15 persons are sitting in a row. In how many ways we can select three of them if adjacent persons are not selected?

Answer

To solve the problem of selecting three persons from a row of 15 such that no two selected persons are adjacent, we can follow these steps:

### Step 1: Understand the Problem
We have 15 persons sitting in a row, and we need to select 3 persons such that no two selected persons are next to each other.

### Step 2: Adjust for Non-Adjacent Selection
When we select a person, we must leave at least one seat empty on either side of that person to ensure that no two selected persons are adjacent. Therefore, if we select 3 persons, we will effectively block out 2 additional seats (one on each side of the selected persons).

### Step 3: Calculate the Effective Seats
If we select 3 persons, we will block out 2 adjacent seats for each selected person. Thus, the total number of seats occupied or blocked will be:
- 3 selected persons
- 2 blocked seats (one for each selected person, except the first and last selected)

This means we will have:
- Total seats = 15
- Blocked seats = 3 (selected) + 2 (blocked) = 5

The number of remaining seats available for selection will be:
\[ 15 - 5 = 10 \]
However, we need to consider that we can only select from the available seats after accounting for the blocked seats.

### Step 4: Reframe the Problem
To simplify the selection, we can think of it as selecting 3 persons from a reduced number of effective positions. Since we have blocked out 5 positions, we can think of the remaining positions as:
- 13 positions (15 - 2 = 13, because we need to account for the two empty seats that must be left adjacent to the selected persons).

### Step 5: Use Combinatorial Formula
Now we can use the combination formula to select 3 persons from these 13 effective positions:
\[ \text{Number of ways} = \binom{n - r + 1}{r} = \binom{13}{3} \]
Where \( n = 15 \) (total persons) and \( r = 3 \) (persons to select).

### Step 6: Calculate the Combination
Now we calculate \( \binom{13}{3} \):
\[ \binom{13}{3} = \frac{13!}{3!(13 - 3)!} = \frac{13!}{3! \cdot 10!} \]
This simplifies to:
\[ = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = \frac{1716}{6} = 286 \]

### Final Answer
Thus, the total number of ways to select 3 persons from 15 such that no two selected persons are adjacent is:
\[ \boxed{286} \]