Let S be a finite set containing $\mathrm{n}$ elements. Then the total number of commutative binary operation on $\mathrm{S}$ is

• Correct answer
  A

  $\mathrm{n}^{\left[\frac{n(n+1}{2}\right]}$
• B

  $\mathrm{n}^{\left[\frac{n(n-1}{2}\right]}$
• C

  $\mathrm{n}^{\left(\mathrm{n}^{2}\right)}$
• D

  $2^{\left(n^{2}\right)}$

Solution

Let $\mathrm{S}=\left\{\mathrm{a}_{\mathrm{i}}\right\}$ where $\mathrm{i}=1.2 \ldots \ldots \mathrm{n}$ From commutative operations, $a_{i} * a_{j}=a_{j} * a_{i} \ldots$ (i) $\forall i, j=1,2,3 \ldots n$ where ${ }^{*}$ represents a binary operation $\therefore$ Number of distinct elements in $\mathrm{S} \times \mathrm{S}$
$=\mathrm{n}\left\{\left(\mathrm{a}_{1}, \mathrm{a}_{1}\right),\left(\mathrm{a}_{1}, \mathrm{a}_{2}\right) \ldots \ldots\left(\mathrm{a}_{1}, \mathrm{a}_{\mathrm{n}}\right)\right.$ $\left(\mathrm{a}_{2}, \mathrm{a}_{2}\right),\left(\mathrm{a}_{2}, \mathrm{a}_{3}\right), \ldots .\left(\mathrm{a}_{2}, \mathrm{a}_{\mathrm{n}}\right)$ $\ldots\left(\mathrm{a}_{\mathrm{n}-1}, \mathrm{a}_{\mathrm{n}-1}\right),\left(\mathrm{a}_{\mathrm{n}-1}, \mathrm{a}_{\mathrm{n}}\right),\left(\mathrm{a}_{\mathrm{n}}, \mathrm{a}_{\mathrm{n}}\right)$ $=\mathrm{n}+(\mathrm{n}-1)+(\mathrm{n}-2)+\ldots .+2+1=\frac{\mathrm{n}(\mathrm{n}+1)}{2}$ $\therefore$ No. of commutative binary operations $=$ No. of functions $f: S \times S \rightarrow$ S subject to (i) $=\text { n.n.n.... } \frac{n(n+1)}{2} \text { times }=n^{\frac{n(n+1)}{2}}$

Hence, the correct answer is option: 'A'.