Question

Let \(S\) be a set containing \(n\) elements. Then, number of binary operation on \(S\) is

• A. \(n^n\)
• B. \(2^{n^2}\)
• C. \(n^{n^2}\)
• D. \(n^2\)

Hint:

Number of binary operations on \(S\) equals number of functions from \(S\times S\) to \(S\). Since \(|S\times S|=n^2\), total is \(n^{n^2}\).

Answer 1

The Correct Option is C

Step 1: Definition of a binary operation.
A binary operation on set \(S\) is a function:
\[ *: S\times S \rightarrow S \] Step 2: Count elements in domain.
If \(|S|=n\), then:
\[ |S\times S| = n^2 \] So there are \(n^2\) ordered pairs.
Step 3: Count number of functions.
For each of the \(n^2\) ordered pairs, the output can be any of the \(n\) elements of \(S\).
So total number of functions is:
\[ n^{n^2} \] Final Answer: \[ \boxed{n^{n^2}} \]