Simple Counting Question How many solutions to the equation $x_1 + x_2 + x_3 = 11$ for positive integers, are there??
Please explain your answer as much as possible. 
 A: The first exercise of the first section of the first chapter of the first volume of George Pólya and Gábor Szegő's masterpiece does it this way: for every positive integers $a_k$ and every nonnegative integer $b$, the number of nonnegative integer solutions of the equation $a_1x_1+\cdots+a_nx_n=b$ is the coefficient of $t^b$ in the series
$$
s(t)=\prod_{k=1}^n\frac1{1-t^{a_k}}.
$$
The proof is by inspection once one notes that, for every positive integer $a$,
$$
\frac1{1-t^{a}}=\sum\limits_{x=0}^{+\infty}t^{ax}.
$$
Hence the product which defines $s(t)$ is
$$
s(t)=\prod_{k=1}^n\sum\limits_{x_k=0}^{+\infty}t^{a_kx_k}=\sum\limits_{x_1,\ldots,x_n}t^{a_1x_1+\cdots+a_nx_n},
$$
and the coefficient of $x^b$ is the number of ways of writing $b$ as a sum $a_1x_1+\cdots+a_nx_n$.
You are looking at the case $n=3$, $a_1=a_2=a_3=1$ and $b=11-3=8$ hence looking for the coefficient of $t^8$ in the series
$$
s(t)=\frac1{(1-t)^3}=\frac12\frac{\text{d}^2}{\text{d}t^2}\frac1{1-t}
=\frac12\frac{\text{d}^2}{\text{d}t^2}\sum\limits_{k=0}^{+\infty}t^k=\frac12\sum\limits_{k=0}^{+\infty}(k+2)(k+1)t^k,
$$
hence the answer is $\frac12(8+2)(8+1)=45$.
A: Consider your question as if you have 11 sites, and you have to put 2 barriers that would split those sites into 3 groups.
Your constraint is that x1,x2,x3 should be positive, i.e. 0 is not a valid value. Means you may not put 2 barriers at the same joint. Also the barrier may not be put at the end. This leads to the following:


*

*There're 11-1 = 10 allowed joints (places where barriers may be put)

*Two barriers may not occupy the same joint

*Barriers are indistinguishable (means - putting barriers at a,b and b,a is the same). 


Hence the answer is C (11-1) over (3-1). Which is (11-1)! / [ (11-3)! * (3-1)! ] = 45
A: This is purely a  Question based upon the combinations and permutations, let me try.
In general we have the standard result $x_1+x_2+.......x_n=k\:$ then number of positive integer solutions is given by 
 $(n+k-1)_{\large C_{k}}$ so
according to your question it's
 $(3+11-1)_{\large C_{11}}$ which is $13_{\large C_{11}}=78$.
I hope I have put forward what I knew,
I hope that it's correct,
but leave it to scholars of this site to decide whether it's correct or not.
A: This is one of the "balls in boxes" problems; you have a collection of k empty boxes in which you want to put n objects. Let's do first the more general case where boxes may be empty (i.e., $x_i=0$ is allowed) , and then we can do your case where $x_i>0$.
So, this is a method used: we put all n objects and k boxes side-by-side. Then the idea is that every partition of these (k+n) objects  into (k-1) parts is a sum $x_1+x_2+...+x_k=n$ (by partition, we mean that we put a total of k dividers after certain of the objects and boxes). How so? If we put , say, the 1st divider after box $x_i$, that is used to mean that we put a total of i objects in the 1st box; if we put the second divider at place j , then that means that there are (j-i) objects in the second box, and so on. So for any assignment of dividers, we have an assignment $x_1,x_2,...,x_k$, with $x_1+x_2+...+x_k=n$. That gives us a total of ${n+k-1} \choose {k-1}$ possible assignments; note we use k-1 instead of k, because once we have chosen k-1 dividers, the number of objects for the $k$th box is just given by all the objects that are left. 
Now, let's consider the case where $x_i>0$. We then throw an object into each of the boxes (for a total of k objects we are throwing in) , so that we have a total of $n+k-1-k=n-1$ places left, to make sure no box is empty, and then we "recur" , by using the method above, i.e., we now find all solutions to $x_1+....+x_k=n-k$ , and we get ${n-1}\choose {k-1}$.
As a small example, take 4 objects to be put into 3 boxes. We then lay out in a line, the 3 objects , followed by the three boxes. Notice that if you put dividers behind objects ,say, 1 ,2,3 respectively . This is then the combination $x_1=x_2=x_3=1$     
