Edit #1: Remember that a, b, and c are positive digits, so their values are between 1 & 9, inclusive
I was thinking casework for this problem, but I got stuck on the last part. I can't figure out whether there are four cases or three.
$20 = 2^2 * 5$, so you need two 2's and a 5.
Case 1: One five, either a four or an eight, and an odd number that is not five
Case 2: Two fives and either a four or an eight
Here's where I get confused. I'm not sure whether it should be:
Case 3: One five and two even numbers
Case 3: One five and two even numbers that are different
Case 4: One five and two even numbers that are the same
Here are the calculations that I have done so far:
Case 1: $2* 4 * 3!$, as the five is fixed, you can choose between the 4 & 8, and four choices for the odd number that is not five. The $3!$ accounts for the various permutations of the three that are chosen.
Case 2: $2 * 3$. The fives are fixed, you choose between the 4 & 8, and there are three different ways to arrange them after you choose your numbers.
Edit #2: I also think that the answer is $102$. I 'cheated' (used Python and wrote a script) to figure that part out, but I want to figure out the answer with math, not programming.
Thank you in advance!