Question

Two different families A and B are blessed with equal number of children. There are 3 tickects to be distributed among 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 $\frac{1}{12}$, then the number of children in each family is :

• A. $3$
• B. $5$
• C. $4$
• D. $6$

Answer 1

The correct option is B $5$

Let ${}^{\prime }{n}^{\prime }$ number of children are there in each family
$\therefore \frac{1}{12}=\frac{{}^n{C}_3}{{}^{2n}{C}_3}\Rightarrow 12n(n-1)(n-2)=2n(2n-1)(2n-2)\Rightarrow n=5$