how many $7$ digit numbers can be formed using $1,2,3,4,5,6,7,8,9,0$ How many seven digit numbers can be made if 
$(a)$ they must be odd and repetition is not allowed
$(b)$ they must be even and repetition is not allowed
$0532129$ is not a seven digit number
So the question is asking, how many 7 digit numbers can be made if zero can't be the first number but it can be the last one.. am i interpreting this correctly ? 
That means there are $5$ evens ${2,4,6,8,0}$ and $5$ odds ${1,3,5,7,9}$.
So for both odds and evens, would it be $6 \times 5 \times 4 \times 5 \times 3 \times 2 \times 1 = 3600$ ?
Any help is appreciated. 

We have $7$ positions to fill:
The $first$ $position$ can be any digit between 1 and 9: $9$ choices
The $second$ $positon$ can be any digit but the first one, however we can use 0 here: $9$ choices.
We have $5$ positions left to fill which can be done in $8 * 7 * 6 * 5 * 4$
Total: $9 * 9 * 8 * 7 * 6 * 5 * 4$
However, this is the total amount of valid numbers we can form. In order to get the amount of even/odd numbers, you need to substract those ending in $0, 2, 4, 6, 8$ or $1, 3, 5, 7, 9$
Can you finish it yourself?
