Question

If n(A intersection B)= 3 and n( A union B)= 10 then find n[p(A∆B)]​

Answer 1

Answer:

We have the identity: n(A∪B)=n(AΔB)+n(A∩B) .

Plugging in the values given in the question, we get n(AΔB)=n(A∪B)−n(A∩B)=10−3=7 .

Proof of the identity

AΔB:={x:(x∈A and x∉B) or (x∈B and x∉A)}

A∩B:={x:x∈A and x∈B}

The simplest way to see the identity (proof by picture) is to draw a Venn diagram of non-trivially intersecting sets A and B , and look at the sets AΔB (symmetric difference) and A∩B (intersection).

But here is the proof in words:

By definition, the two sets AΔB and A∩B are disjoint (have empty intersection).

Now, we claim that (AΔB)∪(A∩B)=A∪B .

If x∈A∪B , then x∈A or x∈B . Suppose x∈A . Then if x∈B , we have x∈A∩B , otherwise we have x∈AΔB . A similar argument applies when x∈B . So, A∪B⊆(AΔB)∪(A∩B) .

We have A∩B⊆A∪B and AΔB⊆A∪B by definition. So, (AΔB)∪(A∩B)⊆A∪B .

Finally, we get:

n(A∪B)=n((AΔB)∪(A∩B))=n(AΔB)+n(A∩B) (since the two sets are disjoint).

Answer 2

Step-by-step explanation:

FORMULA : n(A∆B)=n(AUB)-n(AnB)

n(A∆B)=10-3=7

n[p(A∆B)]= 2^n(A∆B)

2^7= 128