How many 11-digit numbers are there which contain all of the digits 0-9 at least once? How many $11$-digit numbers are there which contain all of the digits $0$-$9$ at least once?
Note that in this question a number cannot begin with a digit of $0$.

My instructor explained this as a case problem, but rushed through it without pausing for questions. The answer ended up being $99 \times (10!/2!)$, but I have no idea how to get there!
My thinking:
When you have 11 items in which 9 are unique and 2 are the same numeral repeated, then that can be arranged in $\frac{11!}{2!}$ ways.  Then, there are 10 different choices for which numeral is repeated.  So we have $\frac{10\times11!}{2!}$.  Finally, one tenth of these arrangements should have 0 as its first digit and those don't count.  So we need to multiply by $\frac{9}{10}$

$$\frac{9}{10}\times\frac{10\times11!}{2!}=99\times\frac{10!}{2!}$$
10!×10×11-9!

10! is permutaion of 0,.......9

10 is the number additionally added

11 is the number of place available to add the number

9! is to eliminate numbers begin with 0