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Question: I was wondering how many $3$-Digits Even Numbers can be formed from $2,4,5$ ?

My Approach: If we take even number $2$ at extreme right _ _ $2$ then 2 permutations can formed from remaining numbers i.e ($4,5$), similarly if we take $4$ at extreme right _ _ $4$ then again 2 permutations will be formed. So,

$2!+2!=6$

Is my answer correct? because according to my teacher it should be 18.

Conclusion: Help will be highly appreciated and i want the clear method to solve questions of this kind in which any number digits either even or odd can be formed using any given numbers.

P.S.(Sorry for Bad English)

Thanks,

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  • $\begingroup$ Are numbers like $222$ allowed? or do all three digits have to be different? $\endgroup$ – user574848 Dec 25 '18 at 10:26
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    $\begingroup$ If you have to use each number exactly once, then aren't the only answers $542,452,524,254$? So, $4$, not $6$. But of course $18$ appears to indicate that you can use each number repeatedly or not at all. $\endgroup$ – lulu Dec 25 '18 at 10:29
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    $\begingroup$ Well, how many options are there for the leftmost digit? How many for the middle digit? How many for the rightmost? $\endgroup$ – lulu Dec 25 '18 at 10:47
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    $\begingroup$ @lulu , $3$ Options for left most digits, $3$ options for middle digits and $2$ options for right most digits. $\endgroup$ – malik727 Dec 25 '18 at 10:49
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    $\begingroup$ Ok, and $3\times 3\times 2=?$ $\endgroup$ – lulu Dec 25 '18 at 10:49
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let's the number is $\,\overline{abc} \,$ with $ a,b,c \in \{ 2,4,5\} $ and $c\in \{2,4\}$ ( because $\,c\,$ is the even number).We have a has three choices, b has three choices and c has two choices. Therefore, we have $3*3*2=18 $ numbers satisfies the problem.

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  • $\begingroup$ Thanks alot! @lulu already solved my problem but your answer made it more clear. $\endgroup$ – malik727 Dec 25 '18 at 12:34

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