# Probability Question: Round table meeting

Six delegations have a round table meeting. The meeting organization randomly arranges the seat for them. It is important to consider who is sitting to the left or right for each delegation. How many ways to arrange the seat?

(A) $720$

(B) $120$

(C) $60$

(D) $30$

I used a cyclic permutation to answer this question, $(6-1)!=120$. Is this correct?

• As the question is posed, you are correct. Of note is that if the question cared only about who sat next to who, and not specifically left or right, it would be $60$. – RandomUser Apr 29 '14 at 18:08
• @RandomUser can you explain why the answer is 60? – Venus Apr 29 '14 at 18:09
• Because person A sitting to your left and person B sitting to your right is the same as person B sitting to your left and person A to your right. However that's not the way the question is worded. It specifically asks about left and right, so the answer is 120 for this question. I thought I would include it as an expansion to the idea. – RandomUser Apr 29 '14 at 18:11
• If you only cared about who was next to who, you would consider ABCDEF clockwise the same as AFEDCB clockwise, giving another division by 2 – Ross Millikan Apr 29 '14 at 18:12
• @RossMillikan So what is the correct answer? I stick to 120 but I'm sure about that. – Venus Apr 29 '14 at 18:15

You have $6$ spots:

_, _, _, _, _, _.

In the first spot, there are $6$ choices, any of the $6$ can fit, then in the next spot, there are only $5$ choices since one of the $6$ is at spot $1$.

In the end, you will have $6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$ choices.

Ignore what I have above, those are for non-round table problem.