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

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Solution

$n (F) = 38 n (B) = 15 n (C) = 20 n (F \cup B \cup C) = 58$
$n (F \cap B \cap C) = 3$
$∵ n (F \cup B \cup C) = n (F) + n (B) + n (C) - n (F \cap B) - n (B \cap C) - n (C \cap F) + n (F \cap B \cap C)$
$n (F \cap B) + n (B \cap C) + n (C \cap F) = n (F) + n (B) + F (C) + n (F \cap B \cap C) - (F \cup B \cup C)$
$= 38 + 15 + 20 + 3 - 58 = 18$

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

n(F)=38n(B)=15n(C)=20n(F∪B∪C)=58n(F∩B∩C)=3∵n(F∪B∪C)=n(F)+n(B)+n(C)−n(F∩B)−n(B∩C)−n(C∩F)+n(F∩B∩C)n(F∩B)+n(B∩C)+n(C∩F)=n(F)+n(B)+F(C)+n(F∩B∩C)−(F∪B∪C)=38+15+20+3−58=18