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I would like to prove the following inequality:

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

Any hints?

2 Answers 2

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

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