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Question
Suppose \[A_1 , A_2 , . . . , A_{30}\] are thirty sets each having 5 elements and \[B_1 , B_2 , . . . , B_n\] are n sets each with 3 elements. Let \[\cup^{30}_{i = 1} A_i = \cup^n_{j = 1} B_j = S\] and each element of S belong to exactly 10 of the \[A_i 's\]and exactly 9 of the\[B_j 's\] then n is equal to
Options
(a) 15
(b) 3
(c) 45
(d) 35
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Solution
It is given that each set Ai \[\left( 1 \leq i \leq contain 5 elements and \[\cup^{30}_{i = 1} A_i = S\]
\[\therefore n\left( S \right) = 30 \times 5 = 150\]
But, it is given that each element of S belong to exactly 10 of the Ai's. \[\frac{150}{10} = 15\] .....(1)
It is also given that each set Bj
\[\left( 1 \leq j \leq n \right)\] contains 3 elements and \[\cup^n_{j = 1} B_j = S\]
\[\therefore n\left( S \right) = n \times 3 = 3n\]
Also, each element of S belong to eactly 9 of Bj's.
∴ Number of distinct elements in S = \[\frac{3n}{9}\]
From (1) and (2), we have
\[\frac{3n}{9} = 15\]
\[ \Rightarrow n = 45\]
Thus, the value of n is 45.
Hence, the correct answer is option (c).
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