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Let $A, B$ be two set such that $n (A) = 4$ and $n (B) = 6$ then the least possible number of elements in the power set of $(A \cup B)$ is

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Solution

$\begin{matrix} L e t A = {1, 2, 3, 4} B = {1, 2, 3, 4, 5, 6} N o w, A \cup B = {1, 2, 3, 4, 5, 6} H e n c e, l e a s t p o s s i b l e n u m b e r o f e l e m e n t i s 4. \end{matrix}$

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Q. Let A,B are two sets such that n(A)=4 and n(B)=6. Then the least possible number of elements in the power set of

(A \cup B)

Q. If A and B are two finite sets such that n(A) > n(B) and the difference of the number of elements of the power sets of A and B is 96, then n(A) – n(B) = ____________.

A, B

are two sets such that

n (A) = 6, n (B) = 8

then the maximum number of elements in

n (A \cup B)

X

be a set containing

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P (X)

be its power set. The sets

A

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n (A) = n (B)

A \neq B,

then total number of ordered pair

(A, B)

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]

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B

are two finite sets such that

n (A) = 3

n (B) = 4

then the number of elements in

A Δ B

.

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