A group of ten people consists of four single women, two single men and 2 couples. The ten people are arranged randomly in a circle.

1. Find the total number of ways to arrange the 10 people in a circle without restrictions.

There are $4 + 2 + 2(2) = 10$ people in total.
Total arrangements in a circle $= (10 - 1)! = 9! = 362\,880$

2. Find the number of ways to arrange the remaining 6 people in a circle.

To ensure the 4 single women are all separated, we must first arrange the other 6 people (2 single men and 4 people from the couples) in a circle.
$\text{Number of ways} = (6 - 1)! = 5! = 120$

3. Find the number of ways to arrange the 4 single women into the gaps between the 6 people.

The 6 people seated in a circle create 6 gaps between them. The 4 single women must be placed into 4 of these 6 gaps.
$\text{Number of ways} = {^6\text{P}_4} = 360$

4. Calculate the final probability.

Total favourable arrangements $= 5! \times {^6\text{P}_4} = 120 \times 360 = 43\,200$
$\text{Probability} = \frac{43\,200}{362\,880} = \frac{5}{42}$

1. Find the probability that the 4 single women are together.

Let $A$ be the event that the $4$ single women are together, and $B$ be the event that the $2$ single men are together.
The total number of unrestricted ways to arrange $10$ people in a circle is $(10-1)! = 9!$.

For event $A$, we treat the $4$ women as $1$ block. Together with the remaining $6$ people, this gives $1 + 6 = 7$ entities to arrange in a circle. The $4$ women can be arranged internally in $4!$ ways:
$P(A) = \frac{(7-1)! \times 4!}{9!} = \frac{6! \times 4!}{9!} = \frac{1}{21}$

2. Find the probability that the 2 single men are together.

For event $B$, treating the $2$ single men as $1$ block gives $1 + 8 = 9$ entities to arrange. The $2$ men can be arranged internally in $2!$ ways:
$P(B) = \frac{(9-1)! \times 2!}{9!} = \frac{8! \times 2!}{9!} = \frac{2}{9}$

3. Find the probability that both groups are together.

For the intersection $A \cap B$, we treat the $4$ single women as $1$ block and the $2$ single men as $1$ block. Along with the remaining $4$ people (from the couples), this gives $1 + 1 + 4 = 6$ entities. Both blocks have their own internal arrangements:
$P(A \cap B) = \frac{(6-1)! \times 4! \times 2!}{9!} = \frac{5! \times 4! \times 2!}{9!} = \frac{1}{63}$

4. Calculate the probability that exactly one of the events occurs (either A or B, but not both).

The probability that exactly one of these events occurs is given by the symmetric difference:
$P(A \text{ or } B \text{ but not both}) = P(A) + P(B) - 2P(A \cap B)$
$= \frac{1}{21} + \frac{2}{9} - 2\left(\frac{1}{63}\right)$
$= \frac{3}{63} + \frac{14}{63} - \frac{2}{63} = \frac{15}{63} = \frac{5}{21}$

1. Determine the number of distinct relative positions for the empty chair.

With 10 chairs arranged in an alternating pattern of sashes (S, N, S, N, ...), there is a 5-fold rotational symmetry for the table.
The single empty chair can be chosen in exactly $2$ distinct ways that cannot be mapped to each other by rotation: either it is a chair with a sash, or a chair without a sash.

2. Calculate the total number of seating arrangements for the 9 people.

Once the type of the empty chair is fixed, the rotational symmetry is broken and the remaining 9 chairs become distinct linear positions.
The 9 distinct people can be seated in these 9 chairs in $9!$ ways.
$\text{Total arrangements} = 2 \times 9! = 2 \times 362\,880 = 725\,760$