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!
 A: 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 $(3)P(6,3)$. 
A: Multiply the answer by $\frac{4}{7}$ since there are 4 odd numbers and $\frac{3}{7}$.
A: 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 =$840-3{6 \choose 3}3!$
