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Here is the problem :

Pak Dengklek is very fond of fiddling with numbers and gives a name for the unique nature of a number. One of the unique characteristics of the numbers by Mr. Dengklek is uphill numbers. A number $X$ is called an Increase Number when the digits of $X$ rise from left to right. Example The uphill number is $122349$. Suddenly Mr. Dengklek is curious, how many uphill numbers are worth less than ($10^{10}$)?

Someone solved this problem by using hockey stick identity, but I don't understand the way he solved it. Could there be another way to solve this problem simply?

N. F. Taussig
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Kadek Surya Mahardika
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1 Answers1

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Write a binary word consisting of $10$ ixes and $9$ separating bars, like so: $$x\ |\ x\ x\ |\ |\ x\ x\ |\ |\ |\ x\ x\ x\ |\ x\ |\ x\ |\ \quad.$$ Now count along this word, starting with $0$. For each separating bar count $1$ ahead, and for each $x$ write down the momentary number. In this way you get a typical "Increase Number". In the above example this is the number $$0\ 1\ 1\ 3\ 3\ 6\ 6\ 6\ 7\ 8$$

It becomes apparent that each such word determines exactly one "Increase Number", and conversely: Each "Increase Number" determines exactly one such code word.

It follows that the number of "Increase Numbers" is ${19\choose9}=92\,378$. As $9$ is not far away from ${19\over2}$ it is possible that there is a combinatorial explanation involving Catalan numbers.

Christian Blatter
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