Jim, May and 8 other friends will be sitting at the round table. How many ways can you arrange them such that there is always exactly one person between Jim and May?
My solution:
I fixed one point to one random guy. There are 9 people left to place. I grouped together Jim May and the person between them as 1. 7!(7 groups) x 7(arrange the 7 people x 2(Jim and May). What did I do wrong?
On
First select one of the person to sit between Jim and May in $\binom {8}{1}$ ways and then form a string of Jim, May and the selected person so that we treat them as a single person. Before starting the circular permutation part, we note that Jim, and May can be permuted among themselves in the string so formed in $2!$ ways. Now we have 8 people to be permuted in circle (Treating the string we formed as a single person) which can be done in $7!$ ways.
Hence the answer would be $$\binom {8}{1}* 2!* 7!=80640$$
There are eight ways to choose the person between Jim and May. There are two ways to seat Jim and May on either side of that person, depending on whether Jim sits to the right or left of that person. Relative to that block of three people, there are $7!$ ways to seat the other seven people as we move clockwise around the circle. Hence, the number of seating arrangements in which there is exactly one person between Jim and May is $$8 \cdot 2 \cdot 7! = 2 \cdot 8!$$