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To be clear, only one group is being formed,and order does not matter. The only thing making the answer NOT 6 choose 3 is that two of the people cannot both be in the group at the same time. How to deal with that is my question.

edit: so I had an idea, since every person has a 1/2 chance of getting in the group, there is a 1/4 chance of any two people being on the group. Meaning that there is a 1/4 chance that the two people who can't be together are, meaning that the answer should be 3/4s of 6 choose 3 right?

Meep
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    Think of the two people as bins and the rest of the people go into those two bins. – CyclotomicField May 25 '22 at 18:35
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    Call those two people $a,,b$. How many trios contain neither? How many only $a$? How many only $b$? Alternatively, subtract the number of trios that do contain both $a,,b$ from $\binom{6}{3}$. (It's probably worth solving both ways as a sanity check.) – J.G. May 25 '22 at 18:35
  • Thanks. Your suggestion to figure out how many trios contained both instead of did not contain both was very helpful. – Meep May 25 '22 at 19:03

2 Answers2

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So I just figured out how to do this and I thought I would post it here so future confused people can find it

Anyway, I am gonna call the two people I don't want together A and B

If I assume A and B are together, and I need to fill the last spot, I can figure out how many ways I can do that with 4 (the people left after choosing A and B) choose 1 (the number of spots left). Since that is 4, I know there are 4 groups in which A and B are together. So the answer is 6 choose 3 minus 4.

Meep
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Your $$\binom{6}{3}-\binom{6-2}{3-2} = \binom{6}{3}-\binom{4}{1}=20-4=16$$ is a correct and simple approach.

Another way is to consider three cases, depending on whether $A$ or $B$ or neither appears: $$\binom{1}{1}\binom{1}{0}\binom{6-2}{3-1}+\binom{1}{0}\binom{1}{1}\binom{6-2}{3-1}+\binom{1}{0}\binom{1}{0}\binom{6-2}{3-0} = 2\binom{4}{2}+\binom{4}{3} = 12+4 = 16.$$

RobPratt
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