There are eight seats at a rectangular table - three at each of the longer sides and one
at each of the shorter sides. Six adults and two children sit around the table, with each
seating configuration equally likely. What is the probability that the two children will
end up seating next to each other at one of the longer sides of the table?

We are asked to determine the probability that two children will sit next to each other on one of the longer sides of a rectangular table with eight seats, where six adults and two children are seated. Let's break down the problem step by step.

### Total Possible Seating Configurations
The total number of ways to arrange all 8 people (6 adults and 2 children) around the table is:

\[
8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320
\]

However, since we are only concerned with which seats are occupied and not with the distinct identities of the adults, the seating arrangements of the adults are indistinguishable among themselves. Therefore, we need to divide by the number of ways to permute the 6 adults:

\[
6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720
\]

Thus, the total number of seating configurations is:

\[
\frac{8!}{6!} = \frac{40320}{720} = 56
\]

### Favorable Configurations (Children Next to Each Other on Longer Side)

Now, we want to find the number of favorable configurations where the two children sit next to each other on one of the longer sides. 

There are three seats on each longer side, which means there are two possible pairs of adjacent seats on each side where the children can sit next to each other (the first and second seat, or the second and third seat). Therefore, for each side, there are 2 possible seating arrangements for the children. Since there are 2 longer sides, the number of ways the children can sit next to each other on a longer side is:

\[
2 \times 2 = 4
\]

Once the two children have taken their seats, the 6 adults will occupy the remaining 6 seats in a unique way because their seating arrangement is indistinguishable. Therefore, there is only **1** way to seat the adults for each favorable seating arrangement of the children.

Thus, the number of favorable seating configurations is:

\[
4 \times 1 = 4
\]

### Probability Calculation

The probability is the ratio of favorable configurations to the total number of configurations:

\[
\text{Probability} = \frac{4}{56} = \frac{1}{14}
\]

### Final Answer

The probability that the two children will end up sitting next to each other on one of the longer sides of the table is:

\[
\boxed{\frac{1}{14}}
\]

Would you like more details or have any questions?

Here are 5 related questions to expand on this concept:
1. What if we wanted to calculate the probability that the children sit on opposite sides of the table?
2. How would the problem change if the table had more or fewer seats?
3. What if we made the children distinguishable from each other? How would that affect the outcome?
4. How would the solution differ if we wanted the children to sit next to each other at any side of the table, not just the longer sides?
5. What if we had more than two children? How would we approach the problem then?

**Tip:** When solving problems involving probability, always first consider the total number of possible outcomes before narrowing down to favorable outcomes.

There are eight seats at a rectangular table - three at each of the longer sides and one at each of the shorter sides. Six adults and two children sit around the table, with each seating configuration equally likely. What is the probability that the two children will end up seating next to each other at one of the longer sides of the table?

Determine the probability that two children sit next to each other on one of the longer sides of a rectangular table with eight seats. This combinatorics problem involves arranging 6 adults and 2 children and calculating the probability based on favorable and total seating configurations. Suitable for Grades 9-12.

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