Question

The number of ways of distributing $50$ identical things among $8$ persons in such a way that three of them get $8$ things each, two of them get $7$ things each, and remaining $3$ get $4$ things each, is equal to

• A. $\frac{\left(50!\right)\left(8!\right)}{\left(8!\right)^{3}\left(3!\right)^{2}\left(7!\right)^{2} \left(4!\right)^{3}\left(2!\right)}$
• B. $\frac{\left(50!\right)\left(8!\right)}{\left(8!\right)^{3}\left(7!\right)^{2} \left(4!\right)^{3}}$
• C. $\frac{\left(50!\right)}{\left(8!\right)^{3}\left(7!\right)^{2} \left(4!\right)^{3}}$
• D. $\frac{\left(8!\right)}{\left(3!\right)^{2} \left(2!\right)}$

Answer 1

Answer: (D)

Number of ways of dividing $8$ persons in three groups,
first having $3$ persons,
second having $2$ persons
and third having $3$ persons = $\frac{8!}{3! 2! 3!}$.
Since all the $50$ things are identical,
so, required number = $\frac{\left(8!\right)}{\left(3!\right)^{2}\cdot\left(2!\right)}$