Suppose A_1, A_2, A_3, dots, A_{30} are 30 se | vedclass

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(C) Let $n(S)$ be the number of elements in the set $S$.Given that each of the $30$ sets $A_i$ has $5$ elements, the total number of elements counted

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(C) Let $n(S)$ be the number of elements in the set $S$.Given that each of the $30$ sets $A_i$ has $5$ elements, the total number of elements counted with multiplicity is $30 	imes 5 = 150$.Since each element of $S$ belongs to exactly $10$ of the sets $A_i$, we have $n(S) = \frac{30 	imes 5}{10} = 15$.Similarly, each of the $n$ sets $B_j$ has $3$ elements, so the total number of elements counted with multiplicity is $n 	imes 3 = 3n$.Since each element of $S$ belongs to exactly $9$ of the sets $B_j$, we have $n(S) = \frac{3n}{9} = \frac{n}{3}$.Equating the two expressions for $n(S)$, we get $\frac{n}{3} = 15$, which implies $n = 45$.

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