What is the number of combinations of the solutions to $a+b+c=7$ in $\mathbb{N}$? My professor gave me this problem: 

Find the number of combinations of the integer solutions to the equation $a+b+c=7$ using combinatorics.

Thank you.
UPDATE
Positive solutions
 A: If N means natural numbers then you need to be clear whether you mean positive or non-negative integers.
In either case, you can use a stars and bars calculation to get ${7-1 \choose 3-1}$ or ${7+3-1 \choose 3-1}$ respectively
It is easier to explain the non-negative case: you have seven stars and two bars (separators) which you can place in any order to get a solution such as:
$$****|**|*$$
so with $9$ elements which can vary.
For the positive case, you must start with a star and the separators are a bar followed by a star such as:
$$(*)***\underline{|*}*\underline{|*}$$
so with $6$ elements which can vary.
A: $$5 \; 1 \;  1\\
-----\\
4  \;  2 \;  1\\
4  \;  1  \; 2\\
-----\\
3  \;  3\; 1\\
3  \;  2  \; 2\\
3  \;  1  \; 3\\
-----\\
2 \;   4  \; 1\\
2 \;   3  \; 2\\
2 \;   2  \; 3\\
2 \;   1 \;  4\\
-----\\
1 \;   5 \;  1\\
1  \;  4 \;  2\\
1  \;  3 \;  3\\
1  \;  2 \;  4\\
1  \;  1  \; 5$$
Note 1: My answer is the same as:
$$7−1 \choose 3−1   $$
Note 2: 0 is a netural int so it does not count.
Note 3: answer above is only for possitive integers.
A: Is this a hoax?
Perhaps I should put that differently. What institution are you studying at?
Was that the whole question? Or was there a part two asking the same thing but =702 or some other rather bigger number?
I ask these questions because the question as posed is absolutely trivial. The only tricky point is deciding what "integers" means. If he/she really wrote just "integers" then your professor is either dumb or careless, because the answer is obviously infinite. If he didn't mean that, then it is ambiguous, as Henry pointed out.
Second, assuming he meant natural numbers, he is a bad setter of questions. Who wants to "use combinatorics" when they can just list the solutions, as Dotanooblet did! Well, if you have it at your fingertips, combinatorics is slightly faster, but in an exam I would list (if "use combinatorics" was not in the question), because it requires less thought allows one to mull over the other questions at the same time.
Third, the approach set out by Henry above is absolutely standard bookwork. There is nothing remotely tricky or hard about it. 
