Let A and B be two sets containing 4 and 2 el | vedclass

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(B) Given that set $A$ has $4$ elements and set $B$ has $2$ elements.The number of elements in the Cartesian product $A 	imes B$ is $n(A 	imes B) =

Vedclass · Accepted Answer

$219$

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(B) Given that set $A$ has $4$ elements and set $B$ has $2$ elements.The number of elements in the Cartesian product $A 	imes B$ is $n(A 	imes B) = n(A) 	imes n(B) = 4 	imes 2 = 8$.The total number of subsets of $A 	imes B$ is $2^n = 2^8 = 256$.We need to find the number of subsets having at least $3$ elements.This is equal to the total number of subsets minus the number of subsets having $0, 1,$ or $2$ elements.Number of subsets with $0$ elements (null set) = $^8C_0 = 1$.Number of subsets with $1$ element = $^8C_1 = 8$.Number of subsets with $2$ elements = $^8C_2 = \frac{8 	imes 7}{2} = 28$.Total subsets with less than $3$ elements = $1 + 8 + 28 = 37$.Number of subsets with at least $3$ elements = $256 - 37 = 219$.

Let $A$ and $B$ be two sets containing $4$ and $2$ elements respectively. Then the number of subsets of the set $A \times B$,each having at least $3$ elements is:

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