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. 

You have 10 numerals. $0,1,2,3,4,5,6,7,8,9$ and want to find the number of orderings of 7 of them so that $0$ is not first, and either $S_O=\{1,3,5,7,9\}$ or $S_E=\{0,2,4,6,8\}$ comes last.
Say we want 0 to come last. We can choose any $6$ of the 9 remaining digits and form a permutation of them, then stick 0 at the end. So there are ${9 \choose 6}\cdot 6!$ valid numbers with 0 last.
Say we want a nonzero numeral to come last. We have $8$ choices for the first number (can't be 0 or the last numeral). To fill in the rest of the number we can choose any 5 of the remaining 8 numerals and form an ordering of them to get: $8\cdot {8 \choose 5}\cdot 5!.$
Thus the number of 7-digit even numbers is ${9 \choose 6}\cdot 6! + 4\cdot 8\cdot {8 \choose 5}\cdot 5!.$
The number of 7-digit odd numbers is $5\cdot 8\cdot {8 \choose 5}\cdot 5!.$
