Let P , Q , and R be three finite sets such that n(P) + n(Q) + n(R) = 110 . If the number of elements in their pairwise symmetric differences are n(P Q) = 40 , n(Q R) = 30 , and n(R P) = 50 , and the number of elements in their intersection is n(P Q R) = 5 , then the number of elements present in exactly one of the sets P , Q , or R is:

Step-by-step solution

The number of elements in the symmetric difference of two sets is given by: n(P Q) = n(P) + n(Q) - 2n(P Q) Summing the three given symmetric differences yields: n(P Q) + n(Q R) + n(R P) = 2(n(P) + n(Q) + n(R)) - 2(n(P Q) + n(Q R) + n(R P)) Let S₁ = n(P) + n(Q) + n(R) and S₂ = n(P Q) + n(Q R) + n(R P) . Substituting the given values: 40 + 30 + 50 = 2(110) - 2S₂ 120 = 220 - 2S₂ 2S₂ = 100 S₂ = 50 The number of elements present in exactly one of the three sets is given by the formula: E₁ = S₁ - 2S₂ + 3S₃ where S₃ = n(P