Let A and B be the sets of those playing hockey and cricket, respectively.

Then n(A) = 28 , n(B) = 33 and n(A ⋃ B) = 60 – 14 = 46.

Number of pupils who play cricket only = n(B) = n(A ⋃ B) – n(A) = 46 – 28 = 18.

Note: The formulae used above can be extended for more than two sets also.

For any three sets A, B, and C,

n(A ⋃ B ⋃ C) = n(A) + n(B) + n(C) – n(A ⋂ C) – n(B ⋂ C) – n(A ⋂ C) + n(A ⋂ B ⋂ C)