How many ordered triples (a, b, c) exist if a, b, and c are positive digits and their product is divisible by 20? 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
or:
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!
 A: We can have two $5$ and one from $\{4,8\}$; gives $2\cdot3=6$ triples.
We can have one $5$ and two from $\{2,4,6,8\}$. We can chose two different ones in ${4\choose2}=6$ ways, and this gives $6\cdot3!=36$ triples of three different digits. When we choose the same even number two times we obtain $4\cdot3=12$ more triples.
We can have one $5$, one from $\{4,8\}$ and one from $\{1,3,7,9\}$. This gives $2\cdot4\cdot3!=48$ triples.
In all there appear $6+36+12+48=102$ admissible triples.
A: Okay, If any of them are $0$ the product is $0$ which is divisible by $20$.
There are $1$ way to have all three being $0$.  $3*9*9$ ways for there to be exactly $1$ zero (3 places to put the zero and $9*9$ choices for the other 2).  There $3*9$ ways for there to be $2$ zeros.
Otherwise you need 5.  You can have 5, 4 or 8, and an odd.  There are $3$ places to put the $5$, there are two remaining places to put the $4$ or $8$. So that is a total of $5*2*2 = 20$ ways.
Or you can have a $5$ and two different evens.  There are $3$ places to put the $5$ and there are $4$ options for the first even and $3$ for the second even.  So there are $3*4*3 = 24$ ways to do that.
Or you can have a $5$ and two equal evens.  There are $3$ places to put the $5$ and there $4$ options for the two equal evens.  That is $5*4 = 20$ ways.
So in total there are $20 + 24+20 + 3*9*9 + 3*9 + 1$ ways to do this. 
A: This is assuming that by positive digit you mean a number in {0, 1, 2,...,9} since that wouldn't give an infinite answer.
Fix 3 digits - 1 unique unordered triple
$(2, 2, 5)$  
Fix 2 digits - 9 + 8 unique unordered triples
$(4, 5, x) \text{ for }x={1,...,9}\\(5, 8, x)\text{ for }x={1,2,3,5,...,9}$
One less in the second to avoid double counting $\because(5,8,4) = (4,5,8)$ in my construct
Fix 1 digit
I claim these are already counted
--> Case 1: choose digit in 2, 4, 5, 8. Then we must choose another digit to ensure divisibility by 20. If divisible, then we are in the second section. If not divisible, we must choose the third digit to ensure divisibility. Then we must be in the first section.
--> Case 2: choose digit not in 2, 4, 5, 8. Then we must fix the other two to ensure divisibility by 20 as in the second section and thus is already counted.
edit: I screwed up the combinations for the final total, whoops
A: Starting over since @fleablood's answer doesn't add up when disregarding the 0's...
The correct answer is 102 as confirmed by my and OP's scripts. 
Let $(a, b, c)$ be our triple with $a,b,c\in{1,...,9}$
$20 = 2^2(5)$. So $5$ must be one element of the triple.
WOLOG set $a=5$. Then $20\mid abc \implies 4\mid bc$.
Thus $bc$ is even, and so at least one of $b$ or $c$ is even.
WOLOG say $b$ is even: 
Cases
1. $(5, b, b)$
2. $(5, b, c)\text{ where }c\text{ is even and } c \ne b$
3. $(5, b, c)\text{ where }c\text{ is odd}$
$\rightarrow$However, $c$ odd means $4\not\mid c$ and together with $4\mid bc$ from above, have $4\mid b$. So $b=4$ or $b=8$.
Counts
1. 3 positions for $5$. $b\in \lbrace 2,4,6,8\rbrace$ means 4 choices for $b$, will fill other 2 slots at once.
$\implies 3*4 = 12$
2. 3 positions for $5$. Fix $b=2$, 3 choices for $c$, 2 perms. Fix $b=4$, 2 choices for $c$, 2 perms. Fix $b=6, c=8$, 2 perms. Can't fix $b=8$ as it's already counted.
$\implies 3*2*(3+2+1+0) = 42$
3. 1 choice for $5$. 2 choices for $b$. 5 choices for $c$.
$\implies 6*(1*2*5) = 60$
But these add to $112 > 102$, which is the correct answer. I'm lost.
Other thing I know: If $a,b,c$ are all distinct, then there are $3!=6$ permutations of (a, b, c). 
I feel like I should be asking the question, not answering. Please someone help, I got sniped hard by this and my combinatorics is weak.
