I have the following question:
Six friends went to a trip. Their seats in the the bus were: rows 1,2,3.
In each row the are seats A,B (next to each other).
In the trip back they got the same seats but no one would sit next to the person
he sat with in the first trip. (So one person won't sit next to the same one).
Order matters, as A1 is different from A2.
In how many ways can the three friends sit in the trip back?
Now what I thought the solution is:
The number of total ways to sit the people in the first trip is: |U|=6!
I want to use the inclusion-exclusion principle, so these are my groups:
Ai = sat together in the first.
Each Ai of these has the ways to be put together like this:
2*3!
This is because you can sit each couple in two ways, and there are 3 choices of "rows".
We have these 3 groups in that question, so I thought the solution to this problem is:
|U|-(A1+A2+A3| = 720- 3*(2*3!) = 684
My question here is: is there any importance to intersection between A1 and A2? Because if there is, the criteria is irrelevant, isn't it so?
Thank you.
Sorry, What I had put before is wrong. here is a correct approach:
note that $2$ incorrect people are together or all of them are together. It is impossible for 4 of them to be together. We use inclusion exclusion:
Exactly $2$ are together. We pick the row in which the couple sits in 3 ways. And then we fix the other 4 persons so none of the incorrect ones are together. Note there are $4$ ways in which they sit incorrectly. So there are $4*3*2-4=24-4=20$ correct ways. So in total $60$ ways in which 2 incorrect persons sit together.
All of them sit in the same pairs as before: then there are $6$ ways to see which couple gets in which row. and then we just need to pick which person gets the left or right seat in each couple (8 ways). So there are 48 ways for all of them to sit incorrectly.
That means out of the $6*5*4*3*2=720$ ways for them to sit there are $720-(108)=602$ so no previous couple sits together