficient of $x^{\beta_{i}}$ in $(1+ x)^{\alpha_{i}}$, while $\binom{\vert \alpha \vert }{\vert\beta \vert}$ is the coefficient of $x^{\vert\beta\vert}$ in $(1+x)^{\vert \alpha \vert}$. The coefficient $\binom{\vert \alpha \vert }{\vert \beta \vert}$ of $x^{\vert \beta \vert }$ in $\prod_{i=1}^{n} (1+x)^{\alpha_{i}}$ is the sum of different products of monomial coefficients which contains $\prod_{i=1}^{n} \binom{\alpha_{i}}{\beta_{i}}$ as summand, and hence the desired inequality follows.

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  • $\begingroup$ Is this Q/A thing done just to get a badge? $\endgroup$
    – Nikunj
    Apr 1, 2016 at 13:54
  • $\begingroup$ No, I just did not see the trivial answer yesterday. $\endgroup$
    – Sebastian
    Apr 1, 2016 at 14:04

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