Given positive interger N, now many non-decreasing sequences of length

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Discussion Overview

The discussion revolves around the problem of determining the number of non-decreasing sequences of a specified length N, with the condition that the entries are positive integers less than N. The scope includes mathematical reasoning and combinatorial analysis.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • Post 1 presents the initial question regarding the count of non-decreasing sequences of length N with entries less than N.
  • Post 2 seeks clarification on whether the entries are positive integers.
  • Post 3 confirms that the entries are indeed positive integers.
  • Post 4 introduces a mathematical approach by defining sets A and B, establishing a bijection between them, and providing a combinatorial formula to count the elements in set B, resulting in the expression \(\binom{2N-2}{N}\).
  • Post 5 expresses appreciation for the mathematical insight but notes that the growth of the sequence count is too rapid for their needs.

Areas of Agreement / Disagreement

Participants generally agree on the mathematical approach presented in Post 4, but there is no consensus on the applicability of the result for specific purposes, as indicated by Post 5's concern about growth rates.

Contextual Notes

The discussion does not address potential limitations or assumptions in the mathematical model, nor does it explore the implications of the growth rate mentioned in Post 5.

Dragonfall
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Given positive interger N, now many non-decreasing sequences of length N are there whose entries are less than N?
 
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are the entries positive integers?
 
Yes .
 


This isn't in the homework section so I'm assuming it isn't homework.

Let A be the set of sequences of the type in the post.
Let B be the set of sequences of N+1 non-negative integers with sum N-2.
A and B are in bijection by associating a sequence [itex]a_1,a_2, \ldots ,a_N[/itex] of A to [itex]b_1,b_2,\ldots,b_{N+1}[/itex] defined by [itex]b_i = a_i - a_{i-1}[/itex] where we let [itex]a_0 = 1, a_{N+1}=N[/itex]. The inverse is given by associating a sequence [itex]b_1,b_2,\ldots,b_{N+1}[/itex] of B to [itex]a_1,\ldots,a_N[/itex] defined by [itex]a_{i+1} = a_{i}+b_{i+1}[/itex] where we let [itex]a_0 = 1[/itex].

Thus we can just count B which has
[tex]\binom{(N+1)+(N-2)-1}{(N+1)-1} = \binom{2N-2}{N}[/tex]
elements.
 


Wow I would never have thought of that. Thanks. Unfortunately it grows too fast for my purposes.
 

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