<s>[INST] A person rolls five fair dice simultaneously. Find the probability that the total number of spots showing is less than or equal to $11$ I am having trouble with this problem. When they say spot I think they are essentially saying the sum, so its the probability that the sum of dice is $11$ or less.
I understand that there are $6^5$ combinations.
I found 6 ways that it can equal to $11$ $(2,3,2,2,2)(3,3,1,1,3),(4,4,1,1,1),(5,2,2,1,1),(6,2,1,1,1)(7,1,1,1,1)$, but I know there has to be an easier way than just counting. Is it like $\binom{6}{5}$? Thanks for the help.
 [/INST] We can solve the problem using generating function.
\begin{align}\left(\frac16\sum_{i=1}^6x^6\right)^5=\frac{1}{6^5}\left( \frac{x(1-x^6)}{1-x}\right)^5 \end{align}
Our goal is to find the sum of coefficients of $x^5$ to  $x^{11}$.
$$(1-x^6)^5 =1-5x^6+\text{higher order terms} $$
By negative binomial series
$$(1-x)^{-5}=\sum_{k=0}^\infty \binom{4+k}{k}x^k$$
Hence, 
\begin{align}\left(\frac16\sum_{i=1}^6x^6\right)^5=\frac{1}{6^5}x^5(1-5x^6+\text{higher order terms})\sum_{k=0}^\infty \binom{4+k}{k}x^k\end{align}
and the sum of coefficients is 
$$\frac1{6^5}\left(\sum_{k=0}^{6}\binom{4+k}{k}-5\binom{4+0}{0} \right)=\frac1{6^5}\left(\sum_{k=0}^{6}\binom{4+k}{4}-5 \right)=\frac{462-5}{6^5}=\frac{457}{7776}$$
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