0
$\begingroup$

$$\sum_{ k\ge 1 } \sum_{ (s_1,...,s_k) } \binom h{ s_1 ,..., s_k }=\binom{m-1}{h-1}$$ where the second summation is taken over all choices of the numbers

$${ s }_{ 1 },...,{ s }_{ k-1 }\ge 0,\quad {s }_{k }\ge 1$$

which satisfy the relations $$\sum _{ i=1 }^{ k }{ { s }_{ i } } =h,\quad \sum _{ i=1 }^{ k }{ i{ s }_{ i } } = m$$

$\endgroup$
6
$\begingroup$

Take a stick of length $m$ and partition it onto $h$ small sticks of positive integer lengths. The number of ways to do it equals $\binom{m-1}{h-1}$, as we choose $h-1$ cut points out of $m-1$. On the other hand, if $s_i$ denotes the number of small sticks of length $i$, and $k$ is the length if the longest small stick, we get the corresponding summand in LHS.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.