What is the probability of the sum of four dice being 22? Question

Four fair six-sided dice are rolled. The probability that the sum of the results being $22$ is $$\frac{X}{1296}.$$ What is the value of $X$?

My Approach
I simplified it to the equation of the form:

$x_{1}+x_{2}+x_{3}+x_{4}=22, 1\,\,\leq x_{i} \,\,\leq 6,\,\,1\,\,\leq i \,\,\leq 4 $

Solving this equation results in:
$x_{1}+x_{2}+x_{3}+x_{4}=22$
I removed restriction of $x_{i} \geq 1$ first as follows-:
$\Rightarrow x_{1}^{'}+1+x_{2}^{'}+1+x_{3}^{'}+1+x_{4}^{'}+1=22$
$\Rightarrow x_{1}^{'}+x_{2}^{'}+x_{3}^{'}+x_{4}^{'}=18$
$\Rightarrow \binom{18+4-1}{18}=1330$
Now i removed restriction for $x_{i} \leq 6$ , by calculating the number of bad cases and then subtracting it from $1330$:
calculating bad combination i.e $x_{i} \geq 7$
$\Rightarrow x_{1}^{'}+x_{2}^{'}+x_{3}^{'}+x_{4}^{'}=18$
We can distribute $7$ to $2$ of $x_{1}^{'},x_{2}^{'},x_{3}^{'},x_{4}^{'}$ i.e$\binom{4}{2}$
We can distribute $7$ to $1$ of $x_{1}^{'},x_{2}^{'},x_{3}^{'},x_{4}^{'}$ i.e$\binom{4}{1}$ and then among all others .
i.e
$$\binom{4}{1} \binom{14}{11}$$
Therefore, the number of bad combinations equals $$\binom{4}{1} \binom{14}{11}  - \binom{4}{2}$$
Therefore, the solution should be:
$$1330-\left( \binom{4}{1} \binom{14}{11} - \binom{4}{2}\right)$$
However, I am getting a negative value. What am I doing wrong?
EDIT
I am asking for my approach, because if the question is for a larger number of dice and if the sum is higher, then predicting the value of dice will not work.

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

The number of configurations that satisfies "the sum is $\ds{22}$" is given by: 

\begin{align}
X & =
\sum_{d_{1} = 1}^{6}\sum_{d_{2} = 1}^{6}\sum_{d_{3} = 1}^{6}\sum_{d_{4} = 1}^{6}
\bracks{z^{22}}z^{d_{1} + d_{2} + d_{3} + d_{4}} =
\bracks{z^{22}}\pars{\sum_{d = 1}^{6}z^{d}}^{4} =
\bracks{z^{22}}\pars{z\,{z^{6} - 1 \over z - 1}}^{4}
\\[5mm] & =
\bracks{z^{18}}{1 - 4z^{6} + 6z^{12} - 4z^{18} + z^{24}\over \pars{1 - z}^{4}}
\\[5mm] & =
\bracks{z^{18}}\pars{1 - z}^{-4} - 4\bracks{z^{12}}\pars{1 - z}^{-4} + 6\bracks{z^{6}}\pars{1 - z}^{-4} - 4\bracks{z^{0}}\pars{1 - z}^{-4}
\\[5mm] & =
{-4 \choose 18}\pars{-1}^{18} - 4{-4 \choose 12}\pars{-1}^{12} +
6{-4 \choose 6}\pars{-1}^{6} - 4 =
{21 \choose 18} - 4{15 \choose 12} + 6{9 \choose 6} - 4
\\[5mm] & =
1330 -4 \times 455 + 6 \times 84 - 4 = \bbx{10}
\end{align}
