Number of ways in which 3 people can throw a normal die to have a total score of 11 
Number of ways in which 3 people can throw a normal die to have a total score of 11      

My approach: 
The answer can be obtained by finding the coefficient of $x^{11}$ in the expansion of $(x+x^2+x^3+x^4+x^5+x^6)^3$. 
General term $T=\frac{3}{a!b!c!d!e!f!} x^{a+2b+3c+4d+5e+6f}$
where $a+b+c+d+e+f=3$ and $a+2b+3c+4d+5e+6f=11$
But I can't figure out how to solve further.

The approach using generating functions is also fine. It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series.

We obtain
  \begin{align*}
\color{blue}{[x^{11}](x^1+x^2+x^3+x^4+x^5+x^6)^3}&=[x^{11}]x^3(1+x+x^2+x^3+x^4+x^5)^3\\
&=[x^8]\left(\frac{1-x^6}{1-x}\right)^3\tag{1}\\
&=[x^8](1-x^6)^3\sum_{n=0}^\infty\binom{-3}{n}(-x)^n\tag{2}\\
&=[x^8](1-3x^6)\sum_{n=0}^\infty\binom{n+2}{2}x^n\tag{3}\\
&=([x^8]-3[x^2])\sum_{n=0}^\infty\binom{n+2}{2}x^n\tag{4}\\
&=\binom{10}{2}-3\binom{4}{2}\tag{5}\\
&=45-3\cdot6\\
&=\color{blue}{27}
\end{align*}

Comment:


*

*In (1) we use the geometric series expansion and apply the rule
\begin{align*}
[x^p]x^qA(x)=[x^{p-q}]A(x)\tag{6}
\end{align*}

*In (2) we apply the binomial series expansion.

*In (3) we expand $(1-x^6)^3$ and observe we do only need two terms of it since we are looking for powers up to $x^8$. We also use the binomial identity
\begin{align*}
\binom{-p}{q}=\binom{p+q-1}{p-1}(-1)^q
\end{align*}

*In (4) we apply the rule (6) twice again.

*In (5) we select the coefficient of $x^8$ and $x^2$ accordingly.
