Let n(A) = 3 and n(B) = 3 (where n(S) denotes | vedclass

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(C) Given $n(A) = 3$ and $n(B) = 3$.The number of elements in the Cartesian product $(A 	imes B)$ is $n(A 	imes B) = n(A) 	imes n(B) = 3 	imes 3 =

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$256$

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(C) Given $n(A) = 3$ and $n(B) = 3$.The number of elements in the Cartesian product $(A 	imes B)$ is $n(A 	imes B) = n(A) 	imes n(B) = 3 	imes 3 = 9$.$A$ subset of $(A 	imes B)$ can have $k$ elements,where $0 \le k \le 9$.The number of subsets having an odd number of elements is given by the sum of combinations: $\binom{9}{1} + \binom{9}{3} + \binom{9}{5} + \binom{9}{7} + \binom{9}{9}$.Using the binomial identity,for any positive integer $n$,the sum of odd-indexed binomial coefficients is $\sum_{k 	ext{ odd}} \binom{n}{k} = 2^{n-1}$.Here,$n = 9$,so the sum is $2^{9-1} = 2^8 = 256$.

Let $n(A) = 3$ and $n(B) = 3$ (where $n(S)$ denotes the number of elements in set $S$). Then,the number of subsets of $(A \times B)$ having an odd number of elements is:

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