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Ques : Let $N$ be the number of functions $f: \{1, 2, 3,\dots, 2018\} \to \{1, 2, 3,\dots, 2018\}$ satisfying $f(j) < f (i) + j - i$ for all integers $i,j$ such that $1 \leq i < j \leq 2018$. Find $N$.

My development so far : difference of elements' value in domain is strictly greater than that in Co domain. Hence every consecutive number should be mapped to a lesser or equal number the previous number is mapped to. Hence the function should be non-strictly decreasing.

user735447
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  • Check again : lesser or less than or equal to? Nonincreasing vs decreasing. – AHusain May 27 '20 at 23:05
  • Reserve the word "permutations" to talk only about bijective functions from a set to itself. This question has nothing to do with permutations and so it should never have been mentioned. – JMoravitz May 27 '20 at 23:05
  • Try proving that $f(j)<f(i)+j-i$ for all $i,j$ such that $1\leq i<j\leq 2018$ is true if and only if $f(j)\leq f(i)$ for all $i,j$ such that $1\leq i<j\leq 2018$. So, you are simply counting the number of monotonic (non-strict) decreasing functions. This should be a problem type you've seen before. – JMoravitz May 27 '20 at 23:11
  • As for "nonincreasing" vs "decreasing"... I dislike the term "nonincreasing." It can be interpreted as the negation of "increasing" meaning simply that there is at least one point where it decreases but possibly others where it increases... not that it always decreases. The condition that $f(j)\leq f(i)$ for all $i,j$ such that $i<j$ is better said as "non-strictly decreasing" in my opinion while the condition that $f(j)<f(i)$ for all $i<j$ is better said as "strictly decreasing." – JMoravitz May 27 '20 at 23:15
  • $f(1) < f(2008) - 2007 \le 1$. $f(2) < f(2008) - 2006 \le 2$. etc... – dust05 May 27 '20 at 23:26
  • @JMoravitz How did you concluded that $f(j)\leq f(i)$. I think inequality must be strict i.e. $f(j)<f(i)$. Correct me if I am wrong – mathophile Jan 28 '22 at 07:52
  • @mathophile consider for example $f$ a constant function. Isn't it true that such an $f$ satisfies $f(j)<f(i)+j-i$ whenever $i<j$ despite the comparison of $f(j)$ and $f(i)$ not being strict? – JMoravitz Jan 28 '22 at 12:21
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    @mathophile Further, it is worth noting that there exists exactly one strictly increasing function $[n]\to[n]$... the identity function. Similarly exactly one strictly decreasing function. That would have made the problem terribly uninteresting – JMoravitz Jan 28 '22 at 14:29

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