How many n-digit numbers with strictly increasing digits do exist?$(n<10)$
We mean numbers like: $13458$,these numbers do not have $0$ as a digit.How can we count them??
I used trees to distinguish and count each case of these numbers,but is there any combinatorial technique to count such numbers???

  • $\begingroup$ is 112233 valid? Repetitions of same digit allowed? Or has to be increasing for every successive place value. $\endgroup$ – sanketalekar Apr 30 '16 at 14:53
  • $\begingroup$ @sanketalekar it says strictly increasing. $\endgroup$ – Henno Brandsma Apr 30 '16 at 17:08
  • $\begingroup$ 112233 is not valid $\endgroup$ – Hamid Reza Ebrahimi Apr 30 '16 at 17:44

Such a number is just a $n$ size subset of $\{1,\ldots,9\}$. Such a subset has a unique increasing order (which is the number representation).

So there are $\binom{9}{n}$ such numbers.

  • $\begingroup$ Can you explain more please? $\endgroup$ – Hamid Reza Ebrahimi Apr 30 '16 at 17:46
  • 1
    $\begingroup$ The digits in the number form a subset of size $n$. And if we have any subset of size $n$, there is only one way to write the subset in increasing order. So there is a bijection between your increasing numbers and subsets of size $n$ (and $0$ is not allowed). The binomial coefficient counts those subsets. $\endgroup$ – Henno Brandsma Apr 30 '16 at 18:22

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