`(n>=5)` persons are sitting in a row. Three of these are selected at random. The probability that no two of the selected persons sit together is

Options

A. `(.^(n-3)P_2)/(.^nP_2)`
B. `(.^(n-3)C_2)/(.^nC_2)`
C. `((n-3)(n-4))/(n(n-1))`
D. `(.^(n-3)C_2)/(.^nP_2)`

Answer

To solve the problem of finding the probability that no two of the selected persons sit together when selecting 3 persons from n (where n ≥ 5) persons sitting in a row, we can follow these steps:

### Step 1: Understand the arrangement
We have n persons sitting in a row. We need to select 3 persons such that no two of them are adjacent.

### Step 2: Create gaps for selection
To ensure that no two selected persons are sitting together, we can visualize the arrangement as follows:
- When we select 3 persons, they will create gaps around them.
- For 3 selected persons, there will be 4 gaps: one before the first selected person, one between the first and second, one between the second and third, and one after the third.

### Step 3: Place the remaining persons
Since we have n persons and we are selecting 3, we have (n - 3) persons left. To ensure that no two selected persons are adjacent, we need to place these (n - 3) persons into the 4 gaps created by the selected persons.

### Step 4: Calculate the number of ways to distribute the remaining persons
The problem now reduces to distributing (n - 3) indistinguishable persons into 4 distinguishable gaps. This can be done using the "stars and bars" combinatorial method.

The number of ways to distribute (n - 3) indistinguishable objects (remaining persons) into 4 distinguishable boxes (gaps) is given by the formula:
\[ \binom{(n - 3) + (4 - 1)}{4 - 1} = \binom{n - 3 + 3}{3} = \binom{n}{3} \]

### Step 5: Calculate total ways to select 3 persons
The total number of ways to select any 3 persons from n persons is given by:
\[ \binom{n}{3} \]

### Step 6: Calculate the probability
The probability that no two of the selected persons sit together is the ratio of the number of favorable outcomes (ways to select 3 persons such that no two are adjacent) to the total outcomes (ways to select any 3 persons):
\[ P(\text{no two together}) = \frac{\text{Ways to select 3 persons such that no two are adjacent}}{\text{Total ways to select 3 persons}} = \frac{\binom{n - 2}{3}}{\binom{n}{3}} \]

### Step 7: Simplify the probability
Using the combinations formula, we can simplify:
\[ P(\text{no two together}) = \frac{\frac{(n - 2)(n - 3)(n - 4)}{6}}{\frac{n(n - 1)(n - 2)}{6}} = \frac{(n - 3)(n - 4)}{n(n - 1)} \]

Thus, the final probability that no two of the selected persons sit together is:
\[ P(\text{no two together}) = \frac{(n - 3)(n - 4)}{n(n - 1)} \]