Question

In how many ways can five people be distributed in three different rooms if no room must be empty?

Answer 1

Calculate the number of ways using permutations

Given that, the total number of people $=5$

And, the total number of rooms $=3$

Since we have to distribute $5$ people in $3$ different rooms such that no room is left empty.

Then, it can be stated that,

Total number of ways $=$ Number of ways in which groups can be formed $\times$ Number of ways groups can be arranged in different rooms

$\Rightarrow$ Total number of ways $=\left[\frac{5!}{3!}\times \frac{1}{2!}\right]3!+\left[\frac{5!}{{\left(2!\right)}^2}\times \frac{1}{2!}\right]3!$

$\Rightarrow$ Total number of ways $=\left[\frac{5\times 4\times 3!}{3!}\times \frac{1}{2\times 1}\right]3!+\left[\frac{5\times 4\times 3\times 2!}{2\times 1\times 2\times 1}\times \frac{1}{2!}\right]3!$

$\Rightarrow$ Total number of ways $=\left[\frac{10}{1}\right]3!+\left[\frac{5\times 1\times 3}{1}\right]3!$

$\Rightarrow$ Total number of ways $=\left[10\right]3!+\left[15\right]3!$

$\Rightarrow$ Total number of ways $=\left[10+15\right]3!$

$\Rightarrow$ Total number of ways $=25\times 3!$

$\Rightarrow$ Total number of ways $=150$

Hence, $5$ people can be distributed in $3$ different rooms such as no room is empty in $150$ ways.