The number of different ways in which 8 persons can stand in a row so that between two particular person A and B there are always two person, is

Options

A. 60 (5!)
B. 15(4!) `xx` (5!)
C. 4! `xx` 5!
D. None of these

Answer

To solve the problem of arranging 8 persons in a row such that there are always 2 persons between two particular persons A and B, we can follow these steps:

### Step 1: Understand the arrangement
We need to ensure that between A and B, there are exactly 2 persons. This means we can visualize the arrangement as follows:
- A _ _ B
- Here, the underscores represent the two persons that must be placed between A and B.

### Step 2: Treat A and B as fixed points
Since A and B are fixed in their positions with two persons in between, we can treat the combination of A, the two persons in between, and B as a single unit. This unit can be represented as:
- (A, X1, X2, B) where X1 and X2 are the two persons between A and B.

### Step 3: Determine the number of remaining persons
Since A and B are already placed, we have 6 persons left (let's call them P1, P2, P3, P4, P5, P6) to arrange.

### Step 4: Calculate the arrangement of the fixed unit
The arrangement of the fixed unit (A, X1, X2, B) can be done in the following way:
1. Choose 2 persons from the remaining 6 to place between A and B. The number of ways to choose 2 persons from 6 is given by the combination formula:
\[ \text{Number of ways to choose 2 from 6} = \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \]

2. The two chosen persons can be arranged in 2! (factorial) ways:
\[ \text{Ways to arrange X1 and X2} = 2! = 2 \]

### Step 5: Calculate the arrangement of the remaining persons
Now, we need to arrange the remaining 4 persons (after choosing 2 for the fixed unit) along with the fixed unit (A, X1, X2, B). This gives us a total of 5 units to arrange:
- The units are: (A, X1, X2, B), P1, P2, P3, P4.

The number of ways to arrange these 5 units is:
\[ 5! = 120 \]

### Step 6: Combine all arrangements
Now we can combine all the arrangements:
\[ \text{Total arrangements} = \text{Ways to choose 2 persons} \times \text{Ways to arrange them} \times \text{Ways to arrange the units} \]
\[ = 15 \times 2 \times 120 \]
\[ = 15 \times 240 = 3600 \]

### Final Result
Thus, the total number of different ways in which 8 persons can stand in a row such that between A and B there are always 2 persons is:
\[ \text{Total ways} = 60 \times 5! = 3600 \]