How many three digit numbers can be formed from the digits $1,2,3,4,5$ and $6$, if each
digit can only be used once? How many of these are odd numbers? How many are
greater than $330$?

I've determined that the total number of combinations formed by $3$ digits is $$6\times 5\times 4$$

Now, for determining how many of these combinations are odd, each of the combinations must end with $1,3$ or $5$. The solution shows that the correct answer is $$3\times(5\times4)$$ 

I'm curious whether this is correct because there are $6$ choices for the first number, not $5$, so wouldn't the answer be $6\times5\times3$?

I also thought to myself, what if the first number chosen was a $1$ and the second number chosen was a $3$, then for the last digit there is only $1$ option $(5)$ since $1$ and $3$ have already been used, so there is no way my solution could be correct.

So maybe the answer is $3 \times 2 \times 3$ since there are: $3$ options for the first number $(2,4,6)$, $2$ options for the second number $(4,6)$ and $3$ options for the last number $(1,3,5)$.

**OR**

I think this is the logic that whoever wrote the solution was using...
If you make sure to always leave one of the odd numbers available: There are $5$ choices for the first digit (all but $1$), $4$ choices for the second digit ... and I don't know how to proceed from here.

First choice possibilities (Assuming we leave the $1$ available) : $2,3,4,5,6$ - $5$ Options   

Second choice possibilities (Still not touching the $1$): $5 - 1$ Options   

Third choice possibilities : $1$ option (the $1$)

Then repeat this for $1, 3$ and $5$. So $$3 \times (5\times4)$$

Can someone set me straight on this? Thanks for any help ahead of time.