To solve the problem, we'll break it down into steps:

Step 1: Treat the four children as a single unit

Since the four children must be seated together, we can treat them as a single block or unit. This changes the problem from arranging 6 people (2 parents and 4 children) to arranging 3 units (2 parents and 1 block of children).

Step 2: Arrange the units

There are $3!$ (3 factorial) ways to arrange the 3 units (2 parents and 1 block of children).

$3! = 3 \times 2 \times 1 = 6$

Step 3: Arrange the children within their block

Within the block of children, the 4 children can be arranged in $4!$ (4 factorial) different ways.

$4! = 4 \times 3 \times 2 \times 1 = 24$

Step 4: Multiply the results

To find the total number of different arrangements, we multiply the number of ways to arrange the units by the number of ways to arrange the children within their block.

$\text{Total arrangements} = 3! \times 4! = 6 \times 24 = 144$

Final Answer:

There are 144 different arrangements in which the children are seated together.

Would you like any more details or have any questions?

Here are some related questions to consider:

1. How many different arrangements are possible if no restrictions are placed on seating?
2. What if the parents must also sit together, how many arrangements are possible?
3. How many different arrangements are possible if exactly two of the children must sit together?
4. If the four children were not all to be seated together, how would you calculate the number of arrangements?
5. How would the answer change if there were three parents and four children?

Tip: When solving combinatorial problems, it's helpful to consider treating groups of items as single units to simplify the problem.