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We have an integer $m \in {1, \dots, n}$ where $n \in \mathbb{N}$.

How many ways can we construct a $m$-length list, where:

  • Each item of the list is in ascending order and distinct
  • Each item $i$ of the list is also in ${1, \dots, n}$
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  • $\begingroup$ What have you tried? I highly recommend doing an example of this, say $n=5$, and then say $m=2$, and see exactly what you're counting. It's not a "complicated" final answer. $\endgroup$
    – ndhanson3
    Feb 9, 2022 at 6:02
  • $\begingroup$ What have you tried? Can there be repeats? $\endgroup$ Feb 9, 2022 at 6:02
  • $\begingroup$ Hint: the way you order(permutations of) the list can not exceed 1 for any combination(I'll leave the why to you). So you need to choose unique m numbers from {1..n}. $\endgroup$ Feb 9, 2022 at 6:10
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    $\begingroup$ Once you choose your elements there is only one qay you can make an admissible list out of them, so you've got $\binom nm$ ways in all. $\endgroup$ Feb 9, 2022 at 6:11

1 Answer 1

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Take the binomial coef. $$\binom{n}{m}$$ from there you will get how many subsets of $m$ elements you can get, now given that these subsets of natural numbers following the principle of good order, you can arrange them from largest to smallest as you are asked, this can only be done once for each subset so you get i.e. there are $$\binom{n}{m}$$ forms

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