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).

vangstosiis 2022-07-22 Answered
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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Answers (1)

Urijah Hahn
Answered 2022-07-23 Author has 13 answers
There are 10 10 strings of length 10 over the alphabet { 0 , 1 , 2 , , 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 1 10 of all remaining strings. The number N of admissible strings therefore is given by
N = 9 10 ( 10 10 10 ! ) = 8 996 734 080   .
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