سوال ۱۱۶ کنکور ارشد علوم کامپیوتر ۱۴۰۰

This is a direct application of Vandermonde's identity:

$\sum_{k=0}^{n}\binom{m}{k}\binom{p}{n-k}=\binom{m+p}{n}.$

Why it holds (combinatorial proof). To choose $n$ objects from a set of $m+p$ items, split the items into a group of $m$ and a group of $p$. If exactly $k$ are taken from the first group, then $n-k$ must come from the second, giving $\binom{m}{k}\binom{p}{n-k}$ ways. Summing over all valid $k$ counts every selection exactly once, so the total is $\binom{m+p}{n}$.

Apply it here. With $m=6$, $p=7$, and $n=5$:

$\sum_{k=0}^{5}\binom{6}{k}\binom{7}{5-k}=\binom{6+7}{5}=\binom{13}{5}.$

Evaluate $\binom{13}{5}$:

$\binom{13}{5}=\frac{13\cdot 12\cdot 11\cdot 10\cdot 9}{5!}=\frac{154440}{120}=1287.$

So the value of the expression is $1287$, matching option 2.

Exam trick: Whenever you see $\sum_k\binom{a}{k}\binom{b}{c-k}$ with the lower indices summing to a constant $c$, it collapses by Vandermonde to a single coefficient $\binom{a+b}{c}$ — no need to expand the sum term by term.