Three friends, Andy, Ben and Charlie, attend a graduation lunch with their parents.

1. Group families into single units and arrange them around the round table

Treat each family as a single unit. There are $3$ families (units) since there are $9$ people in total, meaning each friend is with their $2$ parents.
Number of ways to arrange $3$ units around a circle $= (3 - 1)! = 2! = 2$

2. Arrange the individuals within their respective family units to find the total arrangements

Within each family, the $3$ members can be arranged in $3!$ ways.
Since there are $3$ families, the total number of ways $= 2! \times (3!)^3 = 2 \times 216 = 432$

1. Determine the internal arrangements for Andy's and Ben's families.

Since Andy must stand exactly between his parents, they form a single block of 3: $(P_{A1} A P_{A2})$.
The number of internal arrangements for Andy's block is $2! = 2$.
Similarly, Ben and his parents form another block of 3 with Ben in the middle: $(P_{B1} B P_{B2})$.
The number of internal arrangements for Ben's block is $2! = 2$.

2. Identify the distinct units to be arranged.

Treat Andy's block and Ben's block as 1 unit each. Along with Charlie ($C$), Charlie's father ($F_C$) and Charlie's mother ($M_C$), there are $2 + 3 = 5$ units to arrange in total in the row.

3. Calculate the number of ways Charlie is beside one specific parent.

If $C$ is beside his father $F_C$, we group them as a single block $(C, F_C)$. This block has $2! = 2$ internal arrangements ($C F_C$ or $F_C C$).
With this block, there are now 4 units to arrange: Andy's block, Ben's block, $M_C$, and the $(C, F_C)$ block.
Number of ways $C$ is beside $F_C$ $= 4! \times 2 = 48$.
By identical logic, the number of ways $C$ is beside his mother $M_C$ is also $4! \times 2 = 48$.

4. Calculate the number of ways Charlie is beside both parents.

If $C$ is beside both $F_C$ and $M_C$, he must be exactly in the middle of them, forming a block of 3: $(F_C, C, M_C)$.
This block has $2! = 2$ internal arrangements ($F_C C M_C$ or $M_C C F_C$).
There are now 3 units to arrange: Andy's block, Ben's block, and the $(F_C, C, M_C)$ block.
Number of ways $C$ is beside both parents $= 3! \times 2 = 12$.

5. Use the Principle of Inclusion-Exclusion to find the total valid arrangements.

Using the Principle of Inclusion-Exclusion, the number of ways $C$ is beside at least one parent is:
$n(C \text{ beside } F_C) + n(C \text{ beside } M_C) - n(C \text{ beside both})$
$= 48 + 48 - 12 = 84$

To find the final total number of arrangements, we multiply by the internal ways for Andy's and Ben's blocks:
$\text{Total ways} = 84 \times 2 \times 2 = 336$