<s>[INST] 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
 [/INST] (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.
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