How many numbers are with $5$ different digits in decreasing order?

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.

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

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 $${10 \choose 5} = 252 \,.$$
Alternatively: these numbers correspond one-to-one to $$5$$-element subsets of $$\{0,\dots,9\}$$ (the set of the digits), so their number is $$\binom{10}{5}=252$$.
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.