Would there be any simplification of the following double sum?
$$\sum^{n-1}_{k=1}\sum^n_{j=k}{n-1\choose k-1}{n\choose j}a^{k-1}(1-a)^{n-k}b^j(1-b)^{n-j}$$.
or, equivalently,
$$(1-a)^{n-1}(1-b)^n\sum^{n-1}_{k=1}\big(\frac{a}{1-a}\big)^{k-1}\sum^n_{j=k}{n-1\choose k-1}{n\choose j}\big(\frac{b}{1-b}\big)^j$$.
I'd like to simplify the product of two combinations using Vendermonde's identities like
$$ \sum_{j=0}^{i-1} \binom{i}{j+1}\binom{N-i}{j} = \binom{N}{i-1} $$
but I can't figure out how to deal with my case..