
Nestor R. answered 09/12/19
Mathematically-oriented professional with many years of experience
Let B, S and V stand in for Basketball, Soccer and Volleyball.
n(B) = 11, n(S) = 10 and n(V) = 9.
But it is given that 1 student played both B and S only, 2 played both S and V only, 1 played both B and V only, and 2 played B, S and V, i.e. n(B,S) = 1, n(S,V) = 2, n(B,V) = 1 and n(B,S,V) = 2.
If you add n(B), n(S) and n(V) together the sum is 30. Had each student played only one sport this would answer the question. But the student playing both B and S is counted once for each sport and so is double counted. The same argument applies to the 2 students playing S and V, the one playing B and V and the 2 playing all three sports (these students are triple counted).
Using this information, it can be found that 11 - 1 - 1 - 2 = 7 played B ONLY, 10 - 1 - 2 - 2 = 5 played S ONLY and 9 - 2 - 1 - 2 = 4 played V ONLY.
Therefore 7 + 5 + 4 = 16 students played only 1 sport, 4 students played 2 sports and 2 students played all 3 sports. The answer to the question is that 16+4+2 = 22 students played one or more sport.