Question

If the number of ways in which $8$ different chocolates can be distributed among $3$ children if each gets at least two chocolates is $A,$ then the value of $\frac{A}{10}$ is equal to

Answer 1

Total chocolates $=8$, number of children $=3$
Possible distribution : $(2,2,4),(2,3,3)$
Case $\left(1\right)$ When $8$ chocolates are distributed among $3$ children in group of $2,2,4$
Number of ways $=(\frac{8!}{2!\cdot 2!\cdot 4!}\cdot \frac{1}{2!})\cdot 3!=1260$

Case $\left(2\right)$ When $8$ chocolates are distributed among $3$ children in group of $2,3,3$
Number of ways $=(\frac{8!}{2!\cdot 3!\cdot 3!}\cdot \frac{1}{2!})\cdot 3!=1680$
$\therefore$ Required ways $=1260+1680=2940$