### 1975 IMO SL #7

Prove that from $x + y = 1 \ (x, y \in \mathbb R)$ it follows that

$x^{m+1} \sum_{j=0}^n \binom{m+j}{j} y^j + y^{n+1} \sum_{i=0}^m \binom{n+i}{i} x^i = 1 \qquad (m, n = 0, 1, 2, \ldots ).$

We prove this in a combinatorial manner. Consider an unfair coin that comes up heads with probability $x$ and tails with probability $y$. Note that $x^{m+1}\binom{m+j}{j}y^j$ is the probability that until the moment the $(m+1)$th head has appeared, exactly $j$ tails have appeared. Similarly, $y^{n+1}\binom{n+i}{i}x^i$ is the probability that exactly $i$ tails have appeared before the $(n+1)$-th tails. The equality we have. Hence, the above sum is the probability that either $m + 1$ heads will appear before $n + 1$ tails, or vice versa, and this probability is clearly 1. We are done. $\square$

1. Anonymous1/22/2022

wonderful solution brooooo

extremely elegant
standard sol is to do induction

you interpreted it combinatorially
orz, sol is too op