# Sum involving integer compositions and binomial coefficients

I came across an interesting identity involving binomial coefficients. I'm not sure if I'm looking at the identity the wrong way but I am not aware if this identity is known and if there is an (easy) proof for it.

Take a nonnegative integer $n$ and form two $k$-tuples consisting of integers at most $n$, say, $(a_1,a_2,\ldots,a_k)$ and $(b_1,b_2,\ldots,b_k)$ such that $a_i\geq a_{i+1}$ and $b_i\geq b_{i+1}$. Let $a_0=b_0=n$ and $a_{k+1}=b_{k+1}=0$. Let $j\in\mathbb N$. The sum goes as follows:

$$\sum_{x_1+x_2+\cdots+x_{k+1}=j} ~~\sum_{m=1}^{k+1} \binom{a_{m-1}-a_m+x_m}{a_{m-1}-a_m} = \sum_{x_1+x_2+\cdots+x_{k+1}=j} ~~\sum_{m=1}^{k+1} \binom{b_{m-1}-b_m+x_m}{b_{m-1}-b_m}.$$

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Is it okay to post this on MO? –  Kenjo Jul 17 '12 at 3:22
I think it is ok, but it would be nice to give a link to your MO post also here. –  Martin Sleziak Sep 2 '12 at 19:16