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How many even 2 digit numbers can be formed from the numbers 3,4,5,6,7? The digits cannot repeat (you can't have 44 or 66 for example). I know the answer to this is 8, because I just wrote them all out and then removed the ones that repeated digits, but I need a way to find this using a formula, like P(n,k) or something. What formula could be used to find this?

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    $\begingroup$ We have $2$ choices for the last digit. For every such choice we have $4$ choices for the first digit. $\endgroup$ – André Nicolas May 28 '15 at 0:23
  • $\begingroup$ I think you should be able to choose an answer? $\endgroup$ – Aditya Agarwal May 28 '15 at 14:20
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There is no formula for this, here. Because you have specified the digits and a way to understand this intuitively is-
$__ __ $
Suppose these dashes are the two digits.
So, in order for a number to be even, the last digit should be $0$ or $2,4,6,8$. So second digit can be $4$ or $6$ from your specified numbers. And the digit cannot be repeated, so there are $3+1=4$ digits left for the first place. So by multiplication principle, $$4\cdot 2=8$$ is the answer.

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Since the two-digit number is even (and you are only allowed to use the digits $3, 4, 5, 6, 7$), its last digit is either $4$ or $6$, so you have two choices for the units digit. For each such choice, you can use any of the other four digits as the tens digit. Therefore, there are $4 \cdot 2 = 8$ two-digit even numbers that can be formed using only the digits $3, 4, 5, 6, 7$.

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  • $\begingroup$ Do you have a way to put that into a formula? I need to show that I solved it using some formula. $\endgroup$ – JesW87 May 28 '15 at 1:20
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    $\begingroup$ Aditya and I used the Multiplication Principle. $\endgroup$ – N. F. Taussig May 28 '15 at 9:46

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