An organization awarded 48 medals in event 'A', 25 in event 'B' and 18 in event 'C'. If these medals went to total 60 men and only five men got medals in all the three events, then, how many received medals in exactly two of three events?
Text Solution
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Let A, B and C denote the set of men who received medals in even A, event B and event C respectively.
Then, n(A) = 48, n(B) = 25, n(C) = 18, n(A 
B 
C) = 60 and n(A 
B 
C) = 5,
Now we know that, n(A 
B 
C) = n(A) + n(B) + n(C) - n(A 
B) - n(A 
C) - n(B 
C) + n(A 
B 
C)
60 = 48 + 25 + 18 - n(A 
B) - n(A 
C) - n(B 
C) + 5
n(A 
B) + n(A 
C) + n(A 
C) = 48 + 25 + 18 + 5 - 60 = 36
Therefore, the number of people who received medals in exactly two of the three sports will be 36 -3(A 
B 
C) = 36-15 = 21.
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