Question

The number of ways in which two 10-paise, two 20-paise, three 25-paise and one 50-paise coins can be distributed among 8 children so that each child gets only one coin, is

• A. 1729
• B. 1680
• C. 1570
• D. None of these

Answer 1

The correct option is B 1680

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

$2$ coins are $10$ paise, $2$ are $20$ paise, $3$ are $25$ paise and $1$ is of $50$ paise.

$\therefore$ Required number of ways

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