If n(A)=4 and n(B)=2,then the number of subse | vedclass

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(C) Given $n(A)=4$ and $n(B)=2$. The number of elements in the Cartesian product $A 	imes B$ is $n(A 	imes B) = n(A) 	imes n(B) = 4 	imes 2 = 8$.

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$219$

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(C) Given $n(A)=4$ and $n(B)=2$. The number of elements in the Cartesian product $A 	imes B$ is $n(A 	imes B) = n(A) 	imes n(B) = 4 	imes 2 = 8$. We need to find the number of subsets of $A 	imes B$ having at least $3$ elements. The total number of subsets of a set with $8$ elements is $2^8 = 256$. The number of subsets with $0, 1,$ or $2$ elements are given by $\binom{8}{0} + \binom{8}{1} + \binom{8}{2}$. $\binom{8}{0} = 1$,$\binom{8}{1} = 8$,and $\binom{8}{2} = \frac{8 	imes 7}{2} = 28$. Sum of these subsets $= 1 + 8 + 28 = 37$. The number of subsets with at least $3$ elements $= 2^8 - (\binom{8}{0} + \binom{8}{1} + \binom{8}{2}) = 256 - 37 = 219$.

If $n(A)=4$ and $n(B)=2$,then the number of subsets of the set $A \times B$ each having at least $3$ elements is:

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