Take the tour ×  Let $(n)_k = n(n-1)(n-2)...(n-k+1)$  (Evidently
this is a falling factorial with $0 \le k \le n$).  Need to give
a combinatorial proof of the following  $$(n)_k = \sum_{i=1}^{k}
\binom ki (n-m)_i (m)_{k-i}$$  Please help.  share|improve this
question What is $m$?${}{}$ –  Brian M. Scott Feb 18 at 4:03
Have no idea. Maybe there is a mistake in the question, because
it might ask to prove the combinatorial interpretation of the
falling factorial which is $(x+y)_k = \sum_{i=0}^{k}
\binom{k}{i} x_{i} y_{k-i}$. Only in our question $i=1$, not
$0$... –  John Lennon Feb 18 at 4:22 It turns out not to matter,
at least as long as $0\le m\le n$. And I’m pretty sure that that
lower limit of $1$ is wrong: see the example in my answer. – 
Brian M. Scott Feb 18 at 4:43 add comment  1 Answer  up vote 3
down vote accepted  I prefer the notation $n^{\underline k}$ for
the falling factorial; in that notation the desired identity
becomes  $$n^{\underline
k}=\sum_{i=1}^k\binom{k}i(n-m)^{\underline i}m^{\underline