If `n` objects are arrange in a row, then the number of ways of selecting three of these objects so that no two of them are next to each other is

Options

A. `.^(n-2)C_(3)`
B. `.^(n-3)C_(3)+.^(n-3)C_(2)`
C. `((n-2)(n-3)(n-4))/(6)`
D. `.^(n)C_(2)`

Explanation

To solve the problem of selecting three objects from `n` objects arranged in a row such that no two selected objects are adjacent, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We need to select 3 objects from `n` objects arranged in a row, ensuring that no two selected objects are next to each other.

2. **Define Variables**:
- Let `x` be the number of objects before the first selected object.
- Let `y` be the number of objects between the first and second selected objects.
- Let `w` be the number of objects between the second and third selected objects.
- Let `u` be the number of objects after the third selected object.

3. **Set Up the Equation**: Since we are selecting 3 objects, we can express the total number of objects as:
\[ x + y + w + u + 3 = n \]
Rearranging gives us:
\[ x + y + w + u = n - 3 \quad \text{(Equation 1)} \]

4. **Apply the Condition of Non-Adjacency**:
- Since no two selected objects can be adjacent, we need at least one object between the first and second selected objects, and one object between the second and third selected objects. Thus:
- \( y \geq 1 \)
- \( w \geq 1 \)

5. **Transform the Variables**:
- Let \( a = y - 1 \) (so \( a \geq 0 \))
- Let \( b = w - 1 \) (so \( b \geq 0 \))

Substituting these into Equation 1 gives:
\[ x + (a + 1) + (b + 1) + u = n - 3 \]
Simplifying this results in:
\[ x + a + b + u = n - 5 \quad \text{(Equation 2)} \]

6. **Count the Non-Negative Solutions**:
- We need to find the number of non-negative integer solutions to Equation 2, which is:
\[ x + a + b + u = n - 5 \]
The number of solutions to the equation \( x_1 + x_2 + x_3 + x_4 = r \) in non-negative integers is given by the formula:
\[ \binom{r + k - 1}{k - 1} \]
where \( k \) is the number of variables. Here \( r = n - 5 \) and \( k = 4 \) (for \( x, a, b, u \)):
\[ \text{Number of solutions} = \binom{(n - 5) + 4 - 1}{4 - 1} = \binom{n - 2}{3} \]

7. **Final Answer**: Therefore, the number of ways to select three objects such that no two are adjacent is:
\[ \binom{n - 2}{3} \]