I have a few questions about monotone finite sequences. My motivating problem is the following:

Sixteen players participated in a round-robin tennis tournament. Each of them won a different number of games. How many games did the player finishing sixth win?

I struggled with this problem until I realized that there's only one way to order the corresponding finite sequence (the answer ended up being 10). Is there a way to approach this problem that doesn't require listing out the entire sequence and experimenting until you get to one that adds up to ${16 \choose 2}=120$, or at very least is there a way of knowing whether the sequence is forced like the question above was? Furthermore, do you guys know of any good resources for information about monotone finite sequences?

  • $\begingroup$ I'm not sure I understood the question. Are you interested in counting strictly monotone finite sequences $x_1 < x_2 < \ldots < x_n$, subject to certain bounds $x_1 \geq A, x_n \leq B$, $x_1 + \ldots + x_n = C$? In your proposed problem, $n$ would be $16$, $A = 0, B = 15, C = 240$. $\endgroup$ – Pedro M. Mar 28 '13 at 17:12
  • $\begingroup$ Yes, that's it. It wasn't even clear to me that $B=15$ until I found the sequence. What I'd like to know is a way, if possible, to gain more information about the sequence before I go into testing cases, because it doesn't seem like it would be practical just to test cases generally. $\endgroup$ – Charles Mar 28 '13 at 21:49
  • $\begingroup$ The only thing I recall that's vaguely related to the problem is partitions of a certain number. If $A = 1$, we are essentially counting the number of partitions of $C$ into distinct parts, each one less than or equal to $B$. en.wikipedia.org/wiki/… $\endgroup$ – Pedro M. Mar 29 '13 at 12:56
  • $\begingroup$ I worried that that would be the case. Math's like that sometimes. Thank you for the link and your time. $\endgroup$ – Charles Mar 29 '13 at 15:22

There are $16$ possible winning number for each player i.e. $0,1,\dots,15$. And the number for each player are all different. So it must be: the one in the first place has $15$ wins, and followed by $14$ wins, etc.

  • $\begingroup$ Well, I agree, but could you just see that the only way it'll sum to 120 is if the sequence is like that? I couldn't so it took me a long time to figure it out. I had resort to testing cases, and the answer came from there. My question is about method. $\endgroup$ – Charles Mar 28 '13 at 0:32

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