What is the probability of the sum of four dice being 22?
There aren't too many to just count.
Permutations of $6+6+6+4$: $\binom41=4$
Permutations of $6+6+5+5$: $\binom42=6$
These are the only options, so your numerator must be $4+6=10$
The bad combinations criteria is atleast one $x_i \geq 7$.
The number of bad combinations when:
One of $x_i$ is forced to be greater than or equal to $7$ is $\binom{4}{1}\binom{12 + 4 - 1}{12} = 1820$.
Two of $x_i$'s are forced to be greater than or equal to $7$ is $\binom{4}{2}\binom{6+4-1}{6} = 504$.
Three of $x_i$'s are forced to be greater than or equal to $7$ is $\binom{4}{3}\binom{0+4-1}{0} = 4$.
Four of $x_i$'s are forced to be greater than or equal to $7$ is $0$.
So, total bad combinations $= 1820 - 504 + 4 - 0 = 1320$
I used $n(\cup_{i=1}^{4}A_i) = \sum_{i=1}^{4}n(A_i) - \sum_{i,j, i\neq j}n(A_i\cap A_j) + \ldots$
So, possible combinations $= 1330 - 1320 = 10$.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
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}