I faced this problem on one test. I wrote my solution but then I found out that my solution is wrong, I still cannot find where my mistake is.
The problem says: How many three-digits numbers are there such that they are odd and their digits are all different.
Here is my approach:
We have three digits. Since the number should be odd, the last digit should be one of those numbers $1, 3, 5, 7, 9$. Now the second digits can be one of the digits: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ There are 10 different digits for the second digits, but since the digits should be different we cannot place 10 digits, but we can place 9 digits. And for the first digits we can place digits in the range $1...9$ but we cannot place the digits that are used in the two other digits and we can place only 7 digits.
So my result is $7\cdot9\cdot5 = 315$
However the result is not correct, because there are $320$ odd three-digits numbers with different digits.
Can you point me where is my mistake, thanks in advance.