Question 765170
<pre>
The smallest digit must be 1 because otherwise
the smallest possibility would be 2345 whose sum of digits
is 14, which is too much.
 
The next to smallest digit must be 2 because otherwise
the smallest possibility would be 1345 whose sum of digits
is 13, also too much.

Since the smallest two digits are 1 and 2, which have sum 3,
the remaining two digits must sum to 12-3, or 9. So the
only possibilities for them are 3 and 6 or 4 and 5.

So the set of 4 different digits is either {1,2,3,6} or {1,2,4,5}

There are 4! = 4·3·2·1 = 24 permutations of each of those,

so the answer is 2×24 or 48.

Edwin</pre>