mrdanielkim wrote:
i figured it out after taking a break from studying. thanks anyway!
here's the solution anyway, though the book has the question worded differently:
Q: IN how many ways can 6 people be seated at a round table if one of those seated cannot sit next to two of the other five:
A: Six people can be seated around a round table in 5! ways. there are 2 ways that the two unwelcome people could sit next to the person in question and 3! ways of arranging the other three. This is subtracted from the base case of 5!, giving the result of 108.
This question is different from the one posted above.
There are two versions and the answer would be different in the two cases.
Let me pick this version first:
There are 6 people (say A, B, C, D, E and F). They have to sit around a circular table such that one of them, say A, cannot sit next to D and F at the same time. (This means that A can sit next to D but not while F is on A's other side. Similarly, A can sit next to F too but not while D is on A's other side)
Total number of ways of arranging 6 people in a circle = 5! = 120
In how many of these 120 ways will A be between D and F?
We make DAF sit on three consecutive seats and make other 3 people sit in 3! ways.
or we make FAD sit of three consecutive seats and make other 3 people sit in 3! ways.
In all, we make A sit next to D and F simultaneously in 12 ways.
120 - 12 = 108 is the number of ways in which D and F are not sitting next to A at the same time.
The second version which seemed like the intended meaning of the original poster:
There are 6 people (say A, B, C, D, E and F). They have to sit around a circular table such that one of them, say A, can sit neither next to D nor next to F. (This means that A cannot sit next to D in any case and A cannot sit next to F in any case.)
Here, we say that A has to sit next to two of B, C and E.
Let's choose 2 of B, C and E in 3C2 = 3 ways. Let's arrange them around A in 2 ways (say we choose B and C. We could have BAC or CAB). We make these 3 sit on any 3 consecutive seats in 1 way. Number of ways of arranging these 3 people = 3*2 = 6
The rest of the 3 people can sit in 3! = 6 ways
Total number of ways in which A will sit neither next to D nor next to F = 6*6 = 36 ways
I used another method learnt from one of your posts itself, to solve this question. Please see if the application is correct.
If A can sit neither next to D nor next to F, I calculated the total cases (5!) from which I subtracted all the cases where A and D would necessarily sit together (2*4!) and the ones where A and F would necessarily sit together (2*4!). Now, here, there would be a slight overlap where A would be sitting between D and F (DAF and FAD), so adding that to the equation to remove the overlap (2*3!)