سوال ۱۱۷ کنکور ارشد علوم کامپیوتر ۱۴۰۰

Let $F$, $G$, $E$ be the sets of students interested in French, German, and English. We are given the single-set, pairwise-intersection, and union sizes, and we want the triple intersection $|F\cap G\cap E|$.

Given data:

Inclusion–exclusion for three sets:

$|F\cup G\cup E| = |F| + |G| + |E| - |F\cap G| - |F\cap E| - |G\cap E| + |F\cap G\cap E|.$

Let $x = |F\cap G\cap E|$ and substitute:

$122 = 78 + 32 + 57 - 13 - 21 - 16 + x.$

Sum the singles: $78+32+57 = 167$. Sum the pairs: $13+21+16 = 50$. So

$122 = 167 - 50 + x = 117 + x \;\Longrightarrow\; x = 122 - 117 = 5.$

Thus $\boxed{5}$ students like all three languages, which is option 4.

Note the total of $136$ students is a distractor: $136 - 122 = 14$ students like none of the three languages, but that count is irrelevant to finding the triple intersection.

Pitfall: Apply inclusion–exclusion to the union ($122$), not the total student count ($136$). The total only matters if the problem asked how many like none of the languages. Also, the pairwise figures here are the full pairwise intersections (e.g. $|F\cap E|$), so they enter the formula directly with a minus sign — no need to subtract the triple part out of them first.