Tom K. answered 09/04/20
Knowledgeable and Friendly Math and Statistics Tutor
Note that there are eight equations and eight unknowns (in or not in A, B, and C), so one way to attack this problem would be to set up a system of equations for the eight unknowns and solve.
However, if we want to construct a solution in a different way, note that
n(A'B'C) = n(A'B') - n(A'B'C')
n(A'B') = n(U) - n(A U B) = n(U) - (n(A) + n(B) - n(AB)) = n(U) + n(AB) - n(A) - n(B) = 54 + 11 - 16 - 24 = 25
n(A'B'C') = 25 - 8 = 17