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.
 A: Since people have already given the answer, to explain my approach here:
We make the analogy of this problem to the system $\begin{cases} x_1+x_2+x_3=11\\ 1\leq x_1\leq 6\\ 1\leq x_2\leq 6\\ 1\leq x_3\leq 6\end{cases}$
Letting $S$ be the set of solutions where we don't care about upperbound conditions, and $A_1,A_2,A_3$ the sets of solutions where the upper bound condition is violated for $x_1,x_2,x_3$ respectively, we want to count
$|S\setminus (A_1\cup A_2\cup A_3)|=|S|-|A_1|-|A_2|-|A_3|+|A_1\cap A_2|+|A_1\cap A_3|+|A_2\cap A_3|-|A_1\cap A_2\cap A_3|$
$|S|$ is the number of solutions to the system $\begin{cases}x_1+x_2+x_3=11\\1\leq x_i\end{cases}$ which is $\binom{10}{2}=45$
$|A_1|$ is the number of solutions to the system $\begin{cases}x_1+x_2+x_3=11\\7\leq x_1\\ 1\leq x_2\\ 1\leq x_3\end{cases}$ which is equivalent to the system $\begin{cases} y_1+y_2+y_3=5\\1\leq x_i\end{cases}$ which has $\binom{4}{2}=6$ solutions.  Similarly $A_2$ and $A_3$
The intersection of two or more in this specific problem results in zero solutions since it is impossible for two or more people to roll a $7$ or more each and the sum be $11$.
We get the total is then $\binom{10}{2}-3\binom{4}{2}=45-3\cdot 6 = 45-18=27$

Note: This method easily generalizes to any dice sum, any number of sides to the dice, and any number of people and completely avoids "brute force" which will become horribly tedious with larger number of people and dice sums.
A: 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}
A: 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.
A: An easier way to solve this would probably be as follows:
If Person 1 gets a 1, the Person 2 & 3 need a combined 10, which can be done in 3 ways.
If Person 1 gets a 2, the Person 2 & 3 need a combined 9, which can be done in 4 ways.
If Person 1 gets a 3, the Person 2 & 3 need a combined 8, which can be done in 5 ways.
If Person 1 gets a 4, the Person 2 & 3 need a combined 7, which can be done in 6 ways.
If Person 1 gets a 5, the Person 2 & 3 need a combined 6, which can be done in 5 ways.
If Person 1 gets a 6, the Person 2 & 3 need a combined 5, which can be done in 4 ways.
Adding this up, there are 27 total ways to get 11.
A: My personal approach involves not calculus, but combinatorics and an intuitive definition of dice rolling. This is a slightly more rigorous way to follow browngreen's method that he beat me to the punch with. 
To start, find the probabilities of rolling a total sum T with two people each rolling a normal die: 
$T \leq 1$ - Impossible, as I hope you can see, one person has to roll a negative number or zero, which is impossible on a normal dice.  
$T = 2$ - This is only possible if the first dice roll is a 1 (1/6 chance), and the second roll is a 1 (1/6 chance). Therefore $P_{2}(2)=1/36$
$T = 3$ - This is only possible if the first dice roll is a 1 or 2 (2/6 chance), and the second roll is a specific number (2 or 1 respectively in this case) (1/6 chance). Therefore $P_{2}(3)=2/36$. 
The pattern is clear. 
$T=4$ has a probability of $3/36$ and so on till $T=8$ when the first dice must be $at$ $least$ 2 and so the pattern goes downward with $P_{2}(8)=5/36$
Therefore: $P_{3}(11) = P_{1}(1)P_{2}(10)+P_{1}(2)P_{2}(9)+P_{1}(3)P_{2}(8)+P_{1}(4)P_{2}(7)+P_{1}(5)P_{2}(6)+P_{1}(6)P_{2}(5) = 27/216$
Therefore, 27 out of 216 ways to roll three dice end with a sum of 11. 
Three people rolling one normal dice each can end with a sum of 11 27 out of 216 different ways. 
