How Many Three-Digits Even Numbers Can Be Formed From 2,4,5

Question

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,

Answer 1

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.