Three children, each accompanied by a guardian, seek admission in a school. The principal want to interview all the 6 persons one after the other subject to the condition that no child is interviewed before its guradian. In how many ways can this be done-

Options

A. 60
B. 90
C. 120
D. 180

Explanation

To solve the problem of how many ways the principal can interview three children and their respective guardians while ensuring that no child is interviewed before their guardian, we can follow these steps:

### Step 1: Understand the Problem
We have 3 children (C1, C2, C3) and their respective guardians (G1, G2, G3). The condition is that each guardian must be interviewed before their corresponding child.

### Step 2: Total Arrangements Without Restrictions
If there were no restrictions, the total number of arrangements of 6 people (3 children + 3 guardians) would be calculated using the factorial of the total number of people:
\[ 6! = 720 \]

### Step 3: Apply the Condition
Since each guardian must be interviewed before their respective child, we can consider each pair (G1, C1), (G2, C2), and (G3, C3) as a block where the guardian comes first.

### Step 4: Count the Valid Arrangements
For each pair (G, C), there are 2 arrangements: G before C or C before G. However, since we only want the arrangements where G is before C, we can think of it as fixing the order for each pair.

Thus, for each of the 3 pairs, we only have 1 valid arrangement out of the 2 possible arrangements. Therefore, we can divide the total arrangements by \(2^3\) (since there are 3 pairs):
\[ \text{Valid arrangements} = \frac{6!}{2^3} = \frac{720}{8} = 90 \]

### Step 5: Final Answer
The total number of ways the principal can interview the 6 persons while satisfying the condition is:
\[ \boxed{90} \]