# combinatorial proof of identity involving multisets

I was wondering how to do a combinatorial proof of the following identity: $${n+k-1\choose k-1} = \sum_{j=0}^{\lfloor n/2\rfloor} {k\choose n-2j}{j+k-1\choose k-1}$$?

The LHS is just the number of multisets of $$k$$ types and $$n$$ elements while the RHS seems to represent some sort of Cartesian product. I initially thought it was the number of ways to choose $$n-2j$$ types from $$k$$ types times the number of ways to make a multiset of $$j$$ elements of $$k$$ types, but I can't seem to derive a proper bijection from this.

I have $$k$$ boxes and $$n$$ balls. For some $$j$$ with $$0\le j\le\lfloor n/2\rfloor$$ I distribute $$j$$ balls amongst the boxes, and then I double the number of balls in each box; I can do this in $$\binom{j+k-1}{k-1}$$ ways. At that point I have $$n-2j$$ balls left, and I pick $$n-2j$$ boxes and place one of the remaining balls in each of these boxes; I can do this in $$\binom{k}{n-2j}$$ ways. I claim that each of the $$\binom{n+k-1}{k-1}$$ distributions of the $$n$$ balls amongst the $$k$$ boxes can be obtained in exactly one way through this procedure.
Specifically, take any distribution of the $$n$$ balls to the $$k$$ boxes. Let $$A$$ be the set of boxes containing an odd number of balls. If we remove one ball from each of these boxes, $$n-|A|$$ balls remain the $$k$$ boxes altogether, and the number remaining is even, say $$2k$$ for some $$k$$, where clearly $$0\le k\le\lfloor n/2\rfloor$$. Thus, $$|A|=n-2k$$, and this particular distribution is counted once in the term $$j=k$$. In general, the term $$\binom{k}{n-2j}\binom{j+k-1}{k-1}$$ counts the distributions that have $$n-2j$$ boxes containing an odd number of balls.