# How many different possible ways are there to divide 6 ppl into two groups of 3?

Relevance

From the 6 people, choose 3 to be in the first group:

C(6,3) = (6 x 5 x 4) / (3 x 2 x 1) = 20 ways

Then the remaining 3 people will automatically go in the second group:

C(3,3) = 1 ways

So it would seem there are 20 ways to split them into 2 groups of 3. But the groups both have 3 members, so they aren't distinguishable. For example, if I put Anna, Bob and Carol in the first group and Dave, Edwina and Francisco in the second group, that would look identical to putting Dave, Edwina and Francisco in the first group and Anna, Bob and Carol in the second group, it's really the same grouping.

So you need to divide by 2! (or just 2) to account for the ways to switch the two indistinguishable groups.

10 ways

Specifically:

ABC and DEF

ABD and CEF

ABE and CDF

ABF and CDE

ACD and BEF

ACE and BDF

ACF and BDE

AEF and BCD

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• 6 choose 3 = 6!/3!(6-3)! = (6x5x4)/(3x2x1) = 120/6 = 20. So there are 20 ways to choose one group of 3 from a total of 6 people, once we choose the first group of 3, the second group of 3 is given. So the final answer is also 20.

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• 6! / (3! 3! 2!)

= 720 / 72

= 10

Here's another example o dividing 10 people into two groups of five each.

https://gyazo.com/b002ddb39718f64fcf0fa35fa09ffe64

https://math.stackexchange.com/questions/500544/10...

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