Let the number of elements in sets A and B be | vedclass

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(A) Given that $n(A) = 5$ 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) = 5 	imes

Vedclass · Accepted Answer

$792$

Vedclass · Answer

(A) Given that $n(A) = 5$ and $n(B) = 2$.The number of elements in the Cartesian product $A 	imes B$ is $n(A 	imes B) = n(A) 	imes n(B) = 5 	imes 2 = 10$.We need to find the number of subsets of $A 	imes B$ having at least $3$ and at most $6$ elements.This is equivalent to calculating the sum of combinations: ${}^{10}C_3 + {}^{10}C_4 + {}^{10}C_5 + {}^{10}C_6$.Calculating each term:${}^{10}C_3 = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} = 120$${}^{10}C_4 = \frac{10 	imes 9 	imes 8 	imes 7}{4 	imes 3 	imes 2 	imes 1} = 210$${}^{10}C_5 = \frac{10 	imes 9 	imes 8 	imes 7 	imes 6}{5 	imes 4 	imes 3 	imes 2 	imes 1} = 252$${}^{10}C_6 = {}^{10}C_4 = 210$Summing these values: $120 + 210 + 252 + 210 = 792$.

Let the number of elements in sets $A$ and $B$ be $5$ and $2$ respectively. Then the number of subsets of $A \times B$ each having at least $3$ and at most $6$ elements is:

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