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. 
 A: 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?
A: 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 $5\cdot 8\cdot {8 \choose 5}\cdot 5!.$
A: If even, the final digit should be $0,2,4,6,8$. If it is $0$, then we have 9 digits left with 6 places, and no more restrictions except the non-repeat (as 0 is automatically not on the leftmost position, as it is already on the final one, and no repetitions are allowed). So $9 \times 8 \times 7 \times 6 \times 5 \times 4$ options for that case. Otherwise it's one of $2,4,6,8$ and the $0$ is still "in game". So we pick any of the 8 remaining non-$0$ digits ($8$ options), and then we have $5$ positions left and $8$ more digits, with no more restrictions except the non-repeat again. This has $x$ many options (fill in $x$) and so we have $4 \times 8 \times x$ many options for the second case.
For odd total numbers, the final digit cannot be 0 but is one of 5 odd digits, and the first is non-zero again. So do it like the final case of (a).
