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Question

Suppose $A subscript 1 end subscript comma blank A subscript 2 end subscript comma horizontal ellipsis comma A subscript 30 end subscript$ are thirty sets, each having 5 elements and $B subscript 1 end subscript comma blank B subscript 2 end subscript comma blank horizontal ellipsis comma blank B subscript n end subscript$ are $n$ sets each with 3 elements, let
$not stretchy ⋃ subscript i equals 1 end subscript superscript 30 end superscript A subscript i end subscript equals not stretchy ⋃ subscript j equals 1 end subscript superscript n end superscript B subscript j end subscript equals S$ and each element of $S$ belongs to exactly 10 of the $A subscript i end subscript$ 's and exactly 9 of the $B subscript j end subscript$ 's. Then, $n$ is equal to

115
83
45
None of these

The correct answer is: 45

Given, $A$ 's are 30 sets with five elements each, so
$not stretchy sum from i equals 1 to 30 of n open parentheses A subscript i end subscript close parentheses equals 5 cross times 30 equals 150$ (i)
If the $m$ distinct elements in $S$ and each elements of $S$ belongs to exactly 10 of the $A subscript i end subscript$ 's, then
$not stretchy sum from i equals 1 to 30 of n open parentheses A subscript i end subscript close parentheses equals 10 m$ (ii)
From Eqs. (i) and (ii), $m equals 15$
Similarly, $not stretchy sum from j equals 1 to n of n open parentheses B subscript j end subscript close parentheses equals 3 n blank$ and $blank not stretchy sum from j equals 1 to n of n open parentheses B subscript j end subscript close parentheses equals 9 m$
$therefore 3 n equals 9 m$
$rightwards double arrow blank n equals fraction numerator 9 m over denominator 3 end fraction equals 3 cross times 15 equals 45$

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