How many of these four digit numbers are odd/even? For the following question: 

How many four-digit numbers can you form with the digits $1,2,3,4,5,6$ and $7$ if no digit is repeated?

So, I did $P(7,4) = 840$ which is correct but then the question asks, how many of those numbers are odd and how many of them are even. The answer for odd is $480$ and even is $360$ but I have no clue as to how they arrived to that answer. Can someone please explain the process?
Thanks!

Only the last digit has to be even digit for the number to be even. As there are $3$ even numbers in the list so the last digit can be chosen in $3$ ways. The remaining digits has to be chosen from the remaining $6$  numbers(as one (even) number has already been chosen) in ${6 \choose 3}$ ways and can be permuted among them in $3!$ ways. So the total no. of even numbers equal $3{6 \choose 3}3!$
And the no. of odd numbers =$840-3{6 \choose 3}3!$
