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!

We first count the number of ways to produce an even number. The last digit can be any of $2$, $4$, or $6$. So the last digit can be chosen in $3$ ways. 
For each such choice, the first digit can be chosen in $6$ ways. So there are $(3)(6)$ ways to choose the last digit, and then the first. 
For each of these $(3)(6)$ ways, there are $5$ ways to choose the second digit. So there are $(3)(6)(5)$ ways to choose the last, then the first, then the second. 
Finally, for each of these $(3)(6)(5)$ ways, there are $4$ ways to choose the third digit, for a total of $(3)(6)(5)(4)$. 
Similar reasoning shows that there are $(4)(6)(5)(4)$ odd numbers. Or else we can subtract the number of evens from $840$ to get the number of odds.
Another way: (that I like less). There are $3$ ways to choose the last digit. Once we have chosen this, there are $6$ digits left. We must choose a $3$-digit number, with all digits distinct and chosen from these $6$, to put in front of the chosen last digit. This can be done in $P(6,3)$ ways, for a total of $(3)P(6,3)$. 
