How many solutions are there to $x_{1} + x_{2} + x_{3} + x_{4} = 15$ How many solutions are there to (I think they mean non-negative integer solutions)
$x_{1} + x_{2} + x_{3} + x_{4} = 15$
where
$1\leq x_{1} \leq 4$
$2 \leq x_{2} \leq 5$
$7\leq x_{3}$
$2\leq x_{4}$
My "Solution"
I use the method shown in this video: https://www.youtube.com/watch?v=Y0CYHMqomgI&list=PLDDGPdw7e6Aj0amDsYInT_8p6xTSTGEi2&index=7
$$N(\overline{C_{1}}\overline{C_{2}}\overline{C_{3}}\overline{C_{4}}) = $$
$$ N - (N_{C1}+N_{C2}+N_{C3}+N_{C4}) + N_{C1C2}+ N_{C1C3}+ N_{C1C4}+ N_{C2C3}+ N_{C2C4}+N_{C3C4}  - N_{C1C2C3} - N_{C1C2C4} - N_{C1C3C4} - N_{C2C3C4} + N_{C1C2C3C4}$$
$$N=\binom{18}{15}$$
Now this part (find NC1) i am unsure about. when $$1\leq x_{1} \leq 4$$
$$N_{C1}$$
$$x_{1} + x_{2} + x_{3} + x_{4} = 15$$ $$1\leq x_{1} \leq 4$$
$$ x_{1}{}' + x_{2} + x_{3} + x_{4} = 15 -4 = 11$$ $$x_{1}{}' \geq 0$$
$$N_{C1} =\binom{14}{11}$$
Is this the right way to find $$N_{C1} $$ ?
 A: Substituting 
$$\begin{cases}
y_1=x_1-1\\
y_2=x_2-2\\
y_3=x_3-7\\
y_4=x_4-2
\end{cases}$$
your problem becomes equivalent to finding $y_1,\ldots,y_4$ satisfying $$\begin{split}0&\leq y_1\leq 3,\\0&\leq y_2\leq 3,\\0&\leq y_3,\\0&\leq y_4\end{split}$$ and $$y_1+y_2+y_3+y_4=3.$$ The solution to this are $$y=(y_1,\ldots,y_4)=(3,0,0,0)$$ (and other permutations, making up a total of $4$), $$y=(2,1,0,0)$$ (and other permutations, making up a total of $4!/2!=12$), and $$y=(1,1,1,0)$$ (and other permutations, making up $4$). The number of solutions then is $$4+12+4=20.$$
A: Write $y_1=x_1-1$, $y_2=x_2-2$, $y_3-7$ and  $y_4-2$, then $y_i \geq 0$, $y_1,y_2\leq 3$ and $$y_1+y_2+y_3+y_4 = 3$$
But then $y_3$ and $y_4$ can not be 4 or more. So $y_1,y_2\leq 3$ is not actual restriction here. By stars and bars method we have $${6\choose 3}=20$$ solution. 
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
&\bbox[10px,#ffd]{\sum_{x_{1} = 1}^{4}\sum_{x_{2} = 2}^{5}
\sum_{x_{3} = 7}^{\infty}\sum_{x_{4} = 2}^{\infty}
\bracks{z^{15}}z^{x_{1} + x_{2} + x_{3} + x_{4}}}
\\[5mm] = &\
\sum_{x_{1} = 0}^{3}\sum_{x_{2} = 0}^{3}
\sum_{x_{3} = 0}^{\infty}\sum_{x_{4} = 0}^{\infty}
\bracks{z^{15}}z^{\pars{x_{1} + 1} + \pars{x_{2} + 2} + \pars{x_{3} + 7} + \pars{x_{4} + 2}}
\\[5mm] = &\
\bracks{z^{3}}\sum_{x_{1} = 0}^{3}z^{x_{1}}
\sum_{x_{2} = 0}^{3}z^{x_{2}}
\sum_{x_{3} = 0}^{\infty}z^{x_{3}}\sum_{x_{4} = 0}^{\infty}z^{x_{4}} =
\bracks{z^{3}}\pars{z^{4} - 1 \over z - 1}^{2}
\pars{1 \over 1 - z}^{2}
\\[5mm] = &\
\bracks{z^{3}}\pars{1 - z}^{-4} = {-4 \choose 3}\pars{-1}^{3} =
-{-\pars{-4} + 3 - 1 \choose 3}\pars{-1}^{3}
\\[5mm] = &\ {6 \choose 3} = \bbx{20} 
\end{align}
