TS ICET 24 May 2018 Shift 1 Paper

Question : 54 of 200

Marks: +1, -0

Let A. B. C be sets and S = A∪B∪C. For any T ⊆ S let

\T^1

= S - T and n(T) denote the number of elements in T. If n(A) = 20, n(B) = 21, n(C) = 8 ,n(A∪B∪C ) 35. n( A∩B∩

\C^1

) = 6 . n(B∩C∩

\A^1

) = 2n(C n 4 n B ') = 4
then, n(A∩B∩C)=

Solution:

Let

(A∩B∩C)=x

We know that

n(A ∪ B ∪ C)=n(A)+n(B)+n(C)

-n(A ∩ B)-n(B ∩ C)-n(C ∩ A)+n(A ∩ B ∩ C)

=n(A)+n(B)+n(C)

-\{(n(A ∩ B ∩ C^{'})+(n(A ∩ B ∩ C))\}-n(B ∩ C ∩)

A^{'}+n(A ∩B ∩ C)\}

-\{(n(A ∩ C ∩ B^{'})+(n(A ∩ B ∩ C)\}+(n(A ∩ B ∩ C)

35 = 20 + 21 + 8- (6 + x) -(2 + x)-(4 + x) + x

2 x=49-35-6-2-4

2 x=2

x=1

∴ n(A ∩ B ∩ C)=1

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