Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

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

VITEEEVITEEE 2008Relations and Functions - Part 2

Solution:

Let S = {ai } where i = 1.2.....n
From commutative operations,
ai *a j = a j *ai … (i) $\forall$i, j = 1,2,3....n
where * represents a binary operation
$\therefore $ Number of distinct elements in S × S
i.e.,$\begin{matrix}\left\{a_{i}\right\} \times \left\{a_{j}\right\}\\ i=1,2...n j=1,2...n \end{matrix}$ subject to the condition (i)
= n{(a$_1$,a$_1$),(a$_1$,a$_2$ )......(a$_1$,an ),
(a$_2$, a$_2$ ), (a$_2$,a$_3$),....(a$_2$,a$_n$ ),
...(a$_{n-1}$,a$_{n-1}$),(a$_{n-1}$,a$_n$ ),(a$_n$ ,a$_n$ )
$=n+\left(n-1\right)+\left(n-2\right)+....+2+1=\frac{n\left(n+1\right)}{2}$
$\therefore $ No. of commutative binary operations
= No. of functions f : S × S $\rightarrow$ S subject to (i)
$=n.n.n....\frac{n\left(n+1\right)}{2}times = n^{ \frac{n\left(n+1\right)}{2}}$