How many four digit numbers have no repeat digits, do not contain zero, and have a sum of digits equal to 28?

How many four digit numbers have no repeat digits, do not contain zero, and have a sum of digits equal to 28?

There is given 5 possible answers - one of these contain the right answer:

A. 14

B. 24

C. 28

D. 48

E. 96

• But I don't know how to handle this question. I find it very difficult.
• Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you write what your thoughts are on the problem and include your efforts (work in progress) in this and future posts and in what context you have encountered the problem; this will prevent people from telling you things you already know, and help them give their answers at the right level. – JKnecht Feb 15 '16 at 15:49
• Hi JKnecht. Thanks for your comment :) As stated in the topic, I don't really know how to handle this question. I find it really difficult. – Martin Andersen Feb 15 '16 at 15:50
• Hint: Start by writing 28 as a sum of four non-zero one-digit numbers. – martini Feb 15 '16 at 15:52

Let's start writing 28 as a sum of four different non-zero one-digit numbers. As $28 = 4 \cdot 7$, and the numbers must be different, we start be replacing two 7s by a six and an eight, giving $28 = 6+7+7+8$, by replacing further we get $28 = 5+6+8+9$. Note that we cannot increase the $8$ or $9$ more, the only thing we can do is replacing $5,6$ by $4,7$, i. e. $$28 = 5+6+8+9 = 4+7+8+9$$ Each decomposition of $28$ gives us $4! = 24$ numbers, hence in total there are $2 \cdot 24 = 48$ of them.
• Why not $9,8,7,4$? – GoodDeeds Feb 15 '16 at 15:51