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1. Let A,B, C are the three sets, then n(A)=17, n(B)=17, n(C)=17, n(BnC)=6, n(AnC)=5, and n(AnBnC)=2 then find n(AUBUC).
Solution:
n(AUBUC). = n(A) + n(B) + n(C) - n(AnB)-.
n(BnC) - n(AnC) + n(AnBnC)
= 17 + 17 + 17 - 7 - 6 - 5 + 2
= 17+ 17 + 17 + 2 - 7 - 6 - 5
= 53 - 18
= 35
n(AUBUC) = 35
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