Since there are no restrictions on the three-digit numbers (for example no repetitions), we actually don't care about the values of the digits. All we need to know is that there are
[tex] 5+4+3+2 = 14 [/tex]
possible digits, and that we have to extract a triplet from here.
For problems like this we have the binomial coefficient, defined as
[tex] \displaystyle \binom{n}{k} = \dfrac{n!}{k!(n-k)!},\qquad 0\leq k \leq n [/tex]
This number tells you how many subsets of k elements you can extract from a set of n elements. So, in your case, you want to compute
[tex] \displaystyle \binom{14}{3} = \dfrac{14!}{3!11!} = \dfrac{14\cdot 13 \cdot 12}{3\cdot 2} = 14\cdot 13 \cdot 2 = 364[/tex]
