Let A be a set containing elements. A subset P of the set A is chosen at random. The set A is reconstructed by replacing the elements of P, and another subset Q of A is chosen at random. The probability that `P cap Q` contains exactly `m (m lt n)` elements, is

Options

A. `(3^(n-m))/(4^(n))`
B. `(.^(n)C_(m).3^(m))/(4^(n))`
C. `(.^(n)C_(m).3^(n-m))/(4^(n))`
D. none of these

Explanation

To solve the problem, we need to determine the probability that the intersection of two randomly chosen subsets \( P \) and \( Q \) from a set \( A \) contains exactly \( m \) elements, where \( m < n \) and \( n \) is the total number of elements in set \( A \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- Let \( A \) be a set with \( n \) elements.
- We randomly choose a subset \( P \) from \( A \).
- After choosing \( P \), we reconstruct \( A \) and choose another subset \( Q \) from \( A \).
- We need to find the probability that the intersection \( P \cap Q \) contains exactly \( m \) elements.

2. **Total Number of Ways to Choose Subsets**:
- Each element of \( A \) can either be included or not included in a subset. Therefore, for each of the \( n \) elements, there are 4 possibilities (it can be in \( P \), in \( Q \), in both \( P \) and \( Q \), or in neither).
- Thus, the total number of ways to choose subsets \( P \) and \( Q \) is \( 4^n \).

3. **Choosing Elements in the Intersection**:
- We want \( P \cap Q \) to have exactly \( m \) elements.
- We can select \( m \) elements from the \( n \) elements of \( A \) to be in the intersection \( P \cap Q \). The number of ways to choose these \( m \) elements is given by \( \binom{n}{m} \).

4. **Distributing Remaining Elements**:
- After choosing \( m \) elements for the intersection, there are \( n - m \) elements left.
- Each of these remaining \( n - m \) elements can either be in \( P \) only, in \( Q \) only, or in neither. They cannot be in both because we already selected the intersection.
- Therefore, for each of the \( n - m \) remaining elements, there are 3 choices (in \( P \), in \( Q \), or in neither).
- The number of ways to distribute these \( n - m \) elements is \( 3^{n - m} \).

5. **Calculating the Favorable Outcomes**:
- The total number of favorable outcomes for \( P \cap Q \) having exactly \( m \) elements is the product of the ways to choose the intersection and the ways to distribute the remaining elements:
\[ \text{Favorable Outcomes} = \binom{n}{m} \cdot 3^{n - m} \]

6. **Calculating the Probability**:
- The probability \( P \) that \( P \cap Q \) contains exactly \( m \) elements is given by the ratio of the number of favorable outcomes to the total outcomes:
\[ P(P \cap Q \text{ has exactly } m \text{ elements}) = \frac{\binom{n}{m} \cdot 3^{n - m}}{4^n} \]

### Final Answer:
The probability that \( P \cap Q \) contains exactly \( m \) elements is:
\[ \frac{\binom{n}{m} \cdot 3^{n - m}}{4^n} \]