How many nonnegative integer solutions does the following equation have: $x + y + z = 15$ 
How many nonnegative integer solutions does the following equation
  have: $x + y + z = 15$
(1) if $1 \le x \le 6$
(2) if $x \ge 2$ and $y \le 3$
(3) if $x \ge 3$, $y \ge 2$, and $1 \le z \le 3$

I solved the first question (which is C(3+11-1,11)). Still very stuck on the next two
 A: There are $75$ ways to do it. Using your procedure we get $\binom{17}{15}-\binom{10}{8}-\binom{16}{15}$. The last subtraction is to remove the cases where $x=0$.  (I also did it another way, again getting $75$. Makes $75$ twice as true.)
Added: For the second question, we want the number of solutions of $x+y+z=13$, with the restriction that $y\le 3$.
Count the number of solutions of $x+y+z=13$. Subtract the number of solutions of $x+y+z=9$. That's because to count the number of "bad" solutions of $x+y+z=13$, bad because $y\ge 4$, we give $4$ to $y$ and distribute the remaining $9$ between $x$, $y$, and $z$.
A: (1)  We wish to solve the equation $x + y + z = 15$ in the nonnegative integers subject to the restrictions that $1 \leq x \leq 6$.  Let $w = x - 1$.  Then $w$ is a nonnegative integer.  Substituting $w + 1$ for $x$ in the equation $x + y + z = 15$ yields 
\begin{align*}
w + 1 + y + z & = 15\\
w + y + z & = 14\tag{1}
\end{align*}
A particular solution corresponds to where two addition signs are placed in a row of $14$ ones.  For instance,
$$1 1 1 1 1 1 1 + 1 1 1 + 1 1 1 1 1$$
corresponds to the solution $w = 6$, $y = 3$, and $z = 5$, while 
$$+ 1 1 1 1 1 + 1 1 1 1 1 1 1 1 1$$
corresponds to the solution $w = 0$, $y = 5$, and $z = 9$.  The number of solutions of equation 1 in the nonnegative integers is the number of ways to select which two of the $16$ symbols (the $14$ ones and the two addition signs) will be addition signs, which is 
$$\binom{14 + 2}{2} = \binom{16}{2}$$
However, the restriction that $1 \leq x \leq 6 \Rightarrow 0 \leq w \leq 5$.  Thus, we must remove those solutions in which $w \geq 6$.  Assume $w \geq 6$.  Let $v = w - 6$.  Then $v$ is a nonnegative integer.  Substituting $v + 6$ for $w$ in equation 1 yields
\begin{align*}
v + 6 + y + z & = 14\\
v + y + z & = 8 \tag{2}
\end{align*}
Equation 2 has $$\binom{8 + 2}{2} = \binom{10}{2}$$ solutions in the nonnegative integers.  Hence, the number of solutions of the equation $x + y + z = 15$ in the nonnegative integers subject to the restrictions that $1 \leq x \leq 5$ is $$\binom{16}{2} - \binom{10}{2}$$  
(2) We wish to solve the equation $x + y + z = 15$ in the nonnegative integers subject to the restrictions that $x \geq 2$ and $y \leq 3$.  Let $w = x - 2$.  Then $w$ is a nonnegative integer.  Substituting $w + 2$ for $x$ in the equation $x + y + z = 15$ yields
\begin{align*}
w + 2 + y + z & = 15\\
w + y + z & = 13 \tag{3}
\end{align*}
which is an equation with $$\binom{13 + 2}{2} = \binom{15}{2}$$ solutions in the nonnegative integers.  From these, we must eliminate those solutions in which $y \geq 4$.  Assume $y \geq 4$. Let $v = y - 4$.  Substituting $v + 4$ for $y$ in equation 3 yields
\begin{align*}
w + v + 4 + z & = 13\\
w + v + z & = 9
\end{align*}
which is an equation with $$\binom{9 + 2}{2} = \binom{11}{2}$$ solutions in the nonnegative integers.  Hence, the number of solutions of the equation $x + y + z = 15$ in the nonnegative integers subject to the restrictions that $x \geq 2$ and $y \leq 3$ is 
$$\binom{15}{2} - \binom{11}{2}$$
Using the methods shown above should enable you to solve the third problem.
A: Can use 'stars and bars'/combination 'with repetition':
number of positive integer solutions to $x+y+z=15$ is the same as the number of ways you can place $15$ identical objects in $3$ containers or ${(n-1)+r}\choose r$ with $n=3$ and $r=15$.  gives a total of ${17\choose 15}=136$.  To add restrictions simply subtract all applicable combinations 'with repetition':  
for $1)$ if $x$ is $0$ the fifteen objects must be arranged exclusively in the other $2$ containers: 
$n=2,r=15$: ${{(n-1)+r}\choose r}={16\choose 15}$
if $x\ge 7$:  $r=15-7=8$,$n=3$: ${{(n-1)+r}\choose r}={10\choose 8}$
so solution is: ${17\choose 15}-{16\choose 15}-{10\choose 8}=136-16-45=75$
$2)$ and $3)$ can be solved similarly.
