# 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

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integer solution of this problem is infinite – dato datuashvili Sep 26 '12 at 19:49
it is better to introduce positive space – dato datuashvili Sep 26 '12 at 19:49
yeah sorry, positive solutions. – Umar Jamil Sep 26 '12 at 19:52
Note 1: My Answer is same as: (7−1 3−1 ) Note 2: 0 is a netural int so it does not count. Note 3: answer below is only for possitive int's. – Dotanooblet Sep 26 '12 at 20:37

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.

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– dato datuashvili Sep 26 '12 at 19:55
ok I'm gonna read it – Umar Jamil Sep 26 '12 at 19:56
Thank you, that's how the professor did the problem in class! (the "stars and bars" stuff)... Thank you! – Umar Jamil Sep 26 '12 at 19:57

$$5 \; 1 \; 1\\ -----\\ 4 \; 2 \; 1\\ 4 \; 1 \; 2\\ -----\\ 3 \; 3\; 1\\ 3 \; 2 \; 2\\ 3 \; 1 \; 3\\ -----\\ 2 \; 4 \; 1\\ 2 \; 3 \; 2\\ 2 \; 2 \; 3\\ 2 \; 1 \; 4\\ -----\\ 1 \; 5 \; 1\\ 1 \; 4 \; 2\\ 1 \; 3 \; 3\\ 1 \; 2 \; 4\\ 1 \; 1 \; 5$$

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.

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very instructive :) – enzotib Sep 26 '12 at 20:33
didn't consider zero. – Umar Jamil Sep 26 '12 at 20:34
Very interesting and tough question. This was not a trivial question at all!!! Despite having PHD in Statistics still I was not able to know why answer is (7-1 3-1) without looking at theorems solution! :( – Dotanooblet Sep 28 '12 at 15:42

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.

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I can sort of see where you are coming from, but is it not the case that we learn about mathematical methods by first looking at cases that we could work out, and then once an understanding has been obtained, we may proceed to look at the more general setting. Motivational questions such as these - in my humble opinion at least - are of fundamental importance to a students' understanding of a subject. If a student can gain an intuition behind the maths at an early stage, then surely this can only be of benefit to them. – David Ward Sep 27 '12 at 7:54
What an arrogant and patronising answer. Just because you went to Trinity College Cambridge doesn't make you better than everyone else. – user50229 Dec 18 '12 at 16:58