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Let $[n] = \{1,2,\ldots,n\}$. A function $f: [n] \rightarrow [n]$ is called monotone if for $1 \le i < n, f(i) \le f(i+1)$.

Consider the following monotone function $f: [n] \rightarrow [n]$, such that where $\forall a,b \in [n], f(a) = b$ is at most $m$ times. How may $f$ functions are there? But how to formally proof such a problem?

Although I noticed that $f$ is somewhat a surjective function, it has a constraint on how many images to be mapped. So I've tried to solve the problem by considering a sample case with $n=5,m=2$ and manually count the individual cases, namely for $2 \le k \le n$, I count the different combinations $\displaystyle \binom{n}{k}$ where each image is mapped at most twice. I also noticed that some cases overlap.

0x0584
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  • "Monotone function" usually means something else than what I think you mean (https://en.wikipedia.org/wiki/Monotonic_function). Did you mean to say that for all $b$, there are at most $m$ values of $a$ such that $f(a) = b$? (That's not what you wrote. If $a$ and $b$ are both fixed, then $f(a) = b$ happens either once or zero times.) – Qiaochu Yuan Sep 09 '20 at 02:34
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    Let me know if I understand it right:
    • "f is monotone" - you are basically saying $f(n) \leq f(n+1)$, right?
    • for any value of $b$, there are at most $m$ possible sulutions for $f(x)=b$, right?
     – Anderson Torres Sep 09 '20 at 02:36
  • @AndersonTorres yeah, that's true! – 0x0584 Sep 09 '20 at 02:48
  • Reading my solution to this problem will get you started. However, you will have to use the Inclusion-Exclusion Principle to eliminate those functions in which a particular element in the range appears at least $m + 1$ times. – N. F. Taussig Sep 09 '20 at 15:39

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