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 one another at a round table with n persons, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the 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 eliminate the circular permutations.

**Hint**: Remember that in circular permutations, fixing one position helps simplify the counting.

2. **Choosing 3 Persons**:
We need to choose 3 persons (A, B, C) from these n persons. The total 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 the Remaining Persons**:
After choosing A, B, and C, we are left with \(n-3\) persons. To ensure that A, B, and C are not sitting next to each other, we need to arrange the remaining \(n-3\) persons first. The number of ways to arrange these \(n-3\) persons in a circular manner is \((n-4)!\).

**Hint**: Again, fix one person to account for the circular arrangement.

4. **Creating Gaps for A, B, and C**:
When \(n-3\) persons are arranged, they create \(n-3\) gaps (one gap before each person and one after the last person). We want to place A, B, and C in these gaps such that no two of them are adjacent.

The number of ways to choose 3 gaps from \(n-3\) gaps is \(\binom{n-3}{3}\).

**Hint**: Visualize the arrangement to see how gaps are formed.

5. **Arranging A, B, and C**:
Once we have chosen 3 gaps, A, B, and C can be arranged in these gaps in \(3!\) different ways.

**Hint**: Remember that the number of arrangements of r distinct objects is given by \(r!\).

6. **Calculating the Favorable Outcomes**:
The total number of favorable arrangements where A, B, and C are not sitting next to each other is given by:
\[ (n-4)! \times \binom{n-3}{3} \times 3! \]

**Hint**: Combine the results from the previous steps to find the total favorable outcomes.

7. **Calculating the Probability**:
The probability that A, B, and C are not sitting next to each other is given by the ratio of favorable outcomes to total outcomes:
\[ P = \frac{(n-4)! \times \binom{n-3}{3} \times 3!}{(n-1)!} \]

**Hint**: Simplify the expression by canceling out common factorial terms.

8. **Final Simplification**:
Simplifying the probability expression:
\[ P = \frac{(n-4)! \times \frac{(n-3)(n-4)(n-5)}{3!} \times 6}{(n-1)(n-2)(n-3)(n-4)!} \]
This simplifies to:
\[ P = \frac{(n-4)(n-5)}{(n-1)(n-2)} \]

**Hint**: Make sure to carefully simplify the fractions and factorials.

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