How many different possible ways are there to divide 6 ppl into two groups of 3?
- PuzzlingLv 73 years agoFavorite Answer
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.
ABC and DEF
ABD and CEF
ABE and CDF
ABF and CDE
ACD and BEF
ACE and BDF
ACF and BDE
ADE and BCF
ADF and BCE
AEF and BCD
- 3 years ago
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.
- cidyahLv 73 years ago
6! / (3! 3! 2!)
= 720 / 72
Here's another example o dividing 10 people into two groups of five each.