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I would like to prove the following inequality:

$$ {m+n \choose m} \ge \frac{(n+1)^m}{m!} $$

Any hints?

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3 Answers 3

up vote 2 down vote accepted

$$ \binom{m+n}{m} = \frac{(m+n)!}{m!n!} = \frac{(n+1)(n+2)\dots(n+m)}{m!} $$

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Hint: $$ {m+n \choose m}=\frac{(m+n)(m+n-1)\cdots(n+1)}{m!} $$

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Hint: Put $n + m$ balls in $n + 1$ boxes, one in each except $m$ balls in the last one.

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