Permutations/Integer Solutions to Equations I'm pretty lost on this so I'd appreciate some feedback as to whether or not I'm on the right track.
Find the number of integer solutions of $x_1 + x_2 + x_3 = 15$ subject to the conditions $0 \le x_1 \le 6$, $x_2 \ge 0$, and $x_3 \ge 0$.
Based on other info I've been able to find here, I think this is one way to approach this:
$y_1 + y_2 + y_3 = 12$ where $y_1 = (x_1 - 1)$, $y_2 = (x_2 - 1)$ and $y_3 = (x_3 - 1)$ with restriction $y_1 \gt 6$.
So now, $(y_1 - 6) + y_2 + y_3 = 12 - 6 = 6$
Which now gives us $C(6 + 3 - 2, 3 - 1) = C(8,2) = 28$
Is this even remotely correct or am I completely missing the boat?
 A: You don’t want to replace the variables $x_1,x_2$, and $x_3$ by $y_1,y_2$, and $y_3$: you would do that if your conditions included the inequalities $x_1\ge 1$, $x_2\ge 1$, and $x_3\ge 1$. It’s very specifically a trick to make $0$ the lower bound on all of the variables, and you already have that. Thus, without the restriction that $x_1\le 6$ you get 
$$\binom{15+3-1}{3-1}=\binom{17}2\tag{1}$$
solutions in non-negative integers to the equation $x_1+x_2+x_3=15$.
However, some of these will have $x_1>6$; you want to count those and subtract them from $(1)$. This means counting the solutions in which $x_1\ge 7$, so count non-negative integer solutions to
$$(x_1-7)+x_2+x_3=15-7\;,$$
In other words, count non-negative integer solutions to $y_1+y_2+y_3=8$, and subtract those from $(1)$.
A: You have used a "Stars and Bars" approach. For division of $15$ into $3$ parts, this is perhaps more elaborate than necessary. 
If $x_1=0$, the number of possible pairs $(x_2,x_3)$ is $16$, since $x_2$ can take on any value from $0$ to $15$, and then $x_3$ is determined.
If $x_1=1$, the number of possible pairs $(x_2,x_3)$ is $15$.
Continue. If $x_1=6$, the number of possible pairs $(x_2,x_3)$ is $10$.
That gives a total of $16+15+\cdots +10$. This is an arithmetic progression, $7$ terms which average to $13$, so the sum is $91$. 
