Combinatorics. Falling factorial problem 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.
 A: I prefer the notation $n^{\underline k}$ for the falling factorial; in that notation the desired identity becomes
$$n^{\underline k}=\sum_{i=1}^k\binom{k}i(n-m)^{\underline i}m^{\underline{k-i}}\;.\tag{1}$$
The lefthand side is clearly the number of ways to choose a sequence of $k$ distinct elements of $[n]=\{1,\dots,n\}$. If $a=\langle a_1,\dots,a_k\rangle$ is such a sequence, let $I(a)=\{j\in[k]:a_j\le n-m\}$, and let $i(a)=|I(a)|$. 
Clearly $a$ is completely determined by the set $I(a)$ and the subsequences $a_L=\langle a_j:j\in I(a)\rangle$ and $a_H=\langle a_j:j\in[k]\setminus I(a)\rangle$. There are $(n-m)^{\underline{i(a)}}$ possible choices for $a_L$, $m^{\underline{k-i(a)}}$ possible choices for $a_H$, and $\binom{k}{i(a)}$ possible choices for $I(a)$. Now sum over the possible values of $i(a)$ to get the righthand side.
However, it appears that the lower limit of the summation should be $0$, not $1$, both from the argument given above and from the case $n=m=k=1$: the lefthand side of $(1)$ is $1$, but the righthand side as written is $\binom110^{\underline 1}1^{\underline 0}=0$. The correct identity is then
$$n^{\underline k}=\sum_{i=0}^k\binom{k}i(n-m)^{\underline i}m^{\underline{k-i}}\;.$$
