Originally Posted by

**tangibleLime** Okay, I think I got it.

I was using [tex]\frac{n!}{(n-k)!} because I thought it wanted all of the different combinations of mixing the elements in S with each other.

For example, if S={a, b, c}, then the following binary operations are possible:

a*b

a*c

a*a

b*a

b*c

b*b

c*a

c*b

c*c

Which... now that I look at it, is $\displaystyle n^{n}$, which is what you came up with. Maybe I was just incorrect in the permutation aspect... am I thinking about this the right way as I just explained?

Thanks!

Not really. You are focusing too much on a particular operation. Consider this let $\displaystyle S=\{a,b\}$ consider the following operations

$\displaystyle \begin{array}{c|cc} \star & a & b\\ \hline a & a & a\\ b & a & a\end{array}\quad \begin{array}{c|cc} \ast& a & b\\ \hline a & b & a\\ b & a & b\end{array}\quad\begin{array}{c|cc} \oplus& a & b\\ \hline a & b & b\\ b & a & a\end{array}$

Then, the FUNCTIONS $\displaystyle \ast:S\times S\to S$, $\displaystyle \star:S\times S\to S$, and $\displaystyle \oplus:S\times S\to S$ are examples of binary operations which to reiterate are FUNCTIONS. So, we know that the number of binary operations (...functions...) is at least $\displaystyle 3$ since we have just found three different ones. If counting the functions is hard to you, you can really just count the number of ways to fill those $\displaystyle 2\times 2$ tables I made in. Does that help?