Question

In how many ways can two 10 paise two 20 paisa three 25 paise and one 50 paisa coins be distributed among 8 children so that each student get only one coin

Answer 1

Answer:

final answer

Explanation:

Total number of coins =$2+2+3+2=8$

$2$ coins $10$ paise, $2$ coins $20$ paise, $3$ coins $25$ paise and $1$ coins $50$ paise

Required number of ways

$\begin{aligned}&=\frac{8 !}{2 ! \times 2 ! \times 3 ! \times 1 !}=\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1 \times 2 \times 1 \times 3 \times 2 \times 1 \times 1} \\&=8 \times 7 \times 6 \times 5=1680\end{aligned}$