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).

Question

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).

Answer 1

There are

10^{10}

strings of length 10 over the alphabet

{0, 1, 2, \dots, 9}

, 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

N = \frac{9}{10} (10^{10} - 10!) = 8 996 734 080 .

Number of items	0	1	2	3	4	5
Probability	0.74	0.17	0.04	0.025	0.1	0.015

Answers (1)

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