TITLE: Ways to add up 10 numbers between 1 and 12 to get 70
QUESTION [5 upvotes]: I know this has something to do with factorials, and combinations and permutations. I've been puzzling over this for a little while, and I can't come up with an answer. My question is, How would one figure out how many ways there are to use exactly 10 numbers from 1 to 12 to sum 70. Order is important. Therefore permutations not combinations. Thank You in advance! :D
Edit: They are not necessarilly distinct numbers as in 12,12,12,12,12,2,2,2,2,2 would work.

REPLY [8 votes]: Generating functions would work, I think.
Consider the product 
$$(x + x^2 + \ldots + x^{12})^{10}$$
Then the coefficient of $x^{70}$ in the resulting polynomial is your answer: we have to pick a power from each factor (and we could take the same power several times, in different factors) such that their sum equals 70, and order does matter.
taking out the $x$ from each term we need the coefficient of $x^{60}$ in 
$$(1 + x + \ldots + x^{11})^{10}$$ to get the answer.
We write the term as $\frac{1-x^{12}}{1-x}$, so we need the coefficient of $x^{60}$ in 
$$(1 - x^{12})^{10} (1-x)^{-10}$$ and we can expand the latter using the (generalized) binomial theorem twice:
$$\left(\sum_{k=0}^{10} \binom{10}{k} (-1)^{k}x^{12k}\right)\left(\sum_{k=0}^{\infty} \binom{9+k}{k} x^k\right)\mbox{.}$$
To get $x^{60}$, there are several ways: in the left hand sum we have the powers $1,x^{12},\ldots,x^{60}$ to consider with their coefficients $1,-\binom{10}{1},\binom{10}{2},\ldots ,-\binom{10}{5}$ that we multiply with their complementary powers $x^{60},x^{48},\ldots,x^0$ and their coefficients $\binom{69}{9},\binom{57}{9},\ldots,1$ respectively. 
So we get as answer $$\binom{69}{9} - \binom{10}{1}\binom{57}{9} + \binom{10}{2}\binom{45}{9} - \binom{10}{3}\binom{33}{9} + \binom{10}{4}\binom{21}{9} - \binom{10}{5}$$
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