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.
Best Answer
Problems of this type are often solved more simply if we place the digits in a suitable order, instead of necessarily left to right.
Here we'll place the digits in the order last-first-second, i.e. in the order $Units-Hundreds-Tens$
$5$ choices for the last digit (to make the number odd)
$8$ choices left for the first digit (since $0$ can't be used)
$8$ choices still left for the second digit (since $0$ is now available for use)
Thus, without breaking up into cases, we get $5\cdot8\cdot8 = 320$