How many ways to make a $3$ digits even number with only $2,3,5,6,7$ . And no repeated use of digit.

I think I did something like for the first digit you have $5$ choices, for second digit $4$ choices, last digit will only have $2$ choices, because, we need it to be even, so $40$ in total. Out of these $40$ choices, $2$ of them are not possible to obtain: $26x$ or $62x$. I think I'm still getting the wrong answer.

  • $\begingroup$ I should start by observing that the number is even if and only if the units digit is even. Pick the units digit first, and the others can be anything. You don't say whether repeated digits are allowed. $\endgroup$ – Mark Bennet Oct 30 '14 at 11:34
  • $\begingroup$ 24 as @MarkBennet said $\endgroup$ – Bhaskar Vashishth Oct 30 '14 at 11:38
  • 1
    $\begingroup$ Thanks, repeated digits are not allowed. So should be $3 \times 4 \times 2 = 24$? $\endgroup$ – SamC Oct 30 '14 at 11:39

The ones digit can only be either 2 or 6, so that's 2 choices. For the remaining 2 digits, you have 4 numbers to choose from and they can be permuted. Hence it should be:

4P2 (4 permute 2) * 2 = 12 * 2 = 24

You can also write them all out systematically:



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