Question

Let \(S\) be a finite set containing \(n\) elements. Then the total number of commutative binary operation on \(S\) is

• A. \(n^{\left[\frac{n(n+1)}{2}\right]}\)
• B. \(n^{\left[\frac{n(n-1)}{2}\right]}\)
• C. \((n^2)^n\)
• D. \(2^{(n^2)}\)

Hint:

For commutative binary operation, only upper triangle including diagonal of operation table is independent.

Answer 1

The Correct Option is A

Step 1: Total pairs \((a,b)\) in a binary operation.
Binary operation \( *:S\times S \to S\).
There are \(n^2\) ordered pairs.
Step 2: Condition for commutativity.
Commutative means:
\[ a*b=b*a \]
So values for \((a,b)\) and \((b,a)\) are same.
Step 3: Count independent pairs.
- Diagonal pairs: \((a,a)\): there are \(n\).
- Off-diagonal pairs: \((a,b)\) where \(a\neq b\): they occur in symmetric pairs.
Number of such unordered pairs:
\[ \binom{n}{2}=\frac{n(n-1)}{2} \]
So independent positions:
\[ n+\frac{n(n-1)}{2}=\frac{n(n+1)}{2} \]
Step 4: Assign values from S.
Each independent position can take \(n\) values.
So total commutative operations:
\[ n^{\frac{n(n+1)}{2}} \]
Final Answer:
\[ \boxed{n^{\frac{n(n+1)}{2}}} \]