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?

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.  
