Three married couples are to be randomly seated around a circular table. What is the probability that at least one of the couples will be seated next to one another?

When seating in a circular permutations, there are (n-1)! Ways to do so without restriction.

In this case, 5!=120 ways to seat the 3 couples(6 people) without restriction.

The opposite of 'at least one' is none. So, find the probability of no couples sitting together

and subtract from 1. Or find the number of ways 1 couple can sit together, 2 couples, and

three couples can sit together and add them up.

For two couples together, we can choose 2 pairs in C(3,2)=3 ways and this pair can be

arranged in 2 ways. The next pair in 2 ways and the next pair in two ways.

So, we have 3\cdot{2^{3}}=24 ways for two couples to be seated together.

Following the same logic, there are 3\cdot{2^{4}} ways for 1 couple to be

seated together and 2^{4}=16 ways for all three to be seated together.

Therefore, we have 16+24+48=88 ways for at least one couple to be seated together.

The probability would be 88/120=11/15.