Calculate the number of even three-digit positive integers with no repeated digi

Question

Calculate the number of even three-digit positive integers with no repeated digit by finding the number of such integers that have

a) units digit 0

b) nonzero unit digit

Verify by comparing number of odd 3-digit integers with no repeat digits with the total number of 3-digit positive integers.

Explanation / Answer

a) There are 9 possibilities for the 100-digit. Once it is selected, there are 8 possibilities for the 10-digit. This is because none of them can be 0 (this is the 1-digit) and they have to be different from each other. Total number of such numbers is 9*8=72 b) If the unit digit is not zero, it has 4 possible values: 2,4,6,8, because our number is even. The 100-digit can be any digit but the one used for the units or 0, so it has 8 possibilities. The 10-digit has 8 possibilities as well: it just needs to be different than the other two digits. So, the total number of such numbers is 4*8*8=256 ==> The total number of positive even 3-digit integers with no repeated digit is 72+256=328 -------- Number of odd 3-digit integers is 5 (for unit digit) * 8 (for 100 digit that cannot repeat the unit or be 0) * 8 (for the 10 digit that cannot repeat any of the others) = 5*8*8=320 Total number of 3-digit integers with not repeat: 9*9*8=648 = 320+328

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