A class of problems that requires one to use permutations and combinations for computing probability has at its heart notion of sets and subsets. They are generally an abstract formulation of some concrete situation and require the application of counting techniques.
A is a set containing 10 elements. A subset `P_(1)` of A is chosen and the set A is chosen and the set A is reconstructed by replacing the elements of `P_(1)`. A subset `P_(2)` of A is chosen and again the set A is reconstructed by replacing the elements of `P_(2)`. This process is continued by choosing subsets `P_(1), P_(2), ... P_(10)`.
The probability that `P_(i)capP_(j)=phi" "AA" "i nej,i,j=1,2,....,10` is

Options

A. A. `(11^(10))/(4^(100))`
B. B. `(10^(11))/(2^(100))`
C. C. `(11^(10))/(2^(100))`
D. D, `(10^(11))/(4^(100))`

Answer

To solve the problem, we need to calculate the probability that for subsets \( P_i \) and \( P_j \) chosen from a set \( A \) of 10 elements, the intersection \( P_i \cap P_j \) is empty, given that \( i \neq j \).

### Step-by-Step Solution:

1. **Understanding the Set and Subsets**:
- Let \( A \) be a set containing 10 elements: \( A = \{ a_1, a_2, a_3, \ldots, a_{10} \} \).
- We will choose subsets \( P_1, P_2, \ldots, P_{10} \) from set \( A \).

2. **Choices for Each Element**:
- For each element \( a_k \) in set \( A \), there are three possibilities when forming the subsets:
- \( a_k \) belongs to subset \( P_i \).
- \( a_k \) belongs to subset \( P_j \) (where \( j \neq i \)).
- \( a_k \) belongs to neither subset.
- Therefore, for each element, there are 3 choices.

3. **Total Choices for All Elements**:
- Since there are 10 elements in set \( A \) and each has 3 independent choices, the total number of ways to choose subsets \( P_1, P_2, \ldots, P_{10} \) is:
\[ 3^{10} \]

4. **Total Possible Subset Combinations**:
- Each element can either be included or excluded from any of the subsets. Thus, for each of the 10 elements, there are 2 choices (included or excluded) for each subset. Therefore, the total number of combinations of subsets is:
\[ 2^{10 \times 10} = 2^{100} \]

5. **Calculating the Probability**:
- The probability that \( P_i \cap P_j = \emptyset \) for \( i \neq j \) is given by the ratio of the favorable outcomes to the total outcomes:
\[ P(P_i \cap P_j = \emptyset) = \frac{3^{10}}{2^{100}} \]

6. **Simplifying the Probability**:
- We can express \( 3^{10} \) and \( 2^{100} \) in terms of powers:
\[ P(P_i \cap P_j = \emptyset) = \frac{3^{10}}{2^{100}} = \frac{59049}{1267650600228229401496703205376} \]

### Final Answer:
Thus, the probability that \( P_i \cap P_j = \emptyset \) for \( i \neq j \) is:
\[ P(P_i \cap P_j = \emptyset) = \frac{3^{10}}{2^{100}} \]