Let X be the set of 10-digit numbers that do not contain all digits 0 through 9 in their decimal representation and do not begin with 0. How many elements does the set X have? I think this should be done by counting principle. So I have 10 spots, and 10 options for each spot. However, the first number can't be 0, so the first spot has 9 options. 9 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 9 I think the 10th spot only have 9 options since I can't repeat one number (the representation doesn't have all the digits).

Let X be the set of 10-digit numbers that do not contain all digits 0 through 9 in their decimal representation and do not begin with 0. How many elements does the set X have?
I think this should be done by counting principle. So I have 10 spots, and 10 options for each spot. However, the first number can't be 0, so the first spot has 9 options.
9 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 9
I think the 10th spot only have 9 options since I can't repeat one number (the representation doesn't have all the digits).
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Urijah Hahn
There are ${10}^{10}$ strings of length 10 over the alphabet $\left\{0,1,2,\dots ,9\right\}$, and exactly $10!$ of these use all 10 letters of the given alphabet. From the remaining ${10}^{10}-10!$ strings we should exclude all that have 0 as first digit. By symmetry this amounts to $\frac{1}{10}$ of all remaining strings. The number $N$ of admissible strings therefore is given by