Question

Let A,B,C be finite sets. Suppose that n(A)=10, n(B)=15, n(C)=20, n(A∩ B)=8 and n(B∩ C)=6. Then the possible value of n(A∪ B∪ C) is

• A. 26
• B. 27
• C. 28
• D. Any of the three values 26, 27, 28 is possible

Hint:

When triple intersections are not fixed, multiple values of union are possible.

Answer 1

The Correct Option is D

Using the formula: 

n(A∪ B∪ C)=n(A)+n(B)+n(C)-n(A∩ B)-n(B∩ C)-n(A∩ C)+n(A∩ B∩ C) =10+15+20-8-6-n(A∩ C)+n(A∩ B∩ C) =31-bigl[n(A∩ C)-n(A∩ B∩ C)bigr]

 Since 0≤ n(A∩ B∩ C)≤ n(A∩ C)≤ min(10,20)=10, the quantity n(A∩ C)-n(A∩ B∩ C) can take values 3,4,5. Hence, n(A∪ B∪ C)=31-5,31-4,31-3=26,27,28 So all three values are possible.