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 = 9

Vedclass · Accepted Answer

$256$

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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 $n(A 	imes B) = n(A) 	imes n(B) = 3 	imes 3 = 9$.The number of subsets of a set with $m$ elements having an odd number of elements is given by the sum of combinations $\sum_{k 	ext{ is odd}} \binom{m}{k} = \binom{m}{1} + \binom{m}{3} + \dots + \binom{m}{m}$.For $m = 9$,this sum is $\binom{9}{1} + \binom{9}{3} + \binom{9}{5} + \binom{9}{7} + \binom{9}{9}$.Using the identity $\sum_{k 	ext{ is odd}} \binom{m}{k} = 2^{m-1}$,we get $2^{9-1} = 2^8 = 256$.

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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