round table seating probability

There are $6$ people, let's call them - (a,b,c,d,e,f), to sit at a round table. The number of ways they can arrange themselves is $(6-1)! = 5! = 120$ ways. What is the probability that person 'a' will have person 'b' sat to his immediate left, and person 'c' sat to his immediate right? I'm confused on how to go about this.

• I tried saying 3 people are in fixed position and remaining can arrange themselves 3! ways so 3 x 3! = 18 ways but not sure if this is correct – user3465413 Oct 9 '14 at 22:16
• Are you supposed to consider arrangements which are rotations around the table indistinguishable? For example, are $(a,b,c,d,e,f)$ and $(f,a,b,c,d,e)$ considered the same, since everyone will still be sitting next to the same people in both arrangements? The answer is implicitly yes, it seems because you are claiming that the total number of such arrangements is $(6-1)!$. – Tom Oct 9 '14 at 22:20
• You are supposed to consider rotations around the table to be indistinguishable. There are $5!$ distinct arrangements of six people around the table since there are six identical rotations of each arrangement, giving $$\frac{6!}{6} = 5!$$ arrangements. – N. F. Taussig Oct 9 '14 at 22:22
• @bof, I agree, but his $3!$ arrangements is correct if he answered yes (consistent with the $(6-1)!$ arrangements in the problem; however, the $3\times 3!$ answer is not (for a couple reasons) – Tom Oct 9 '14 at 22:26
• Not enough information given. We don't know the spacing of the people seated at the round table and for person 'a' to have person 'b' sat to his left, 'b' doesn't necessarily have to be seated next to 'a' so the answers given here are questionable (as in possibly wrong). – David Oct 10 '14 at 1:40

You have already counted the number of arrangements (with rotations being equivalent) correctly as $$\frac{6!}{6} = 120$$

Now you need to count the number of arrangements (with rotations being equivalent) in which $a$ has $b$ to his left and $c$ to his right.

To do this, treat the group of $a$, $b$, and $c$ as one person, and count the number of arrangements of the four people as $$\frac{4!}{4}=6$$

Your final probability is the number of "successes" ($a$ has $b$ to the left and $c$ to the right) divided by the total number of possibilities, or $$\frac{6}{120}=\frac{1}{20}=\boxed{0.05}$$

Actually, the way you had the question worded originally, there is no way to solve this question without more information and/or making assumptions. For example, there could be several seated positions that are to 'a's left but we don't know for sure because we don't know the spacings of how the people are seated (such as are they equally spaced around the round table?).

Given that $a$, $b$ and $c$ are placed in the order $bac$, with $a$ in a given position, there are $3!=6$ ways to arrange the remaining three people relative to these. The probability of such a placement ocuring is then just $\frac{6}{120}=\frac{1}{20}$.

• If there are "$6$ different positions for $a$", then the denominator has to be $6!=720$, not $5!=120$. Anyway, the correct answer is $\frac1{20}$. To get it more simply, with $a$ seated, there is one chance in $5$ that $b$ sits on his left, and then there is one chance in $4$ that $c$ sits on his right, so $\frac15\cdot\frac14=\frac1{20}$. – bof Oct 9 '14 at 22:21
• In other words, there are $5\cdot4=20$ choices for the two people sitting on $a$'s left and right; all $20$ are equally likely, so the probability of $(b,c)$ is $\frac1{20}$. – bof Oct 9 '14 at 22:26
• how can the denominator be 6! when in circular probability it has to (6-1)!?? – user3465413 Oct 9 '14 at 22:39