<s>[INST] 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.
 [/INST] Here's a pen-and-paper approach.
First, let's compute the number of ways to throw $2$ through $12$ with two dice.
\begin{eqnarray}
N(2)&= 1 &: \{1+1\}\\
N(3)&= 2 &: \{1+2,2+1\}\\
N(4)&= 3 &: \{1+3,2+2,3+1\}\\
N(5)&= 4 &: \{1+4,2+3,3+2,4+1\}\\
N(6)&= 5 &: \{1+5,2+4,3+3,4+2,5+1\}\\
N(7)&= 6 &: \{1+6,2+5,3+4,4+3,5+2,6+1\}\\
N(8)&= 5 &: \{2+6,3+5,4+4,5+3,6+2\}\\
N(9)&= 4 &: \{3+6,4+5,5+5,6+4\}\\
N(10)&= 3 &: \{4+6,5+5,6+4\}\\
N(11)&= 2 &: \{5+6,6+5\}\\
N(12)&= 1 &: \{6+6\}\\
\end{eqnarray}
Then for three dice, these are all the possibilities, where $N(11-d)$ is the case where the first person rolls $d$.
\begin{eqnarray}
N(11-1) &+& N(11-2) + N(11-3) + N(11-4) + N(11-5) + N(11-6)\\
&=& N(10) + N(9) + N(8) + N(7) + N(6) + N(5)\\
&=& 3+4+5+6+5+4\\
&=& 27
\end{eqnarray}
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