There are 9 people in a lift of 12 storey house.they have to leave the lift in groups of 2,3 and 4 at different storeys..no.of ways? ( Lift does not stop at second storey)..No.of ways of selecting 2,3and4 people out of 9 people is (9C2*7C3*4C4)...no.of ways of selecting 3 storeys out of 11 storeys is 11C3...total no of ways by which people can leave the lift should be (9C2*7C3*4C4)*11C3*3!...but my answer is wrong...I don't know where I am making mistake?
 A: So, three group departure patterns available:


*

*case $1$: $\{4,3,2\}$


*

*floor choice for the various groups is $10\cdot 9\cdot 8$, assuming that they don't get off at the first or second storeys, since didn't they just get on at the first? - then choosing who is in the groups, $\binom {9}{4,3,2} = \frac {9!}{4!\,3!\,2!}$


*case $2$: $\{3,3,3\}$


*

*choose $3$ floors from the $10$ available, $\binom {10}{3}$, then assign the people into those groups, $\binom {9}{3,3,3}   = \frac {9!}{3!\,3!\,3!}$


*case $3$: $\{3,2,2,2\}$


*

*choose a floor for the group of $3$ then choose $3$ floors for the groups of $2$, $10\cdot \binom 93$, and choose the people by  $\binom {9}{3,2,2,2}   = \frac {9!}{3!\,2!\,2!\,2!}$



Overall:
$$\frac{10!}{7!} \frac {9!}{4!\,3!\,2!} + \frac{10!}{7!\,3!} \frac {9!}{3!\,3!\,3!} + \frac{10!}{6!\,3!} \frac {9!}{3!\,2!\,2!\,2!}
$$

Alternatively, the question - instead of restricting the possible group sizes as I read it  - could be asking for the scenarios where the departures are exactly one group of $4$, one group of $3$ and one group of $2$, in which case the answer is only as in case $1$, $\frac{10!}{7!} \frac {9!}{4!\,3!\,2!}$, or if there are $11$ floor departure choices,  $\frac{11!}{8!} \frac {9!}{4!\,3!\,2!}$  - which matches your answer, once you multiplied in the $3!$ to shift from ${}_{11}C_3{}$ to ${}_{11}P_3{}$. 
So either the number of available exit floors is $10$ or the question did mean that you should consider all possible scenarios that include only groups of $2,3,4$ leaving per floor (or both). Do you have the book answer, in numeric, factorial or combinatorial form?
