Q:

Permutations- the number of six digit integers that are even

Determine the number of six digit integers in which no digit may be repeated and the integers are even.
I understand how to do this when we are repeating digits 
$9*10^4*5$
When repetition is not allowed, the total number of integers is given by
$\frac{9*9*8*7*6*5}{2}$
Would it be correct to divide the above by $2$. I basically figured that because the total is even, dividing by $2$ will give me the total number of even integers. But, is there another way to get to the right answer using permutation formulas?

A:

Hint:
Determine $N_0,N_2,N_4,N_6,N_8$ where $N_i$ denotes the number of five digit integers in wich no digit may be repeated and digit $i$ is not one of the used digits.
Then you are looking for $N_0+N_2+N_4+N_6+N_8$.
Further on it is evident that $N_2=N_4=N_6=N_8$ so it is enough to find $N_0$ and $N_2$.
The idea is that every five digit integers in which no digit may be repeated and in which e.g. digit $2$ is not allowed ($N_2$ exist) will be made a six digit even integer by adding a $2$ on the right.

