Two married couples, two single adults and two children formed a team of 8 to take part in a series of games.

1. Determine the total number of entities to arrange by grouping the couples.

Since each married couple must be seated together, we treat each couple as a single unit.
This gives us:

• 2 couple units
• 2 single adults
• 2 children
  Total entities $= 2 + 2 + 2 = 6$ units.

2. Find the number of ways to arrange the units in a circle.

The number of ways to arrange $n$ units in a circle is $(n - 1)!$.
For 6 units: $(6 - 1)! = 5! = 120$ ways.

3. Multiply by the internal arrangements of the couples to find the total number of ways.

Within each couple unit, the two people can switch places in $2!$ ways. Since there are $2$ couples, we multiply by $2!$ twice.
Total number of arrangements $= 5! \times 2! \times 2! = 120 \times 2 \times 2 = 480$

1. Find the total number of ways to choose a group of 3 from 8 people without restrictions.

Total unrestricted groups = $\binom{8}{3}$
$\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$

2. Find the number of ways to choose a group that contains a married couple.

To form a group of 3 with a married couple, we must choose 1 of the 2 couples (which takes up 2 spots), and then choose 1 person from the remaining 6 people.
Number of such groups = $\binom{2}{1} \times \binom{6}{1}$
$2 \times 6 = 12$

3. Subtract the number of restricted groups from the total to find the number of valid groups.

Number of groups without a married couple = $56 - 12 = 44$

1. Select 4 numbers for the children and single adults, and determine their valid assignments.

We first choose $4$ numbers out of the $8$ available for the $2$ children and $2$ single adults: $\binom{8}{4}$ ways.

For any set of $4$ chosen numbers, the $2$ largest must go to the children and the $2$ smallest must go to the single adults.
The children can be assigned the $2$ largest numbers in $2!$ ways.
The single adults can be assigned the $2$ smallest numbers in $2!$ ways.

Ways to assign these $4$ people $= \binom{8}{4} \times 2! \times 2! = 70 \times 2 \times 2 = 280$

2. Assign the remaining numbers to the married adults and calculate the total.

The remaining $4$ numbers can be freely assigned to the $4$ married adults in $4!$ ways.

Total number of ways $= 280 \times 4! = 280 \times 24 = 6720$