If $15$ boys of different ages are distributed into $3$ groups of $4, 5$, and $6$ boys randomly then the probability that three youngest boys are in different groups is

Total number of ways of distributing $15$ boys in three groups is

$=\left({}^{15}C_4\right)\left({}^{11}C_5\right)\left({}^{6}C_6\right)=\frac{15!}{4!5!6!}$

The number of ways of distributing 12 remaining boys (excluding three youngest) is $\left({}^{12}C_3\right)\left({}^{9}C_4\right)\left({}^{5}C_5\right)=\frac{12!}{3!4!5!}$
and the number of ways of distributing three youngest boys

is ${}^{3}P_3=3!$

$\therefore$ the required probability

$=\frac{3!(12!)}{3!4!5!}\times \frac{4!5!6!}{15!}=\frac{24}{91}$