The answer key on the back of Combinatorics A Problem Based Approach by Pavle Mladenovic says it's 512 but I don't think that's right.
Here's my thought process:
A 5 digit number divisible by 5 means the ones digit is 0 or 5
Case last Digit is 5:
1st Digit: (1-9) choices but can't have 5 because it would repeat the last digit so possible choices are 8
2nd Digit: (0-9) choices but can't have 5 and can't have the previous digit so 8 choices
3rd Digit: (0-9) Choices but can't have 5 and the previous digit so 7 choices
4th Digit: (0-9) choices but can't have 5 and previous 3 digits so 6 choices
5th Digit: Must have 5 so 1 choice
Product: 8 * 8 * 7 * 6 * 1 = 2688 possible numbers
Case last Digit is 0:
1st Digit: (1-9) choices so 9
2nd Digit: (0-9) choices but last digit is 0 and cant repeat previous digit so 8 choices
3rd Digit: (0-9) choices but last digit is 0 and can't repeat previous 2 so 7 choices.
4th Digit: (0-9) choices but last digit is 0 and can't repeat previous 3 choices so only 6 choices.
5th Digit Last Digit is 0 so only 1 choice.
Product: 9 * 8 * 7 * 6 = 3024
Total Choices is 2688 + 3024 = 5712.
How do you get 512 from this question?
Best Answer
Let $a_i$ be the $i$th digit of a $5$-digit number divisble by $5$.
If $a_5=5$, then $a_1$ can be $1,\dots,4,6\dots,9$, so there are $8$ choices for $a_1$ in this case.
If $a_5=0$, then $a_1$ can be $1,\dots,9$, so there are $9$ choices for $a_1$ in this case.
Therefore, there are $8+9=17$ chocies for $(a_1,a_5)$ pairs.
Now, $(a_2,a_3,a_4)$ can be any permutation constructed from any digits that are not $a_1$ and not $a_5$.
Therefore, we have to determine the number of permutations of $3$ digits from $10-2=8$ digits, which is $8\cdot 7\cdot 6=336$.
Finally, any $5$-digit number divisible by $5$ can be obtained from any combination of $(a_1,a_5)$ and $(a_2,a_3,a_4)$, and any such combination also results in a $5$-digit number divisible by $5$, so the answer is $17\cdot 336=5712$.