Counting up to a sum. I understand the common and simple counting problems, those involving baseball line ups and poker hands, but this one has me a little puzzled.

How many distinct $6$-tuples of positive integers $(n_1, ... , n_6)$ are
  there such that $n_1 + ... + n_6 = 17$?

First, it seems obvious every number must be $1 \leq N \leq 12$. Because every number must be positive, and the max only happens when the other numbers are at their lowest, $1$, therefore the max is $17 - 5 = 12$.
My first instinct is to say that something along the lines of $12! + 11! + ... + 7!$. I have no logical path that led me to this point, so I suppose my question is, How do I approach these problems, and further, am I on the right track with this one?
 A: Think of this like distributing 17 identical balls into 6 compartments. As all the integers must be positive, place one ball in each compartment; we are then left with 11. The general expression for distributing $n$ balls into $r$ compartments (where a compartment may be empty) is $^{n+r-1}C_{n}$. The answer to your problem is therefore $^{16}C_{11}$.
A: You want a total of 17 distributed among 6 positive integer. This is a partition problem. For example, one partition would be
X | X | X | X | X | X X X X X X X X X X X X
corresponding to
1 + 1 + 1 + 1 + 1 + 12 = 17.
Another would be
X X | X X X | X X | X | X X X X X | X X X X
corresponding to
2 + 3 + 2 + 1 + 5 + 4 = 17
The 5 partitions must be in different places between the X's because the integers must be positive. The order in which the partitions are chosen is irrelevant. You have to choose 5 distinct partition positions from among the 16 possible partition positions. Because order is unimportant in choosing the partition positions, this is "16 choose 5" corresponding to 17 and 6. The answer
$$
   \frac{(16)(15)(14)(13)(12)}{(1)(2)(3)(4)(5)}=\frac{16!}{(16-5)!5!}={{16}\choose{5}}
$$
