Show me what you've got.
edit: Yay, I proved it to myself. I think my reasoning still made just as much sense with three characters or more because it'll be just like normal combination if any characters that match are treated as different ones. Since we can't increase the size of k, we increase the size of n. I learned the hard way many times that words don't mean much when it comes to proofs in math, but this intuition convinced me.
I worked out an actual proof that n multichoose k equals n+k-1 choose k after the tourney on Sunday. Here's a sketch:
Let X be the set of k-tuples of integers from 1 to n such that (n_1, ..., n_k) is in X iff n_1<=n_2<=...<=n_k.
Let Y be the set of k-tuples of integers from 1 to n+k-1 such that (n_1, ..., n_k) is in Y iff n_1<n_2<...<n_k.
It's straightforward to show that the size of X is n multichoose k and the size of Y is n+k-1 choose k.
Let f: X -> Y be a function that maps (n_1, ..., n_k) to ((n_1)+0, (n_2)+1, (n_3)+2, ..., (n_k)+(k-1)). It's similarly straightforward to show that f is well-defined and that f is a bijection, which yields the desired result.
