Question

The number of ways a father can distribute $50$ different coins equally among his $5$ childrens is

• A. $\frac{50!\cdot 5!}{(10!{)}^5}$
• B. $\frac{50!}{(10!{)}^5\cdot 5!}$
• C. $\frac{50!}{(10!{)}^5}$
• D. $\frac{50!}{(10!)\cdot 5!}$

Answer 1

The correct option is C $\frac{50!}{(10!{)}^5}$

Given : Total different coins $=50$,
Each child gets $10$ coins exactly.
We need to make $5$ groups of $10$ coins each and arrange them in $5!$ ways
$\therefore$ Total number of ways
$=\frac{50!}{(10!{)}^5\cdot 5!}\times 5!=\frac{50!}{(10!{)}^5}$