# How many n-digit numbers with strictly increasing digits do exist?$(n<10)$

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???

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

## 1 Answer

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.

• Can you explain more please? – Hamid Reza Ebrahimi Apr 30 '16 at 17:46
• 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. – Henno Brandsma Apr 30 '16 at 18:22