In a $12$ storey house, $9$ people enter a lift cabin. It is known that they will leave the lift in groups of $2$, $3$ and $4$ people at different storeys. In how many ways can they do so if the lift does not stop at the second storey?
$9$ people can be grouped in groups of $2$, $3$ and $4$ people as follows:
$1$ group of two people $+$ $1$ group of three people $+$ $1$ group of four people $\;$ i.e. $\;$ $3$ groups
$0$ group of two people $+$ $3$ groups of three people $+$ $0$ group of four people $\;$ i.e. $\;$ $3$ groups
$3$ groups of two people $+$ $1$ group of three people $+$ $0$ group of four people $\;$ i.e. $\;$ $4$ groups
$\therefore \;$ Total number of possible groups $= 3 + 3 + 4 = 10$ groups
Now, $3$ groups can be selected from $10$ possible groups in $P^{10}_{3}$ ways.
$\therefore \;$ $9$ people entering the lift can leave in groups of $2$, $3$ and $4$ people in $P^{10}_{3}$ ways.