Discrete Mathematics - Is my answer correct? $x_1 + x_2 + x_3 = 15$ where $x_1$ and $x_2$ and $x_3$ are non negative integers.
How many solutions are there when $1\le x_1\le 6$?
the solution i came up with is =  $\binom{15+3-1}{15} - \binom{6+3-1}{6}$
Is this correct?
 A: An expression like your answer is correct.
We want to distribute $15$ identical candies between $3$ kids, so that Kid 1 gets between $1$ and $6$ candies. We use Stars and Bars ideas that are familiar to you.   
There are $\binom{15+3-1}{15}$ ways to give out the candies without restriction. We want to remove the ways in which Kid 1 gets $0$, and also the ways she gets more than $6$.
Giving Kid 1 $0$ means we are distributing $15$ candies among $2$ kids. The number of ways is $\binom{15+2-1}{15}$. 
Now we count the number of ways Kid 1 gets more than $6$. Give her $7$. Now we have $8$ candies, to be distributed among $3$ kids. The number of ways is $\binom{8+3-1}{8}$.
The total is therefore $\dbinom{15+3-1}{15}-\dbinom{15+2-1}{15}-\dbinom{8+3-1}{8}$. 
Remark: Stars and Bars is overkill for $3$ kids, but would work equally simply for a problem with the same structure, but more candies and kids.  
A: I am unsure of how you reached your answer, however, let's look at the problem this way:
Let's look at a simple case. Say $x_1$ is $1$. Then you are asking the question how many non-negative integer solutions are there to $x_2+x_3=14$. It is not too hard to verify that there is exactly fifteen solutions (careful with those OBOEs!). 
Now, there's five more $x_1$'s to go through. Each of them is going to give you $15-x_1+1$ solutions. That makes up for the sum $$ \sum^6_{x_n=1}{15-x_1+1}=15+14+13+12+11+10=75$$
Which is different from your answer $136-28=108$
A: The number of compositions of $n$ with $k$ parts is $\binom{n+k-1}{n}=\binom{n+k-1}{k-1}$, so your solution subtracts from all solutions the solutions where all three variables sum up to $6$, which is not what you are looking for. 
Generalizing Dahn Jahn's answer, we find
$$\sum_{i=1}^m\binom{n-i+k-1-1}{k-1-1}=\sum_{i=1}^{m}\binom{n-i+k-2}{k-2}$$
as the solution whenever $x_1$ satisfies $1\le x_1\le m$ (in your example $k=3,n=15$).
