I confused about how to find the number of combinations with limits

Question

How many 7 digit numbers end in an odd number and none of the digits can repeat? The first digit cannot be 0.

Answer 1

The problem is that for the last digit you might not have $5$ choices if you have used some odd digits before. In fact you must have used at least one. You also do not have $10$ choices for the second digit as you have already used one for the first. It is better to start with the number of ways to choose the last digit, then the number of ways to choose the first because of the zero exclusion, then how many choices for the second and each other in turn.

Answer 2

This route isn't as quick as Ross Millikan's, but it's another way to figure it out:

How many of these numbers follow the criteria and are of the form $1abcdef$ ? Well we have four options for $f$ $(3,5,7, $ and $9)$. Then for what $a$ can be, we can pick any number as long as it isn't the same as $1$ or $f$. This leaves $8$ choices. Keep this process up for the options for $b,c,d,$ and $e$. This should give $4*8*7*6*5*4$

Now about $2abcdef$? See if you can figure this out.

Continue onward up to $9abcdef$ and add all the cases together to get the final answer.