I have to find the total number of three-digit even numbers where no digit can be repeated.
I tried and got answer $9 \times 9 \times 5$, but it is wrong. There is something weird with $2$ digits. I mean if an even number is there in the second place, then in the third place $4$ types of even can come, but if odd is there in the second place, then five even numbers can come in the third place. Thanks in advance.
On
The last digit must be even but one of the first two digits we chose may be an even number so it reduces the number of possibilities for the last digit. We will have to break this up into cases.
Case $1$: First digit is odd
Sub case $1_1$: second digit is odd
Sub case $1_2$: second digit is even
In Sub case $1_1$ the number of possibilities for first, second and third digit are 5, 4 and 5. So the total no. of possibilities is $5\times4\times5=100$.
In Sub case $1_2$ the number of possibilities for first, second and third digit are 5,4 and 3. So the total no. of possibilities is $5\times4\times3=60$.
So total no. of possibilities of Case 1 is $100+60=160$.
Case 2: First digit is even
Sub case $2_1$: second digit is odd
Sub case $2_2$: second digit is even
The no. of possibilities of Sub case $2_1$ is $3 \times 4 \times 3=36$. The first number is 3 not 4 because 0 cannot be the first digit.
The no. of possibilities of Sub case $2_2$ is $3 \times 3 \times 2=18$
So the no. of possibilities of case 2 is $36+18=54$
So totally the number of cases is the sum of case 1 and case 2 which is $160+54=214$.
The answer is $\boxed{214}$.
The last digit can be determined by $5$ different ways. If it is $0$, then the two other digits are obtained in $9\times 8$ different cases.
If the last digit is something other than $0$, then the first digit can be determined by $8$ different cases (other than the last digit and $0$) and the middle one in $8$ different cases (excluding the first and last digits). Hence the total cases are $$9\times 8+4\times 8\times 8$$