$5$ kids toss one die each. Find the number of ways that the sum of dice points is $22$. I have already done these following steps and I'm already lost. Can someone please help me?

$x_1 + x_2 + x_3 + x_4 + x_5 = 22$
  when $1 \leq x_n \leq 6$ and $n = 1,2,3,4,5$.
  Then, 
  \begin{align*}
f(x) & = (x+x^2+x^3+x^4+x^5+x^6)^5\\
& = \left(\frac{x(1-x^6)}{1-x}\right)^5\\
& = x^5 \cdot (1-x^6)^5 \cdot (1-x)^{-5}
\end{align*}

 A: As with the other answers, here is an expansion of the generating function.
$$
\begin{align}
x^5\left(\frac{1-x^6}{1-x}\right)^5
&=x^5\sum_{j=0}^5\binom{5}{j}\left(-x^6\right)^j\sum_{k=0}^\infty\binom{-5}{k}(-x)^k\tag{1}\\
&=\sum_{j=0}^5\sum_{k=0}^\infty(-1)^{j+k}\binom{5}{j}\binom{-5}{k}x^{k+6j+5}\tag{2}\\
&=\sum_{j=0}^5\sum_{k=0}^\infty(-1)^j\binom{5}{j}\binom{k+4}{k}x^{k+6j+5}\tag{3}\\
&=\sum_{j=0}^5\sum_{k=6j+5}^\infty(-1)^j\binom{5}{j}\binom{k-6j-1}{k-6j-5}x^k\tag{4}\\
&=\sum_{k=5}^\infty\color{#C00}{\sum_{j=0}^{\left\lfloor\frac{k-5}6\right\rfloor}
(-1)^j\binom{5}{j}\binom{k-6j-1}{4}}x^k\tag{5}
\end{align}
$$
Explanation:
$(1)$: Binomial Theorem
$(2)$: combine and rearrange
$(3)$: $\binom{-5}{k}=(-1)^k\binom{k+4}{k}$
$(4)$: $k\mapsto k-6j-5$
$(5)$: change order of summation and $\binom{n}{k}=\binom{n}{n-k}$
For $k=22$, we get from the red part of $(5)$:
$$
\begin{align}
\sum_{j=0}^2
(-1)^j\binom{5}{j}\binom{21-6j}{4}
&=\binom{5}{0}\binom{21}{4}-\binom{5}{1}\binom{15}{4}+\binom{5}{2}\binom{9}{4}\\[6pt]
&=420
\end{align}
$$
