9 person randomly enter 3 different rooms 9 person randomly enter 3 different rooms. What is the probability that
a)the first room has 3 person?
b)every room has 3 person?
c)the first room has n person, second room has 3 persons, third room 2 persons?
What i want to know is that which probability techniques i need to use when dealing with this type of questions, i have only learnt few techniques like combination and permutation only. 
 A: "$9$ persons randomly enter $3$ different rooms" will be normally interpreted as
"$9$ persons randomly choose (uniformly and indipendently) the room to enter among $3$ different (distinguishable) rooms available".
In this case we consider to have $3^9$ equi-probable results given by the $9$-tuple individuating the choice of every
person. The persons are distinguished by assigning their position in the $9$-tuple, practically like as if they are taking
the choice one after the other.
This is equivalent to the scheme of throwing $9$ (undistinguishable) balls into $3$ (distinguishable) bins.
We can see that each possible $9$-tuple corresponds to a term of the expansion of the multinomial
$$
\eqalign{
  & \left( {r_{\,1}  + r_{\,2}  + r_{\,3} } \right)^{\,9}  =   \cr 
  &  =  \cdots  + r_{j_{\,1} } \,r_{j_{\,2} } \,r_{j_{\,3} } \, \cdots r_{j_{\,8} } \,r_{j_{\,9} } \,
 +  \cdots \quad \left| {\;j_{\,k}  \in \left\{ {1,2,3} \right\}} \right. =   \cr 
  &  = \sum\limits_{k_{\,1}  + k_{\,2}  + k_{\,3} \; = \,9}
   {\;\left( \matrix{  9 \cr k_{\,1} ,k_{\,2} ,k_{\,3}  \cr}  \right)r_{\,1} ^{\,k_{\,1} } r_{\,2} ^{\,k_{\,2} } r_{\,3} ^{\,k_{\,3} } }  \cr} 
$$
and the multinomial coefficient gives the number of ways to have the same occupancy hystogram : $k_1$ persons in room $1$, etc.
So the answer to question a) is 
$$
\eqalign{
  & \sum\limits_{k_{\,2}  + k_{\,3} \; = \,6} {\;\left( \matrix{  9 \cr   3,k_{\,2} ,k_{\,3}  \cr}  \right)}
  = \sum\limits_{k_{\,2}  + k_{\,3} \; = \,6} {\;{{9!} \over {3!k_{\,2} !k_{\,3} !}}}  =   \cr 
  &  = \left( \matrix{  9 \hfill \cr   3 \hfill \cr}  \right)\sum\limits_k
 {\;\left( \matrix{  6 \hfill \cr   k \hfill \cr}  \right)}
  = \left( \matrix{  9 \hfill \cr   3 \hfill \cr}  \right)2^{\,6}  \cr} 
$$
which has an obvious interpretation as
(No of ways to choose the three occupants of 1st room) * (No. of ways the remaining $6$ people can choose $2$ rooms).
Then it is clear how to go for the other questions (re @abc..' answer).
Instead, the case of undistinguished persons is normally rendered as
"$9$ persons are randomly assigned to $3$ different rooms".
This problem is related to the Compositions
 of $9$ into $3$ parts, or equivalently with stars and bars scheme.
Here each triple $(n_1,n_2,n_3)$
$$
\left( {n_{\,1} ,n_{\,2} ,n_{\,3} } \right)\quad \left| \matrix{
  \;0 \le n_{\,k}  \le 9 \hfill \cr 
  \;n_{\,1}  + n_{\,2}  + n_{\,3}  = 9 \hfill \cr}  \right.
$$
is considered equi-probable.
There are 
$$
\binom{9+3-1}{3-1} = \binom{11}{2} = 110
$$
possible accomodations (with empty rooms included), of which 
$$
\binom{6+2-1}{2-1} = 7
$$
will have $3$ persons in the first room.
A: a) $\binom93*2^6=5376$
b) $\binom93*\binom63=1680$
c) $\binom94*\binom53=1260$
