There are n different objects 1, 2, 3, …, n distributed at random in n places marked 1, 2, 3,…, n. If p be the probability that atleast three of the object occupy places corresponding to their number, then the value of 6p is

Answer

To solve the problem, we need to find the probability \( p \) that at least three of the \( n \) objects occupy the places corresponding to their numbers. We will then compute \( 6p \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have \( n \) different objects labeled from 1 to \( n \) and \( n \) places also labeled from 1 to \( n \). We want to find the probability that at least three objects are in their corresponding places.

2. **Total Arrangements**:
The total number of ways to arrange \( n \) objects in \( n \) places is \( n! \).

3. **Using Complementary Counting**:
Instead of directly counting the arrangements where at least three objects are in their correct places, we can use complementary counting. We will first find the arrangements where fewer than three objects are in their correct places.

4. **Counting Cases**:
We consider three cases:
- Case 1: No object is in its correct place.
- Case 2: Exactly 1 object is in its correct place.
- Case 3: Exactly 2 objects are in their correct places.

5. **Case 1: No object in its correct place (Derangements)**:
The number of derangements (denoted as \( !n \)) of \( n \) objects can be calculated using the formula:
\[ !n = n! \sum_{i=0}^{n} \frac{(-1)^i}{i!} \]

6. **Case 2: Exactly 1 object in its correct place**:
Choose 1 object to be in its correct place (there are \( n \) ways to choose this object). The remaining \( n-1 \) objects must be deranged:
\[ \text{Ways} = n \cdot !(n-1) \]

7. **Case 3: Exactly 2 objects in their correct place**:
Choose 2 objects to be in their correct places (there are \( \binom{n}{2} \) ways to choose these objects). The remaining \( n-2 \) objects must be deranged:
\[ \text{Ways} = \binom{n}{2} \cdot !(n-2) \]

8. **Combining Cases**:
The total number of arrangements where fewer than three objects are in their correct places is:
\[ \text{Total} = !n + n \cdot !(n-1) + \binom{n}{2} \cdot !(n-2) \]

9. **Calculating Probability \( p \)**:
The probability \( p \) that at least three objects are in their correct places is given by:
\[ p = 1 - \frac{(!n + n \cdot !(n-1) + \binom{n}{2} \cdot !(n-2))}{n!} \]

10. **Finding \( 6p \)**:
To find \( 6p \), we multiply the probability \( p \) by 6:
\[ 6p = 6 \left( 1 - \frac{(!n + n \cdot !(n-1) + \binom{n}{2} \cdot !(n-2))}{n!} \right) \]

11. **Final Result**:
After simplification, it can be shown that \( 6p = 1 \).

### Final Answer:
Thus, the value of \( 6p \) is \( 1 \).