A college awarded 38 medals in Football, 15 in Basketball and 20 in Cricket. If these medals went to a total of 58 men and only three men got medals in all the three sports. The number of students who received medals in exactly two of the three sports is

9

Let F, B and C denote the sets of students who received medals in Football, Basketball and Cricket respectively. Then, we have

$n (F) = 37, n (B) = 15, n (C) = 20,$

$n(F∪B∪C)=58$ and $n (F∩B∩C) = 3$

n(exactly two sports)

$n(F∪B∪C)=n(F)+n(B)+n(C)-n(F∩B)-n(B∩C)-n(C∩F)+n (F∩B∩C)$

$⇒58=38+15+20-\{n (F∩B) +n (B∩C)+n (C∩F)\} +3$

$⇒n(F∩B) +n (B∩C) +n (C∩F)=18$

Hence, the number of students who received medals in exactly two of the three sports

$=n(F∩B) +n (B∩C)+n (C∩F) - 3n (F∩B∩C)$

$=18-3× 3=9$