Question

If n(anb)=3and n(aub)=10 then find n(p(a∆b))

Answer 1

The symmetric difference of two sets A and B denoted by A Δ B is the set which does not contain the elements that are common in both A and B (A ∩ B).

Means the symmetric difference of A and B is the union of the differences of these sets.

AΔB = (A - B) ∪ (B - A)

We can also find the symmetric difference by finding the difference of the intersection of them from the union of them.

A Δ B = (A ∪ B) - (A ∩ B)

We apply this here to find the answer.

n(A Δ B) = n(A ∪ B) - n(A ∩ B)

$\cline{1-}$

Given,

$\bullet\ n(A\cap B)=3\\ \\ \bullet\ n(A\cup B)=10$

We have to find,

$n(P(A\Delta B))$

So,

$\begin{aligned}&n(A\Delta B)=n(A\cup B)-n(A\cap B)\\ \\ \Longrightarrow\ \ &n(A\Delta B)=10-3\\ \\ \Longrightarrow\ \ &n(A\Delta B)=7\end{aligned}$

Now,

$n(P(A\Delta B))=2^7=\mathbf{128}$

Hence 128 is the answer.

Answer 2

Answer:

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