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?
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
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.
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.
