Suppose
Question:

Suppose $A_{1}, A_{2}, \ldots, A_{30}$ are thirty sets each having 5 elements and $B_{1}, B_{2}, \ldots, B_{n}$ are $n$ sets each with 3 elements. Let $\bigcup_{i=1}^{30} A_{i}=\bigcup_{j=1}^{n} B_{j}=S$ and each element of $S$ belong to exactly 10 of the $A_{i}$ ‘s and exactly 9 of the BjsBj’s, then n is equal to

(a) 15                               

(b) 3                                

(c) 45                               

(d) 35                        

Solution:

It is given that each set $A_{i}(1 \leq i \leq 30)$ contains 5 elements and $\bigcup_{i=1}^{30} A_{i}=S$.

$\therefore n(S)=30 \times 5=150$

But, it is given that each element of S belong to exactly 10 of the Ai‘s.

$\therefore$ Number of distinct elements in $S=\frac{150}{10}=15$      ….(1)

It is also given that each set $B_{j}(1 \leq j \leq n)$ contains 3 elements and $\bigcup_{j=1}^{n} B_{j}=S$.

$\therefore n(S)=n \times 3=3 n$

Also, each element of S belong to eactly 9 of Bj‘s.

$\therefore$ Number of distinct elements in $S=\frac{3 n}{9}$    ….(2)

From (1) and (2), we have

$\frac{3 n}{9}=15$

$\Rightarrow n=45$

Thus, the value of n is 45.

Hence, the correct answer is option (c).

 

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