Question

In how many ways can $6$ prizes be distributed equally among $3$ persons?

Answer 1

Step 1: Calculate the number of ways of price distribution for each person

It is given that, the total number of prizes $=6$

And, the number of people $=3$

Since the $6$ prizes have to be distributed equally among $3$ persons. This means that each person gets $2$ prizes.

For the first person, we have a total of $6$ prizes.

So, the number of ways to give the prize to the first person $={}^{6}C_2$

Now, for the second person, we have a total of $4$ available prizes.

So, the number of ways to give the prize to the second person $={}^{4}C_2$

And, for the third person, we have a total of $2$ available prizes.

So, the number of ways to give the prize to the third person $={}^{2}C_2$

Step 2: Calculate the total number of ways

Now, the total number of ways to distribute the prizes $={}^{6}C_2\times {}^{4}C_2\times {}^{2}C_2$

$\Rightarrow$ the total number of ways to distribute the prizes $=\frac{6!}{2!\left(6-2\right)!}\times \frac{4!}{2!\left(4-2\right)!}\times \frac{2!}{2!\left(2-2\right)!}$ $\left[\because {}^{n}C_r=\frac{n!}{r!\left(n-r\right)!}\right]$

$\Rightarrow$ the total number of ways to distribute the prizes $=\frac{6\times 5\times 4!}{2\times 1\times 4!}\times \frac{4\times 3\times 2!}{2\times 1\times 2!}\times \frac{2!}{2!\times 0!}$

$\Rightarrow$ the total number of ways to distribute the prizes $=\left(3\times 5\right)\times \left(2\times 3\right)\times \frac{1}{1\times 1}$ $\left[\because 0!=1\right]$

$\Rightarrow$ the total number of ways to distribute the prizes $=90$

Hence, $6$ prizes can be distributed equally among $3$ persons in $90$ ways.