Question

8 teachers and 4 students are sitting around a circular table at random. The probability that no two students sit together is:

• A. \( \frac{7}{88} \)
• B. \( \frac{14}{33} \)
• C. \( \frac{8}{33} \)
• D. \( \frac{7}{33} \)

Hint:

In circular permutations, fix one element to remove symmetry in counting.

Answer 1

The Correct Option is D

Step 1: Arranging the teachers Since the arrangement is in a circular table, we fix one teacher’s seat and arrange the remaining 7 teachers in a circular manner. \[ \text{Ways to arrange 8 teachers} = (8-1)! = 7! = 5040 \]

Step 2: Placing the students To ensure that no two students sit together, we place them in the gaps between the teachers. Since there are 8 teachers, there are 8 available gaps. We need to select 4 out of these 8 gaps and arrange the students in them. - Choosing 4 gaps out of 8: \[ \text{Ways to choose 4 gaps} = \binom{8}{4} = \frac{8!}{4!(8-4)!} = \frac{8!}{4!4!} = 70 \] - Arranging the 4 students in these chosen gaps: \[ \text{Ways to arrange 4 students} = 4! = 24 \]
Step 3: Computing total arrangements Total ways to arrange 12 people (teachers and students) randomly in a circular seating: \[ (12-1)! = 11! = 39916800 \]
Step 4: Compute probability The probability of no two students sitting together is: \[ P = \frac{\text{Ways to arrange teachers and students in required manner}}{\text{Total possible arrangements}} \] \[ P = \frac{(7!) \times \binom{8}{4} \times 4!}{11!} \] \[ P = \frac{5040 \times 70 \times 24}{39916800} \] \[ P = \frac{7}{33} \] Thus, the correct answer is option (4).

Answer 2

To determine the probability that no two students sit together at a circular table where 8 teachers and 4 students are seated randomly, follow these steps:

1. Fix the position of one teacher to remove rotational symmetry in a circular permutation. This leaves us with arranging the remaining 7 teachers and 4 students.

2. Arrange 7 teachers around the table. The number of ways to do this is \(7!\).

3. These 7 teachers create 7 gaps between them where the students can sit. To ensure no two students sit together, each student must sit in a separate gap.

4. Choose 4 gaps out of these 7 to place the students. The number of ways to choose 4 gaps out of 7 is given by the combination formula \(\binom{7}{4}\).

5. Arrange the 4 students in these chosen gaps. The number of ways to arrange 4 students is \(4!\).

6. Compute the total number of favorable arrangements: \(7! \times \binom{7}{4} \times 4!\).

7. Compute the total number of possible arrangements of the 12 people around the table: \(11!\) (since one person's position is fixed).

8. The probability that no two students sit together is the ratio of favorable arrangements to total arrangements:

\[\frac{7! \times \binom{7}{4} \times 4!}{11!}=\frac{5040\times 35\times 24}{39916800}=\frac{7}{33}\]

Thus, the probability that no two students sit together is \(\frac{7}{33}\).