Let N denote the number of ways in which n boys can be arranged in a line so that 3 particular boys are seperated then N is equal to

Options

A. `""^(n-2)P_(3)(n-3)!`
B. `((n-2)!(n-3)!)/(n-5!)`
C. `""^(n-2)C_(3)3!(n-3)!`
D. `(n-3)(n-4)(n-2)!`

Answer

To find the number of ways in which \( n \) boys can be arranged in a line such that 3 particular boys are separated, we can follow these steps:

### Step 1: Calculate Total Arrangements Without Restrictions
First, we calculate the total number of arrangements of \( n \) boys without any restrictions. This is given by \( n! \).

### Step 2: Calculate Arrangements Where 3 Particular Boys Are Together
Next, we treat the 3 particular boys (let's call them A, B, and C) as a single unit or block. This means we now have \( n - 2 \) units to arrange (the block of 3 boys plus the remaining \( n - 3 \) boys).

The number of arrangements of these \( n - 2 \) units is \( (n - 2)! \).

Within the block, the 3 boys can be arranged among themselves in \( 3! \) ways. Therefore, the total arrangements where A, B, and C are together is:
\[ (n - 2)! \times 3! \]

### Step 3: Calculate Arrangements Where 3 Particular Boys Are Separated
To find the arrangements where the 3 boys are separated, we subtract the arrangements where they are together from the total arrangements:
\[ N = n! - (n - 2)! \times 3! \]

### Step 4: Simplify the Expression
We can factor out \( (n - 2)! \) from the second term:
\[ N = n! - 6(n - 2)! \]
Since \( n! = n \times (n - 1) \times (n - 2)! \), we can rewrite \( N \) as:
\[ N = n(n - 1)(n - 2)! - 6(n - 2)! \]
Factoring out \( (n - 2)! \):
\[ N = (n - 2)! \left[n(n - 1) - 6\right] \]

### Final Result
Thus, the number of ways in which \( n \) boys can be arranged in a line such that 3 particular boys are separated is:
\[ N = (n - 2)! \left[n(n - 1) - 6\right] \]