If one quarter of all three element subsets of the set $A = a_{1}, a_{2}, a_{3} . . ., a_{n}$ contains the element $a_{3}$ then n $=$

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Solution

The correct option is C 12
Since each set must contain $a_{3}$ , other two elements taken from remaining $(n - 1)$ can be selected in $^{n - 1} C_{2}$ ways
Given, $\frac{1}{4} .^{n} C_{3} =^{n - 1} C_{2}$
$=> n = 12$

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