Two different families A and B are blessed with equal number of children. There are 3 tickets to be distributed amongst the children of these families so that no child gets more than one ticket. If the probability that all the tickets go to the children of the family B is 1/12, then the number of children in each family is :

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answer is B.

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Detailed Solution

Let the number of children in each family be $x$ .So, total number of children in both families are $2 x$ .

According to question, $\frac{1}{12} = \frac{{}^{x}C_{3} \cdot 3!}{{}^{2 x}C_{3} \cdot 3!} \Rightarrow \frac{{}^{x}C_{3}}{{}^{2 x}C_{3}} = \frac{1}{12}$

$\Rightarrow \frac{x (x - 1) (x - 2) (x - 3)! (2 x - 3)!}{(x - 3)! \times (2 x) (2 x - 1) (2 x - 2) (2 x - 3)!} = \frac{1}{12}$

$\Rightarrow \frac{x \cdot (x - 1) \cdot (x - 2)}{(2 x) (2 x - 1) (2 x - 2)} = \frac{1}{12}$

$\Rightarrow \frac{(x - 2)}{4 (2 x - 1)} = \frac{1}{12} \Rightarrow 3 (x - 2) = 2 x - 1 \Rightarrow x = 5$

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