Combinatorics - multi-binomial special case 2 variable in n multiple parentheses and r summed locations I came across the following combinatorial question and lost my way, wonder if you can assist ??? I think it might be a sub case of a multi-binomial theorem, maybe more related to set/multiset theorem. 
I have 2 variables for example  assume {a,b} 
I have n parentheses and r locations in each parentheses - 
inside each parentheses variables are summed and the parentheses are are multiplied with each other. 
example of n = 3 , r = 3  - ( _ + _ + _ ) * (_ + _ + _ ) * (_ + _ + _ ) 
since {a.b} are just 2 variables, this mean it can be repeated many times within the parentheses.
I made some progress that you might be able to see in this example - 
I will spread here all options for r=2, n=2 : 
{a+a} * {a+a} * 1 + 
{b+b} * {b+b} * 1 + 
{a+b} * {a+a} * 4 + ( 4 due to {a+b} * {a+a} = {b+a} * {a+a} = {a+a} * {b+a} = {a+a} * {a+b} )
{a+b} * {b+b} * 4 + 
{a+b} * {a+b} * 4 + 
{a+a} * {b+b} * 2 
for any n, r variable a variable b - I would like to find a formula that generate the sum of all possible options. 
Some progress I made is the following - 
$ \sum_{x1=0}^r(x1*a+(r-x1)*b)$ stands for all possible sums within a single parentheses.  
if I would like to add to it the number of occurrences it will be as follows 
$ \sum_{x1=0}^r(x1*a+(r-x1)*b) * {r \choose x1}  $
This now should be multiplied n times based on the number of repetitions, I feel I am getting closer...
Thanks for the corrections, hope now it will be clearer.
I am not a native English speaker, and would appreciate further corrections if needed. 
Thanks, 
Ran.
 A: Let there be $r$ summands in each factor, and $s$ factors in all. If the operators are all conmutative and associative, then there are exactly $r + 1$ possible values for each factor (0 times $a$, 1 time $a$, ..., $r$ times $a$). Now we have $r + 1$ possible factors, and we want to know how many integer solutions the equation $x_1 + x_2 + \ldots x_{r + 1} = s$ has, where $0 \le x_i$. Representing each $x_i$ as a sequence of $*$, this means splitting a string of $s$ times $*$ by placing $|$ in it $r$ times . This is distributing $r$ bars over $r + s$ places, and that can be done in $\binom{r + s}{r}$ ways.
A: Denote the requested sum by $S(n,r)$, with a tacit dependence on $a$ and $b$. I claim that
$$S(n,r)=\bigl( r 2^{r-1}(a+b)\bigr)^n\qquad(n\geq0,\ r\geq1)\ .\tag{1}$$
Proof. $\ S(0,r)$ is the empty product and has the value $1$. Therefore assume that $(1)$ is true for $n$. The sum $S(n,r)$ is a sum of a great many terms $T_\iota$ (in fact $2^{nr}$ of them, see below). Each $T_\iota$  consists of $n$ factors, each of the latter being an $r$-nom. Each such $T_\iota$ now gets an additional factor of the form $$(x_1+x_2+\ldots x_r),\qquad x_k\in\{a,b\}\quad(1\leq k\leq r)$$ to the right. 
Such factors can be set up in the following way: Choose a $j$ with $0\leq j\leq r$, then choose $j$ places in $[r]$ with $x_k=a$, and at the remaining $r-j$ places take $x_k=b$.  The  factor built in this way has value $ja+(r-j)b$. For given $j$ this can be done in ${r\choose j}$ ways. It follows that the sum of all the factors so generated is
$$\sum_{j=0}^r{r\choose j}\bigl(ja+(r-j)b\bigr)=(a+b)\sum_{j=0}^r{r\choose j} j= r2^{r-1}(a+b)\ .$$ From this we conclude that each term $T_\iota$ in $S(n,r)$ gives rise to $2^r$ terms in $S(n+1,r)$, and that the sum of these terms is $r2^{r-1}(a+b)\cdot T_\iota\ $. Altogether we obtain $(1)$ with $n+1$ in place of $n$.
