The Ang family, the Bao family and the Chew family have three, five and four members respectively. The three families are invited to a wedding dinner. Due to venue constraints, only 2 members from each family can attend the dinner.

1. Set up the expression to choose 2 members from each family

Number of ways = ${}^3C_2 \times {}^5C_2 \times {}^4C_2$

2. Calculate the total number of ways

$= 3 \times 10 \times 6$
$= 180$

1. Find the total number of ways to arrange the 6 distinct attendees at a round table.

$\text{Total arrangements} = (6 - 1)! = 5! = 120$

2. Find the number of arrangements where the 2 Ang family members are seated next to each other.

Treat the 2 Ang family members as a single unit. There are now 5 units in total (1 Ang unit + 4 other distinct people).

$\text{Arrangements of 5 units in a circle} = (5 - 1)! = 4! = 24$

The 2 Ang family members can swap positions within their unit in $2!$ ways.

$\text{Arrangements where they are together} = 24 \times 2! = 48$

3. Calculate the probability that they are not next to each other using the complement rule.

$\text{Probability (together)} = \frac{48}{120} = \frac{2}{5}$

$\text{Probability (not together)} = 1 - \frac{2}{5} = \frac{3}{5}$

1. Identify the possible family member breakdowns that sum to 7, with at least 2 from each of the 3 families.

Since each of the 3 families must have at least 2 members attending, this accounts for $2 + 2 + 2 = 6$ members.
To select 7 members in total, exactly one family must have 3 members attending, while the other two have 2.

The possible cases are:
Case 1: 3 Ang, 2 Bao, 2 Chew
Case 2: 2 Ang, 3 Bao, 2 Chew
Case 3: 2 Ang, 2 Bao, 3 Chew

2. Calculate the number of ways to select the members for each case.

Case 1: ${}^3C_3 \times {}^5C_2 \times {}^4C_2 = 1 \times 10 \times 6 = 60$ ways
Case 2: ${}^3C_2 \times {}^5C_3 \times {}^4C_2 = 3 \times 10 \times 6 = 180$ ways
Case 3: ${}^3C_2 \times {}^5C_2 \times {}^4C_3 = 3 \times 10 \times 4 = 120$ ways

3. Sum the possibilities across all valid cases to find the total number of ways.

Total number of ways $= 60 + 180 + 120 = 360$

1. Deduce the family group sizes of the selected attendees.

Since 7 attendees are selected with at least 2 from each of the 3 families, the only possible distribution of attendees per family is $3$, $2$, and $2$.

2. Calculate the number of ways to arrange the attendees so that family members are together.

To keep members of the same family together, we treat each family group as a single block.
Number of ways to arrange the 3 family blocks $= 3! = 6$
Number of ways to arrange members within their respective family blocks $= 3! \times 2! \times 2! = 24$
Total favourable arrangements $= 3! \times (3! \times 2! \times 2!) = 144$

3. Find the total number of arrangements for the 7 attendees and compute the probability.

Total number of unrestricted arrangements for 7 people in a row $= 7! = 5040$

$\text{Probability} = \frac{144}{5040} = \frac{1}{35}$