Question

Consider all of the positive integers with exactly 8 digits. How many of them are even and have all 8 digits distinct?

Answer 1

For the number to be even, the last digit has to be one of the following: 0, 2, 4, 6, 8 (five possibilities). Having chosen the last digit, the digit before that has 9 possibilities if no numbers are to be repeated. The digit before that has 8 possibilities, ....
(the first digit cannot be a zero as that would make it less than an eight digit number).

Answer 2

Hold on... I have a slightly better approach.

Answer 3

Yes I am at that point. However, the last number might not contain 5 possibilities because an even number could have been used in the previous 7 numbers. That's where I'm stuck because I could technically only have 2 or 3 possibilities for my final digit.

Answer 4

Assume the last digit (the eight digit) is a 0.
The first digit has 9 possibilities
The second digit has 8 possibilities
The third digit has 7 possibilities
The fourth digit has 6 possibilities
The fifth digit has 5 possibilities
The sixth digit has 4 possibilities
The seventh digit has 3 possibilities
Total: 9x8x7x6x5x4x3

Assume the last digit is NOT a 0.
The last digit has 4 possibilities because it has to be an even number.
The first digit has 8 possibilities (no zero and no repeat of the last digit)
The second digit has 8 possibilities (can't repeat the first or the last digit)
The third digit has 7 possibilities
The fourth digit has 6 possibilities
The fifth digit has 5 possibilities
The sixth digit has 4 possibilities
The seventh digit has 3 possibilities
Total: 4x8x8x7x6x5x4x3

Add up the two totals.

Answer 5

That's the right answer. Thank you so much!!

Answer 6

You are welcome.