Is there a formula for finding the number of ways 3 dice throws can add to 6? Consider 3 dice throws: t1,t2,t3
Let E be the event that t1+t2+t3 add up to 6
What is the probability of event E?
I know the denominator is an example of ordered repetition. So with n=6 and k=3 there are $n^k$=$6^3$ ways. 
But I can't figure out how to get the numerator, other than by listing all the ways.
[Update]
That is I can list the ways as
{1,1,4},{1,4,1}{4,1,1} 
{1,2,3},{1,3,2}{2,1,3}{2,3,1}{3,1,2}{3,21} 
{2,2,2} 
i.e 3 ways that use a one a one and a four 
plus 6 ways that use a one, a two and a three 
plus 1 way that uses a two, a two and a two 
Is there a pattern that can be used to calculate the ways, other than by listing them?
I can see that the 6 ways is given by $3!$ and the 1 way is given by $1!$
 A: Here's another way to solve this - it's simply the total number of ways you can throw 2, 3, 4, or 5 using two dice. As the number of ways you can throw a total of $n$ with two dice is $n-1$ (for $n \le 7$), this is 1, 2, 3, or 4 respectively, hence $1+2+3+4=10$.
For some trivia - this is the 4th triangular number and the number of ways of throwing 3 to 8 with 3 dice is given by the first 6 triangular numbers. We can get these from the third diagonal of Pascal's triangle:
$$
\begin{array}{c}
\text{total}&&&&&&&&&\rlap{\text{1 throw}}\\
1&&&&&&&&{1}&&\rlap{\text{2 throws}}\\
2&&&&&&&{1}&&{1}&&\rlap{\require{enclose}\enclose{circle}{\text{3 throws}}}\\
3&&&&&&{1}&&{2}&&{1}&&\rlap{\text{4 throws}}\\
4&&&&&{1}&&{3}&&{3}&&{1}\\
5&&&&{1}&&{4}&&{6}&&{4}&&{1}&\rlap{...}\\
\enclose{circle}6&&&{1}&&{5}&&{\enclose{circle}{10}}&&{10}&&{5}&&{1}{}\\
7&&{\require{cancel}\cancel1}&&{6}&&{15}&&{20}&&{15}&&{6}&&{1}\\
8&{\cancel1}&&{\cancel7}&&{21}&&{35}&&{35}&&{21}&&{7}&&{1} \end{array}
$$
The other diagonals give the number of combinations for each total for other numbers of throws. This only works for the first 6 possible totals with each number of throws though, which is why I've crossed out some of the numbers.
