Let bigcup_{i=1}^{50} X_{i} = bigcup_{i=1}^{n | vedclass

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(D) Let $n(T)$ be the number of elements in set $T$.Given that $\bigcup_{i=1}^{50} X_{i} = T$ and each $X_{i}$ has $10$ elements,the sum of the number

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$30$

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(D) Let $n(T)$ be the number of elements in set $T$.Given that $\bigcup_{i=1}^{50} X_{i} = T$ and each $X_{i}$ has $10$ elements,the sum of the number of elements in all $X_{i}$ is $50 	imes 10 = 500$.Since each element of $T$ belongs to exactly $20$ sets of $X_{i}$,the number of distinct elements in $T$ is $n(T) = \frac{500}{20} = 25$.Similarly,for the sets $Y_{i}$,we have $\bigcup_{i=1}^{n} Y_{i} = T$ and each $Y_{i}$ has $5$ elements. The sum of the number of elements in all $Y_{i}$ is $n 	imes 5 = 5n$.Since each element of $T$ belongs to exactly $6$ sets of $Y_{i}$,the number of distinct elements in $T$ is $n(T) = \frac{5n}{6}$.Equating the two expressions for $n(T)$,we get $\frac{5n}{6} = 25$.Solving for $n$,we get $5n = 150$,which implies $n = 30$.

Let $\bigcup_{i=1}^{50} X_{i} = \bigcup_{i=1}^{n} Y_{i} = T$,where each $X_{i}$ contains $10$ elements and each $Y_{i}$ contains $5$ elements. If each element of the set $T$ is an element of exactly $20$ of the sets $X_{i}$ and exactly $6$ of the sets $Y_{i}$,then $n$ is equal to:

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