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Question

Difficulty: MediumSets and Set Operations

In a sports academy of 120 athletes, 70 play football, 60 play basketball, and 50 play tennis. If 10 athletes play none of these three sports and 15 athletes play all three sports, how many athletes play exactly two of these sports?

Answer: 40 athletes

Answer

40 athletes play exactly two of the sports.
The correct answer is 40. Subtracting the 10 athletes who play no sports from the total of 120 leaves 110 athletes playing at least one sport. Using inclusion-exclusion, the sum of pairwise intersections is S2=70+60+50+15−110=85S_2 = 70 + 60 + 50 + 15 - 110 = 85. Since S2S_2 contains the region of all three sports counted three times, subtracting 3×15=453 \times 15 = 45 gives 40 athletes who play exactly two sports.

Step-by-Step Solution

1
Determine the cardinality of the union of all three sets
n(F∪B∪T)=120−10=110n(F \cup B \cup T) = 120 - 10 = 110
Athletes who play none of the three sports are excluded from the total universal set.
2
Apply the Principle of Inclusion-Exclusion for three sets to find the sum of 2-set intersections
S2=n(F)+n(B)+n(T)+n(F∩B∩T)−n(F∪B∪T)=70+60+50+15−110=85S_2 = n(F) + n(B) + n(T) + n(F \cap B \cap T) - n(F \cup B \cup T) = 70 + 60 + 50 + 15 - 110 = 85
The formula relates the total union, individual set cardinalities, pairwise intersections, and triple intersection.
3
Subtract three times the triple intersection from S2S_2 to isolate regions corresponding to exactly two sports
Exactly two sports = S2−3×n(F∩B∩T)=85−3(15)=40S_2 - 3 \times n(F \cap B \cap T) = 85 - 3(15) = 40
Each pairwise intersection sum S2S_2 includes the triple intersection region three times.

Key Concept

Three-set inclusion-exclusion principle and region cardinality decomposition
Estimated Time:1m 30s
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Topics

Sets and Set Operations

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