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 both =  n( A ⋂ B) = n(A) + n(B) – n(A ⋃ B) = 28 + 33 – 46 = 15
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) + (A ⋂ B ⋂ C)