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 $(3)P(6,3)$.