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

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.

Thanks,

• Are numbers like $222$ allowed? or do all three digits have to be different? – user574848 Dec 25 '18 at 10:26
• 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. – lulu Dec 25 '18 at 10:29
• Well, how many options are there for the leftmost digit? How many for the middle digit? How many for the rightmost? – lulu Dec 25 '18 at 10:47
• @lulu , $3$ Options for left most digits, $3$ options for middle digits and $2$ options for right most digits. – malik727 Dec 25 '18 at 10:49
• Ok, and $3\times 3\times 2=?$ – lulu Dec 25 '18 at 10:49

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.