<s>[INST] Posible combinations of available rooms I have $x$ people to divide over different type of rooms.
Possible rooms: 


*

*$1$ person bedroom normal view; 

*$1$ person bedroom sea view;

*$2$ person bedroom


For $x = 4$
Possible combinations are
$$(0,0,2),  
(0 ,2, 1),  
(2 ,0, 1),  
(1 ,1, 1),  
(4 ,0, 0),  
(3 ,1, 0),  
(2 ,2, 0),
(1 ,3 ,0),
(0 ,4, 0)$$
Note : Plenty of rooms are available. 
How can I get all posible combinations?
 [/INST] Are you interested in the number of compositions, or the list of compositions itself?
The number of compositions can be obtained using generating functions.  I assume, as in your example, that a two-person bedroom cannot be occupied by a single person.  The number of compositions for $n$ people is the coefficient of $y^n$ in the expansion of
$$
\frac{1}{(1-y)^2(1-y^2)}=\frac{(1+y)^2}{(1-y^2)^3}=(1+2y+y^2)\sum_{j=0}^\infty\frac{(-3)(-2)\ldots(-j-2)}{j!}(-y^2)^j.
$$
The expression on the left comes from interpreting $\frac{1}{1-y}$ as $1+y+y^2+\ldots$ and $\frac{1}{(1-y^2)}$ as $1+y^2+y^4+\ldots$ and realizing that the coefficient of $y^n$ when the three factors are multiplied out is exactly the number of whole-number solutions to $a+b+2c=n.$  The middle expression comes from writing $\frac{1}{1-y}$ as $\frac{1+y}{(1-y)(1+y)}.$  The expression on the right is obtained using the binomial theorem.
With some further manipulations, the generating function becomes
$$
(1+2y+y^2)\sum_{j=0}^\infty\binom{j+2}{j}y^{2j}=(1+2y+y^2)\sum_{j=0}^\infty\binom{j+2}{2}y^{2j}.
$$
You can see that when $n$ is even, the coefficient of $y^n$ is $\binom{n/2+2}{2}+\binom{n/2+1}{2}$.  When $n$ is odd, the coefficient is $2\binom{(n-1)/2+2}{2}.$
These answers can be understood using stars-and-bars: if $n$ is even, pair up the people.  So there are $n/2$ pairs.  Either we have an even number of people in all three types of room, in which case there are $\binom{n/2+2}{2}$ compositions, or there is an odd number of people in the single-type rooms, in which case we split one of the pairs, putting one member of the pair in a normal room and the other in a sea-view room, and then find that there are $\binom{n/2+1}{2}$ compositions for the remaining pairs.
If $n$ is odd, we have $(n-1)/2$ pairs, and one single person.  That single person goes either in a normal room or a sea-view room, and there are $\binom{(n-1)/2+2}{2}$ compositions for the pairs.
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