$12$ people need to be split up into equal teams for a quiz. How many ways can this be done?

The answer may initially seem to be $\displaystyle \frac{12!}{6!6!}$.

but, since a single grouping will be represented twice (i.e. the red balls in a given position = the blue balls in the same position), the actual answer is: $\displaystyle \frac{1}{2}\frac{12!}{6!6!}$.

I guess this corrective term arises from considering the basic symmetric notion that a selection of a given subset equals the selection of all elements except for the elements from the said subset; yet, when the subset contains half the number of elements found within the set as a whole, a selection will occur twice, as the "position" of subsets can be perfectly swapped.

It seems this property extends to whenever $2$ or more equal subsets are chosen for a set and we are counting the combinations.

What is the name for this type of property (and what further generalizations are there)?

  • 1
    $\begingroup$ I think the word you seek is "symmetry". Two sets of size $6$ are symmetric in a way that a set of size $5$ together with a set of size $7$ are not. $\endgroup$ – vadim123 May 24 '14 at 14:23

The general way to see this is that you are counting a set which isn't in a bijection with the set of interest, the map is $n$ to one. So you divide by $n$.

  • $\begingroup$ Wow, very well put! And I guess n can be found by considering the products of factorials of the "equi-sized" subsets. So for a case where there are 15 people, and subsets of 2,2,2,3,3,3, we have n = 3!3! (or is this faulty)? $\endgroup$ – Just_a_fool May 24 '14 at 15:18
  • $\begingroup$ @Just_a_fool, it isn't that simple. You have to check that each element of the set of interest is counted exactly the same number of times. Nothing guarantees that a formula will give all cases. $\endgroup$ – vonbrand May 24 '14 at 15:30

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