Counting the arrangements of 8 people around a square table?

I am trying to solve this problem of counting the number of arrangements of 8 people around a square table, as shown in the figure below, To solve this problem you can consider arrangements obtained from rotation to be similar -: The first part of the question asks how many possible arrangements of 8 people are there on this square table, my reasoning for coming up with an answer is as follows, each of the circular arrangement of 8 people around the square table corresponds to 4 linear arrangements so by this reasoning the answer I came up with $$\frac{8!}{4} = 10080$$ square arrangements.

The second part of the question asks me in how many square arrangements do A and B don't sit together, here is how I approached the problem, I first counted the number of linear arrangements in which A and B sit together $7! \cdot 2!$ and using this I counted the number of square arrangements in which A and B sit next to each other as $\dfrac{7! \cdot 2!}{4 \cdot 2!} = 1260$ square arrangements in which A and B sit next to each other and then I subtract this number from the total number of square arrangements $10080 - 1260 = 8820$ arrangements in which A and B don't sit next to each other, I am not sure if my answer is correct but I think it should be, it would be great if someone could confirm this.

• If I understand the first question right, the first answer is correct, but for an arbitrarily arrangement, chances for B to be beside A is 2/7, the second answer should be 10080 / 7 * 5 – Coolwater Jul 7 '14 at 11:15
• How, can you explain it a little bit? – AnkitSablok Jul 7 '14 at 11:18
• Using your approach there are 8 places for A, 2 for B, 6 for C, 5 for D, etc. so A is beside B in 8*2*6*5*4*3*2*1 / 4 = 2880 cases – Coolwater Jul 7 '14 at 11:21
• how are there 2 places for B?, okk I got it the number of arrangements in which A and B are together, yeah B can be either to the right of A or to left of A, but then how do you come up with 7200 as the answer? – AnkitSablok Jul 7 '14 at 11:23
• You know from first part there is 10080. Subtract the 2880 and you get the right answer.There are 7 other people than A in a random arrangement, 2 of them are beside A, hence 100% - 2/7 = 5/7 of the 10080 cases is also right answer. – Coolwater Jul 7 '14 at 11:34

The exclusion of arrangements that can be obtained from rotation comes to the same as the extra condition that $A$ is seated e.g. at the upper side. This because in any case there is exactly one rotation that brings him there. Then there are $2$ possibilities for $A$. The first part then gives $2\times7!=10080$ possibilities, confirming your own answer. The second part gives $2\times6\times6!=4320$ possibilities (if I understand well that $A$ and $B$ are not sitting next to eachother here). The factor $6$ corresponds with the possibilities for $B$.