An organization awarded 48 medals in event 'A', 25 in event 'B' and 18 in event 'C'. If these medals went to a total of 60 men and only five men got medals in all three events, then how many received medals in exactly two of the three events?
Solutions · 1
Official solution
Let t=5 be the triple intersection. By inclusion--exclusion:
\[60 = 48+25+18 - (\text{sum of pairwise intersections}) + t\]
\[\text{sum of pairwise intersections} = 48+25+18+5-60 = 36\]
Number who received exactly two medals = (sum of pairwise intersections) −3×(triple intersection):
36−3(5)=36−15=21