Can anyone confirm formula (combinations)

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    Combinations Formula
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Discussion Overview

The discussion revolves around the formula for partitioning n distinct objects into k sets, each containing m objects. Participants explore the validity of the proposed formula and its application to specific examples, including organizing games among players.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes a formula for the number of distinct partitionings: n! / (k! * (m!)^k, suggesting it might be correct.
  • Another participant finds the proposed formula plausible but seeks clarification on whether the sets are indistinguishable.
  • A participant confirms that the order of sets does not matter in the proposed formula, providing reasoning based on the nature of partitioning.
  • One participant mentions a specific example involving 12 people and 6 games of chess, indicating they can prove the formula works for the case where m=2 but expresses uncertainty for m>2.

Areas of Agreement / Disagreement

Participants generally agree on the plausibility of the proposed formula but have not reached a consensus on its correctness for all values of m, particularly for m>2. The discussion remains unresolved regarding the broader applicability of the formula.

Contextual Notes

There are unresolved questions about the assumptions regarding indistinguishability of sets and the specific conditions under which the formula is valid, particularly for different values of m.

uart
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Hi, I've been scratching around trying to figure out a formula for the following problem and I've got one that I think is correct. Just wondering if anyone can confirm it for certain (like maybe you have it in a textbook or know it well etc). Thanks.

Problem : You need to partition n=k*m distinct objects into k sets each containing m objects. How many ways can you do this?



Proposed Answer :

Number of possible distinct partitionings = n! / ( k! * (m!)^k )

(I think it's correct).
 
Last edited:
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Sounds plausible.
 
Are the sets into which we partition indistinguishable? Ie if we partition n into n sets there are n! ways of doing this if we consider order, or just 1 if we say that they are all equivalent. I'm guessing fromyour formula order doesn't matter.

So there are nCm ways of picking the first set, mutliplied by (n-m)Cm for the second and so on, but we need to divide by k! to forget the ordering which is, I suspect, exactly what your formula is.
 
matt grime said:
Are the sets into which we partition indistinguishable? Ie if we partition n into n sets there are n! ways of doing this if we consider order, or just 1 if we say that they are all equivalent. I'm guessing fromyour formula order doesn't matter.

Your guess is correct Matt, the way I set it up is that order doesn't matter. So in the example of partitioning n items into n sets of one item each then yes there is only one way to do it, not n! ways.

An example of the type of problem that I wanted to solve is : say you have 12 people meet to play 6 games of chess, how many distinct ways can you organize that round of 6 games.

BTW, I can prove for certain that the formula works for the m=2 case (like in the chess example) but I was just a little unsure if it was correct for m>2.
 
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