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 was trying to do the second (intended meaning's calculation, albeit in a different way. I was getting a different answer. Can you tell me where I am going wrong?
Cases where A & F are together: (4!) * 2 (Considering that AF and FA are different) = 48
Cases where A & D are together: (4!) * 2 (Considering that AD and DA are different) = 48
Now the above 2 cases will each include identical cases where a D was immediately after or before AF &FA or an F was immediately after or before AD and DA. Each case will be counted twice. Number of such cases counted twice is ADF taken together and permutated i.e. 3! * 3! (first factorial for the circular permutation, and the second for the permutation among ADF) = 36
now 120-48-48+36 = 60 should be the answer according to me.
I know this is wrong, but can you please help me understand where is it I am going wrong?