Out of n persons sitting at a round table, three, A, B, C are chosen at random. The chance that no two of these are sitting next to one another is

Options

A. `((n-4)(n-5))/(2xx(n-1))`
B. `((n-4)(n-5))/(2xx(n-1)(n-2)(n-3))`
C. `((n-4)(n-5))/((n-1)(n-2))`
D. `(n(n-4))/((n-1)(n-2))`

Answer

To solve the problem of finding the probability that no two of the chosen persons A, B, and C are sitting next to each other at a round table with n persons, we can follow these steps:

### Step-by-Step Solution:

1. **Total Arrangements**:
In a circular arrangement of n persons, the total number of ways to arrange them is given by (n-1)!. This is because one person can be fixed to break the circular symmetry.

**Hint**: Remember that in circular permutations, we fix one position to avoid counting rotations as different arrangements.

2. **Choosing 3 Persons**:
We need to choose 3 persons from the n persons. The number of ways to choose 3 persons from n is given by \( \binom{n}{3} \).

**Hint**: Use the combination formula \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \) to calculate the number of ways to choose r persons from n.

3. **Arranging Remaining Persons**:
After choosing 3 persons, we have \( n-3 \) persons left. We can arrange these \( n-3 \) persons in a circular manner, which can be done in \( (n-4)! \) ways.

**Hint**: Again, remember to adjust for circular arrangements by using \( (n-4)! \) instead of \( (n-3)! \).

4. **Identifying Gaps**:
When \( n-3 \) persons are seated, they create \( n-3 \) gaps between them (including the gap between the last and the first person). We need to place the 3 chosen persons in these gaps such that no two of them are adjacent.

**Hint**: Visualize the gaps created by the seated persons to understand how to place the chosen persons without them being adjacent.

5. **Choosing Gaps**:
We can choose 3 gaps from the \( n-3 \) available gaps to place our chosen persons. The number of ways to choose 3 gaps from \( n-3 \) is \( \binom{n-3}{3} \).

**Hint**: This is another application of the combination formula.

6. **Arranging Chosen Persons**:
The 3 chosen persons can be arranged among themselves in \( 3! \) ways.

**Hint**: Factorial of the number of persons gives the number of ways to arrange them.

7. **Calculating Favorable Outcomes**:
The total number of favorable arrangements where no two chosen persons are adjacent is given by:
\[ \text{Favorable outcomes} = (n-4)! \times \binom{n-3}{3} \times 3! \]

**Hint**: Combine the arrangements of remaining persons, the choices of gaps, and the arrangements of chosen persons.

8. **Calculating Probability**:
The probability \( P \) that no two of the chosen persons are sitting next to each other is given by:
\[ P = \frac{\text{Favorable outcomes}}{\text{Total arrangements}} = \frac{(n-4)! \times \binom{n-3}{3} \times 3!}{(n-1)!} \]

**Hint**: This is a direct application of the probability formula \( P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} \).

9. **Simplifying the Probability**:
Substitute the combination and simplify:
\[ P = \frac{(n-4)! \times \frac{(n-3)(n-4)(n-5)}{3!}}{(n-1)(n-2)(n-3)(n-4)!} = \frac{(n-4)(n-5)}{(n-1)(n-2)} \]

**Hint**: Cancel out common terms in the numerator and denominator to simplify the expression.

### Final Answer:
The probability that no two of the chosen persons A, B, and C are sitting next to one another is:
\[ P = \frac{(n-4)(n-5)}{(n-1)(n-2)} \]