How many odd three-digits numbers are there whose all three digits are different
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.
$\endgroup$ 44 Answers
$\begingroup$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 $5\cdot8\cdot8 = 320$
$\endgroup$ $\begingroup$The mistake is you might have used $0$ for the middle digit. So there are two cases.
Case ($1$): Using $0$ for the middle digit:
$\qquad$The count is $5\,{\times}\,1\,{\times}\,8=40$.
Case ($2$): Not using $0$ for the middle digit:
$\qquad$The count is $5\,{\times}\,8\,{\times}\,7=280$.
So you get a total count of $40+280=320$.
$\endgroup$ 2 $\begingroup$Since this problem is only created by one single digit i.e. zero as you can't put zero on hundred's place like 013 which is not 3 digit number so when you put zero on the ten's place i.e. in the middle you will get 8 remaining digits for hundred place. Therefore we have two cases which are: Case (1): Using 0 for the middle digit.
The count is 5×1×8=40.
Case (2): Not using 0 for the middle digit.
The count is 5×8×7=280.
So you get a total count of 40+280=320.
$\endgroup$ 1 $\begingroup$Three digit odd numbers implies the numbers would only be made of digits 1 , 3 , 5 , 7 , 9
With repetition of digits we would have had 5 * 5 * 5 = 125
But for every hundreds, maximum of 4 tens is possible (avoiding the duplicate of digit used in hundreds)
And each of these 4 tens, maximum of 3 units is possible (avoiding the duplicate of digits used in tens)
5 * 4 * 3 = 60 is my answer
$\endgroup$ 1
