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 3 persons from 15 seated in a row such that no two selected persons are adjacent, we can follow these steps:

### Step 1: Understand the problem
We need to select 3 persons from a total of 15, ensuring that no two selected persons are next to each other. This means that if we select a person, we cannot select the person immediately to their left or right.

### Step 2: Represent the selected persons
Let's denote the selected persons as \( P_1, P_2, \) and \( P_3 \). Since they cannot be adjacent, we can think of placing "gaps" between the selected persons.

### Step 3: Create gaps
If we select 3 persons, we will have 2 gaps between them (since there are 3 selected persons). Additionally, we will have gaps before the first selected person and after the last selected person. Thus, we can visualize the arrangement as follows:

- Let \( x_1 \) be the number of unselected persons before \( P_1 \).
- Let \( x_2 \) be the number of unselected persons between \( P_1 \) and \( P_2 \) (at least 1 since they cannot be adjacent).
- Let \( x_3 \) be the number of unselected persons between \( P_2 \) and \( P_3 \) (at least 1 since they cannot be adjacent).
- Let \( x_4 \) be the number of unselected persons after \( P_3 \).

### Step 4: Set up the equation
The total number of persons is 15. Therefore, we have:
\[ x_1 + (x_2 + 1) + (x_3 + 1) + x_4 + 3 = 15 \]
This simplifies to:
\[ x_1 + x_2 + x_3 + x_4 = 11 \]
where \( x_2 \) and \( x_3 \) must be at least 1.

### Step 5: Substitute variables
Let \( y_2 = x_2 - 1 \) and \( y_3 = x_3 - 1 \). Then \( y_2 \) and \( y_3 \) can be 0 or more. The equation now becomes:
\[ x_1 + (y_2 + 1) + (y_3 + 1) + x_4 = 11 \]
which simplifies to:
\[ x_1 + y_2 + y_3 + x_4 = 9 \]

### Step 6: Count the non-negative integer solutions
The number of non-negative integer solutions to the equation \( x_1 + y_2 + y_3 + x_4 = 9 \) can be found using the "stars and bars" theorem. The number of solutions is given by:
\[ \binom{n + k - 1}{k - 1} \]
where \( n \) is the total number (9) and \( k \) is the number of variables (4). Thus, we have:
\[ \binom{9 + 4 - 1}{4 - 1} = \binom{12}{3} \]

### Step 7: Calculate the binomial coefficient
Now we calculate \( \binom{12}{3} \):
\[ \binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = \frac{1320}{6} = 220 \]

### Final Answer
Therefore, the total number of ways to select 3 persons from 15 such that no two selected persons are adjacent is **220**.

---