# How many total monotonous numbers are there?

Question: Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or decreasing sequence. For example, 3, 23578, and 987620 are monotonous, but 88, 7434, and 23557 are not. How many monotonous positive integers are there?

I attempted to solve the following problem by finding out how many 2-digit and 3-digit monotonous numbers there are and then finding the pattern within them and applying them all the way to the max number of digits in a monotonous number - 9. I figured that when the tens digit is 0, there are 9 possible combinations for the ones digit, when the tens digit is 1, there are 8 possible combinations, and so forth. Thus with 1+2+3+4+5+6+7+8+9 I arrived at 45 increasing monotonous numbers for 2-digit numbers. Then as there is one decreasing monotonous number for every increasing monotonous number, I multiplied it by 2 to get 90 total 2-digit monotonous numbers. However I can not figure out how to accomplish this for 3-digit numbers. Can someone please help me with this, or if there is a more efficient method, could you please explain it? Thanks a lot!

• Why is $23557$ non monotonous? Isn't it increasing from left to right? Also, why is $323578$ a monotonous number? – S.C.B. Feb 16 '17 at 1:46
• It cannot have repetitions of any digits. Additionally the second example does not count as a monotonous number as it doesn't decrease continuously with all the digits. – celestialwayfarer Feb 16 '17 at 1:48
• I don't have a solution but I can see that it would be useful to note that all strictly monotonous 3 digit numbers have strictly monotonous 2 digit prefixes. – Q the Platypus Feb 16 '17 at 1:52
• "I figured that when the tens digit is 0, there are 9 possible combinations for the ones digit,": Are you counting $01, 02,03$, etc., as separate from $1,2,3,$ etc.? – Jonas Meyer Feb 16 '17 at 2:00
• No those would be counted the same. That is why I started counting with the 2-digit numbers as the 1-digit numbers would already be included within this set. – celestialwayfarer Feb 16 '17 at 2:02

Consider the string $9876543210$. We can choose or ignore each digit from this string to get decreasing monotonous numbers for a total of $2^{10}=1024$ options. Since that includes ignoring all of the digits and only choosing $0$, there are $1022$ valid choices that yield positive numbers.

Similarly we can choose from the string $123456789$ to get $2^9-1=511$ options for increasing monotonous numbers.

Finally, noting that the $9$ single-digit numbers belong to both groups, the total count of different monotonous numbers is $1022+511-9=1524$.

There are $10$ digits and $10 \choose n$ ways to choose $n$ distinct digits (which must be distinct to preserve strict monotonicity). There is exactly one way to arrange those $n$ digits in decreasing order. Thus there are \begin{align*} {10\choose1}+\dots+{10\choose10}&={10\choose0}1^01^{10-0}+\dots+{10\choose10}1^{10}1^{10-10}-1\\ &=(1+1)^{10}-1\\ &=2^{10}-1\\ &=1023\\ \end{align*} strictly decreasing non-negative integers. This counted the number $0$, so subtract one from our result for $1022$ strictly decreasing positive integers.

Next consider increasing integers. Anything with $1$ digit has already been counted, so we only want numbers with $2$ or more digits. Additionally, none of these integers will have the digit $0$, since it would be the first digit, yielding a result with a smaller number of digits. We similarly find \begin{align*} {9\choose2}+\dots+{9\choose9}&=2^9-10\\ &=502\\ \end{align*} additional strictly increasing positive integers.

Adding these two results together yields $\boxed{1524}$ total monotonous positive integers.

• $1023+502=1525$ so why is the answer $1524$? – Old Peter Jun 12 '18 at 18:53
• The original question asked about positive integers, and the $1023$ count included the number $0$. Just afterward, you'll notice we subtracted $1$ to get a total of $1022$ instead, yielding the desired result of $1022+502=1524$. – Hans Musgrave Jun 12 '18 at 20:36