Question

in n(a)=20,n(b)=15and n(AuB)=25 then n(AnB)?
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Answer 1

ANSWER

n(A△B)=n(A∪B)−n(A∩B)

for n (A \triangle B)tobemax.n (A \cap B)=0$$

We know, that

n(A∪B)=n(A)+n(B)−n(A∩B)=15+25−0=40

⇒n(A△B)

max

=40−0=40

For minimum value of n(A△B)

n(A∪B) should be min, n(A∩B) should be max.

n(A△B) min =25−15=10

So. value of

n(A△B)=n(A∪B)−n(A∩B) lies om the set

10,11,12,......,3,9,40

Now, when n(A△B) is max. i.e. when

n(A∪B)=40 & n(A∩B)=0

If we decrease n(A∪B) by 1 then n(A∩B)

Will increase by 1

n(A△B)=39−1=38

Similarly on for the decrease of 1 you will get in (A△B) as 36 and 30 so on.

Hence

Range of n(A△B)=10,12,14,16,18,20,......,38,40

=16 values