How prove this$\sum_{i=0}^{m-1}\binom{n-1+i}{i}x^ny^i+\sum_{j=0}^{n-1}\binom{m-1+j}{j}x^my^j=1$ let $m,n$ be positive numbers,and $x,y>0$ such $x+y=1$,show that
$$\sum_{i=0}^{m-1}\binom{n-1+i}{i}x^ny^i+\sum_{j=0}^{n-1}\binom{m-1+j}{j}x^jy^m=1$$
My try:
$$\sum_{i=0}^{m-1}\binom{n-1+i}{i}x^ny^i=\sum_{i=0}^{m-1}\binom{n-1+i}{i}x^n(1-x)^i=\sum_{i=0}^{m-1}\binom{n-1+i}{i}x^n\sum_{k=0}^{i}(-1)^k\binom{i}{k}x^k$$

The exponents are reversed in the second summation: it should be
$$\sum_{j=0}^{n-1}\binom{n-1+j}jx^jy^m\;.$$
(You can see that the version in the problem is wrong by actual calculation with $m=n=2$ and $x=\frac14$, for instance.)
Think of $x$ as the probability of a success and $y$ as the probability of failure in a Bernoulli trial. The experiment consists of $m+n-1$ independent Bernoulli trials each with success probability $x$. Then
$$\binom{n-1+i}ix^ny^i$$
is the probability of getting the $n$-th success on the $(n+i)$-th trial (why?), so 
$$\sum_{i=0}^{m-1}\binom{n-1+i}ix^ny^i$$
is the probability of getting at least $n$ successes. Similarly,
$$\sum_{j=0}^{n-1}\binom{n-1+j}jx^jy^m$$
is the probability of getting at least $m$ failures. To complete the proof, show that exactly one of these two events must occur.
