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I mean that $54321$ is a valid number, but $16755$ is not.

I was thinking that there are $6^5$ possible numbers because there are 6 possibilities for each position in the number, but I don't know exaclty.

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    $\begingroup$ Why are there $6$ possibilities for each position? $\endgroup$
    – Math Lover
    Mar 26, 2021 at 11:23
  • $\begingroup$ Do leading zeroes count ? $\endgroup$
    – user65203
    Mar 26, 2021 at 11:42
  • $\begingroup$ @YvesDaoust Leading zeroes don't matter since no decreasing sequence can begin with a zero. $\endgroup$
    – smalldog
    Mar 26, 2021 at 11:43
  • $\begingroup$ @chaos: ooops, quite right ! Thanks. $\endgroup$
    – user65203
    Mar 26, 2021 at 11:43
  • $\begingroup$ You should clarify whether you mean strictly decreasing or non-increasing like 66531 $\endgroup$ Mar 26, 2021 at 11:59

3 Answers 3

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Each number between $0$ and $9$ can appear exactly once or not at all (since there can be no repetitions). Given a set of five such numbers (say $\{1, 3, 0, 5, 8\}$) there is exactly one corresponding decreasing sequence (85310). So the number of decreasing sequences is the number of ways to choose 5 elements from a set of 10 digits. This is just $$ {10 \choose 5} = 252 \,. $$

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  • $\begingroup$ @Joffan Thanks for that, I missed the word in the title. $\endgroup$
    – smalldog
    Mar 26, 2021 at 12:15
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Alternatively: these numbers correspond one-to-one to $5$-element subsets of $\{0,\dots,9\}$ (the set of the digits), so their number is $\binom{10}{5}=252$.

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You can write $10\cdot9\cdot8\cdot7\cdot6\cdot5$ distinct numbers made of $6$ digits. Divide this amount by $5!$ to only allow the decreasing permutations.

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