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 sets $$X_i$$'s and exactly 6 of sets $$Y_i$$'s then $$n$$ is equal to:

Let us denote the number of distinct elements in the union by $$|T|$$. We begin with the family $$\{X_1,X_2,\dots ,X_{50}\}$$.

Each set $$X_i$$ contains 10 elements, so the total of all ordered pairs “(set, element in that set)” contributed by the 50 sets equals

$$\sum_{i=1}^{50}|X_i| \;=\;50\times 10 \;=\;500.$$

However, the statement tells us that every single element of $$T$$ lies in exactly 20 of the $$X_i$$’s. Hence, if we count the same ordered pairs by first choosing an element of $$T$$ and then choosing the set $$X_i$$ that contains it, we obtain

$$20\times |T|.$$

Because both computations refer to the same collection of ordered pairs, they are equal. Therefore

$$20\times |T| = 500 \;\;\Longrightarrow\;\; |T|=\frac{500}{20}=25.$$

Now we turn to the family $$\{Y_1,Y_2,\dots ,Y_n\}$$. Each $$Y_i$$ has 5 elements, so the total number of ordered pairs “(set, element in that set)” here is

$$\sum_{i=1}^{n}|Y_i| \;=\;n\times 5 \;=\;5n.$$

Again, the problem states that every element of $$T$$ appears in exactly 6 of the $$Y_i$$’s. Counting by first picking an element and then one of the 6 sets that contain it gives

$$6\times |T| = 6\times 25 = 150.$$

Equating the two counts for the $$Y_i$$ family, we have

$$5n = 150 \;\;\Longrightarrow\;\; n = \frac{150}{5}=30.$$

Hence, the correct answer is Option D.