A binary operation is chosen at random from the set of all binary operations on a set A containing n elements. The probability that the binary operation is commutative, is

Options

A. `(n^(n))/(n^(n^(2)))`
B. `(n^(n//2))/(n^(n^(2)))`
C. `(n^(n//2))/(n^(n^(2//2)))`
D. none of these

Answer

To solve the problem, we need to determine the probability that a randomly chosen binary operation on a set \( A \) with \( n \) elements is commutative.

### Step-by-Step Solution:

1. **Understanding Binary Operations**:
A binary operation on a set \( A \) is a function that combines any two elements of \( A \) to form another element of \( A \). If \( A \) has \( n \) elements, the binary operation can be represented as a function \( f: A \times A \rightarrow A \).

2. **Total Number of Binary Operations**:
The total number of binary operations on a set \( A \) with \( n \) elements can be calculated as follows:
- The set \( A \times A \) has \( n^2 \) pairs (since there are \( n \) choices for the first element and \( n \) choices for the second).
- For each of these \( n^2 \) pairs, we can choose any of the \( n \) elements in \( A \) as the output.
- Therefore, the total number of binary operations is \( n^{n^2} \).

3. **Counting Commutative Binary Operations**:
A binary operation \( f \) is commutative if \( f(a, b) = f(b, a) \) for all \( a, b \in A \).
- For commutative operations, we only need to define the operation for pairs \( (a, b) \) where \( a \leq b \).
- The number of such pairs is given by the formula for combinations: \( \binom{n}{2} + n \) (the \( n \) accounts for the pairs where both elements are the same).
- This simplifies to \( \frac{n(n+1)}{2} \).
- For each of these \( \frac{n(n+1)}{2} \) pairs, we can choose any of the \( n \) elements in \( A \) as the output.
- Therefore, the number of commutative binary operations is \( n^{\frac{n(n+1)}{2}} \).

4. **Calculating the Probability**:
The probability \( P \) that a randomly chosen binary operation is commutative is given by the ratio of the number of commutative binary operations to the total number of binary operations:
\[ P = \frac{\text{Number of commutative binary operations}}{\text{Total number of binary operations}} = \frac{n^{\frac{n(n+1)}{2}}}{n^{n^2}} = n^{\frac{n(n+1)}{2} - n^2} \]
Simplifying the exponent:
\[ \frac{n(n+1)}{2} - n^2 = \frac{n^2 + n - 2n^2}{2} = \frac{-n^2 + n}{2} = \frac{n(n-1)}{2} \]
Thus, the probability becomes:
\[ P = n^{\frac{n(n-1)}{2}} \]

### Final Answer:
The probability that a randomly chosen binary operation on a set \( A \) with \( n \) elements is commutative is:
\[ P = n^{\frac{n(n-1)}{2}} \]