Question

In how many ways can five people be arranged in three different rooms if no room must be empty and each room has $5$ seats in a single row.

Answer 1

$5$ People are to arranged $3$ rooms given that no room is empty.

There are $2$ basic possibilities $:$

$1.(2,2,1)$ $2.(3,1,1)$

There are $3$ rearrangements (room rearrangement) for each possibility.

For $1.$ Since there are $5$ seats in each room

$\Rightarrow$ Permutations $={}^5{P}_2\times {}^5{P}_2\times {}^5{P}_1$

$=5^3\times 4^2$

$=4000.$

$\Rightarrow$ For $2.$ $={}^5{P}_3\times {}^5{P}_1\times {}^5{P}_1$

$=5^3\times 4\times 3$

$=3000.$

$\Rightarrow$ Total $=3\times (4000+3000)$

$=21000$ ways.

Hence, the answer is $21000.$