Probability problem! Combinations with repetition guys ! Who can explain this problem?I know that I have to use combinations with repetitions,but also I have to  extract some cases.Any ideeas ? : 
In how many ways can Santa Claus distribute 10 presents to three children for Christmas, supposing that
each child will receive at least one present?
 A: If you know the stars and bars type method(explained as:) for identical gifts:
$$\sum^n x_i=m;x_i\geq1\implies \binom{m-1}{n-1}$$
In your case it's $\binom{9}{2}=36$

Note: See method explanation

If gifts are not identical its total ways minus the ways in  which any child goes empty handed: $$c^{t}-\sum_{\phi\ne J\subseteq\{i\mid i\leq c,i\in\mathbb N\}}(-1)^{|J|-1}n\left(\bigcap_{j\in J}A_j\right)\quad c=3,t=10$$
where $n(A_j)$ is the number of ways in which the $j^{th}$ child go empty handed and $c,t$ are the no. of children and toys respectively.
Also no. of elements in union of $c_0$ sets is:
$$\binom c{c_0}(c-c_0)^t$$
So:
$$3^{10}-\left[\binom312^{10}-\binom321^{10}+\binom330^{10}\right]=55980$$

Note: See method explanation
A: Just use the formula for distributions, which is distributing n objects to r people, each person recieving 1 or more with a number of ways of (n-1)C(r-1). So, in this case, it would be 9C2=36 ways.
A: Assuming that the gifts are not identical, the question is asking for the number of surjections (onto functions) from a set of $10$ elements (gifts) to a set of $3$ elements (children).
See for instance Counting the number of surjections. 
