Question

Number of ways in which $5$ different mobiles can be distributed among $3$ boys such that each gets atleast one mobile and none gets three is

• A. $90$
• B. $360$
• C. $180$
• D. $120$

Answer 1

The correct option is A $90$

The only way to divide $5$ mobiles among them according to required condition is $2,2,1$
there is $5!$ ways to arrange the mobiles but in the group of $2,1,1$
So total ways will be = $\frac{5!}{2!2!1!}$
to distribute among $3$ children total ways are $\frac{5!}{2!2!1!}\times 3!$ but two distribution will be same (having two mobiles each)
So required number of ways are $\frac{5!}{2!2!1!}\times \frac{3!}{2!}=90$