How to get the combinatory of the numbers? I have a statement that says:

Let the digits be: $1, 2, 3, 4, 5, 6$ So, how many even numbers of
  three figures  different can be formed?

So, the number should be like: $\overline{abc}$, where $a$ can be $1, 3 , 5$ that is $3$ options, $b$ can be $2$ options, because an odd number has already been used for $a$ and can not be repeated, and $c$ can be $3$ options, that are {$2, 4, 6$} because it must be a even number.
So, they should be formed: $3 * 2 * 3 = 18$, according to the multiplicative principle.
But my answer isn't correct, i would like to know why it is not correct, and how it should be done
 A: If the number is $\square\square\square$ then:
$$\square\quad\square\quad\underbrace{\square}_{\text{3 pos}}$$
Now, complete the rest:
$$\underbrace{\square}_{\text{5 pos (all except the number put at the end)}}\quad\underbrace{\square}_{\text{4 pos (all except the number put at the end and the number put on the left)}}\quad\underbrace{\square}_{\text{3 pos}}$$
So, you have $5\times4\times3=60$ possibilities.
A: Your mistake is that if the three-digit number $\overline{abc}$ is even, $c$ can be $2$, $4$, $6$ then $a$ can be any of the remaining numbers (including the remaing two even numbers in the set $(2;4;6)$).


*

*Case $1$: $c=6$, there are $5$ choices remaining for $a$ and $4$ choices remaining for $b$ after picking $a$.

*Case $2$: $c=4$, there are $5$ choices remaining for $a$ and $4$ choices remaining for $b$ after picking $a$.

*Case $3$: $c=2$, there are $5$ choices remaining for $a$ and $4$ choices remaining for $b$ after picking $a$.
Note that if you choose $c$ first, there will be $5$ choices remaining for $a$, not $3$.
A: If there were no restriction we have 6x5x4or 120 ways....now unit digit can either end up with even or odd , and both cases are equally likely , so the required no of cases 60.
