Q.A college awarded 38 medals in volley ball, 15 in football and 20 in basket ball. The medals awarded to a total of 58 players and only 3 players got medals in all three sports. How many received medals in exactly two of the three sports?

Let V, F, B be volleyball, football, basketball medal-winners, with $n(V)=38$, $n(F)=15$, $n(B)=20$, total medal-winning players $n(V\cup F\cup B)=58$, and triple overlap $n(V\cap F\cap B)=3$. By inclusion-exclusion, $n(V\cup F\cup B) = n(V)+n(F)+n(B) - [n(V\cap F)+n(F\cap B)+n(V\cap B)] + n(V\cap F\cap B)$. Substituting: $58 = 38+15+20 - S + 3$ where $S$ is the sum of the three pairwise overlaps, so $58 = 76 - S$, giving $S = 18$... let's recompute carefully: $38+15+20=73$, so $58 = 73 - S + 3 = 76 - S$, hence $S = 76-58 = 18$? Recomputing precisely: $73 - S + 3 = 58 \Rightarrow 76 - S = 58 \Rightarrow S = 18$. So the sum of the three pairwise intersection counts is 18. Now, each pairwise intersection count includes the triple-overlap players; the number who got 'exactl …

Use n(V)+n(F)+n(B) - S + (triple overlap) = total, to solve for S (the sum of the three pairwise-overlap counts); then subtract 3×(triple overlap) from S, since each triple-medal player is counted once in ea …

A frequent mistake is to report S itself as the 'exactly two' answer — S is the sum of pairwise overlaps, which still includes triple-medal players three times over …