determine the number of even six-digit integers (no leading zeroes) in which no digit may be repeated
_ _ _ _ _ _
you have six places
for an even number, the end place must end in 2,4,6,8
therefore there are four numbers that can go there
place the zero. number of ways to place zero is 4C1 = 4
once that number goes there, there are only 8 numbers left to place
in the second last place, 8 numbers to place
in the third last place, 7 numbers
in the fourth last place, 6 numbers
in the fifth last place, 5 numbers
in the sixth last place, 4 numbers
total 4 x 5 x 6 x 7 x 8 x 4 x 4 = 107520
I think....
The last digit can be five numbers - 0,2,4,6,8. Let's split this into two cases:
(i) The last digit is 0. Then, we need to palce five of the remaining 9 digits in the 5 spots; there are $\displaystyle \binom{9}{5}\cdot 5! = 15120 ~ ~ (*)$ ways of doing that.
(ii) The last digit is not 0. Then, we can put 8 digits in the first digit (since it can not be 0 and it can also not be the one we placed in the last digit), and now we only need to pick 4 of the 8 remaining digits. There are $\displaystyle 4 \cdot 8 \cdot \binom{8}{4}\cdot 4! = 53760 ~ ~ (**)$ ways of doing that.
Overall, we get: $\displaystyle (*) + (**) = 15120 + 53760 = 68880$