<s>[INST] How many integer solutions does the following system have? $$y_1 + y_2 + y_3 = 20$$
$$1 \le y_1 \le 5$$
$$y_2 \ge 5$$
$$y_3 \ge 5$$
I know how to solve it if it were:
$$y_1 + y_2 + y_3 = 20$$
$$y_1 \ge 1$$
$$y_2 \ge 5$$
$$y_3 \ge 5$$
then I would do:
$$\begin{align}
x_1 + x_2 + x_3 &= (y_1 - 1) + (y_2 - 5) + (y_3 - 5) \\[2ex]
 &= (y_1 + y_2 + y_3) - 11 \\[2ex]
 &= 20 - 11 \\[2ex]
 &= 9
\end{align}$$
and then the answer is $\binom{9+3-1}{3-1}=\binom{11}{2}$
but from here I don't understand how I can solve it when $(1 \le y_1 \le 5)$.
 [/INST] You wish to solve the equation 
$$y_1 + y_2 + y_3 = 20 \tag{1}$$
in the integers subject to the restrictions $1 \leq y_1 \leq 5$, $y_2 \geq 5$, $y_3 \geq 5$.  By making the substitutions of $x_1 + 1$ for $y_1$, $x_2 + 5$ for $y_2$, and $x_3 + 5$ for $y_3$, you obtained
$$x_1 + x_2 + x_3 = 9 \tag{2}$$
and correctly noted that if there were no restrictions, then equation 2 would have $$\binom{9 + 2}{2} = \binom{11}{2}$$ solutions.  Note that the substitutions you made impose the restrictions that 
$0 \leq x_1 \leq 4$, $0 \leq x_2$, and $0 \leq x_3$.  Thus, we must exclude those solutions of equation 2 in which $x_2 \geq 5$. To do so, let
\begin{align*}
w_1 & = x_1 - 5\\
w_2 & = x_2\\
w_3 & = x_3
\end{align*}
Substituting for $w_1 + 5$ for $x_1$, $w_2$ for $x_2$, and $w_3$ for $x_3$ in equation 1 yields
$$w_1 + w_2 + w_3 = 4$$
which is an equation in the nonnegative integers with 
$$\binom{4 + 2}{2} = \binom{6}{2}$$
solutions.   Hence, the number of solutions of equation 1 subject to the restrictions $1 \leq y_1 \leq 6$, $y_2 \geq 5$, and $y_3 \geq 5$ is $$\binom{11}{2} - \binom{6}{2}$$
</s>