Digits: 1, 2, 3, 4, 5

Q: how many even 3 digit numbers can be made without repeating them?

In total, I worked out that there's 60 three digit numbers that can be made without repeating (5C1 x 4C1 x 3C1) = 60.

But, I have no idea about the even bit. Could somebody talk me through it so I can understand?


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    $\begingroup$ the title of the question is malformed just a hint. $\endgroup$ – user451844 Sep 6 '17 at 13:35

For a number to be even it must end in a even digit $2$ or $4$ so $2$ choices. The second digit can be any of the $4$ digits which wasn't used as the last digit, and the first digit can be any except those $2$ which were used as the second and third digit hence $3$ choices, all in all we get $2\cdot 4\cdot 3=24$.

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  • $\begingroup$ Thanks! Could you clarify something? For digit one, I pick one of the even numbers (e.g. 2). For digit 2, I can pick any digit that I didn't just use? So, 1, 3, 4 or 5? (e.g. 5). And for the third digit, I use any that I haven't used previously? I.e. 1, 3, 4? Because in this situation, 2 x 5 x 4 = 40, which is different to 24. $\endgroup$ – user477985 Sep 6 '17 at 13:55
  • $\begingroup$ @MathsHelp That's right, though keep in mind that digit one is the last digit (or the first one from right). How do you get 2 x 5 x 4? You have $2$ choices for the first(last) digit and then 4 digits for the second and 3 for the last. $\endgroup$ – kingW3 Sep 6 '17 at 14:00
  • $\begingroup$ Oh, sorry, I think I misinterpreted your words. I thought you meant from my set of numbers that I could literally pick any. I get it now, thanks! If I were to change the question slightly and I had 1, 2, 3, 4, 5 and 6, would it be 3 x 5 x 4 instead of 2 x 4 x 3? $\endgroup$ – user477985 Sep 6 '17 at 14:05
  • $\begingroup$ @MathsHelp Exactly it would be $3\times 5\times 4$. $\endgroup$ – kingW3 Sep 6 '17 at 14:07
  • $\begingroup$ just to be sure, 24 is the final answer isn't it? You don't deduct this from 60 to get 36 as the final answer? $\endgroup$ – user477985 Sep 13 '17 at 5:29

kingW3 answer is absolutely correct. I'm not debating on his answer, just answering in the way the question has been asked.

  1. Selecting either 2 or 4 from the Set for the unit place as the no. should be even: 2C1

  2. Selecting the digit from the left over digits for the tens place: 4C1

  3. Selecting the digit from the left over digits for the hundreds place: 3C1

Therefore we get, 3C1 x 4C1 x 2C1 = 24

The thing you have done 5C1 x 4C1 x 3C1 would have been correct if you were asked to make any 3 digits no from the given set without repeating.

I apologize if the repeating of an answer is not allowed.

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