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.
[Math] round table seating probability
permutationsprobability
Best Answer
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 $$\frac{6}{120}=\frac{1}{20}=\boxed{0.05}$$