Can anyone confirm formula (combinations)

In summary, the problem is to partition n distinct objects into k sets, each containing m objects. The proposed answer is n! / (k! * (m!)^k), which is correct. The sets are considered indistinguishable, so order does not matter. An example of this type of problem is organizing a round of chess games with 12 people into 6 games. The formula has been proven to work for the m=2 case, but may need further confirmation for m>2.
  • #1
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).
 
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  • #2
Sounds plausible.
 
  • #3
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.
 
  • #4
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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1. What is the formula for combinations?

The formula for combinations is nCr = n! / (r! * (n-r)!), where n represents the total number of items and r represents the number of items being selected.

2. How do I know when to use the combination formula?

You would use the combination formula when you want to know the number of ways to select a certain number of items from a larger group, regardless of the order in which they are selected.

3. Can the combination formula be used for both ordered and unordered selections?

No, the combination formula is only used for unordered selections. For ordered selections, you would use the permutation formula.

4. Are there any real-world applications for the combination formula?

Yes, the combination formula is often used in probability and statistics to calculate the likelihood of certain events occurring. It can also be used in economics and finance to determine the number of ways to allocate resources.

5. How can I confirm that I am using the correct formula for combinations?

You can confirm the formula by checking if your result matches the expected outcome and by double-checking your calculations. Additionally, you can consult reliable sources or consult with a mathematician or statistician for confirmation.

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