# binary operation on a set

#### oldguynewstudent

Given a set S, a function f: S X S $$\displaystyle \longrightarrow$$ S is called a binary operation on S. If S is a finite set, then how many different binary operations on S are possible?

I have no clue on this one? Could someone point me in the right direction?

I realize that if S has n elements then SxS would have $$\displaystyle n^2$$ possible ordered pairs. However, I'm not sure what is meant by how many different binary operations. Would we have addition, subtraction, multiplication, division, exponentiation, etc. times $$\displaystyle n^2$$ binary operations?

#### Drexel28

MHF Hall of Honor
Given a set S, a function f: S X S $$\displaystyle \longrightarrow$$ S is called a binary operation on S. If S is a finite set, then how many different binary operations on S are possible?

I have no clue on this one? Could someone point me in the right direction?

I realize that if S has n elements then SxS would have $$\displaystyle n^2$$ possible ordered pairs. However, I'm not sure what is meant by how many different binary operations. Would we have addition, subtraction, multiplication, division, exponentiation, etc. times $$\displaystyle n^2$$ binary operations?
It's asking how many functions are there from a set with $$\displaystyle n^2$$ elements to a set with $$\displaystyle n$$ elements.

oldguynewstudent

#### oldguynewstudent

Given a set S, a function f: S X S $$\displaystyle \longrightarrow$$ S is called a binary operation on S. If S is a finite set, then how many different binary operations on S are possible?
With the help of Drexel28, we have a set S with n elements which gives $$\displaystyle n^2$$ mapped to n. This should give us $$\displaystyle n^3$$ binary operations, correct?

#### aman_cc

With the help of Drexel28, we have a set S with n elements which gives $$\displaystyle n^2$$ mapped to n. This should give us $$\displaystyle n^3$$ binary operations, correct?

Shouldn't it be (n)^(n^2)

#### rcopher6

Yes I believe that (n)^(n^2) is correct

#### Deveno

MHF Hall of Honor
in general, the number of functions f:A→B is:

|B||A|.

it is easiest to see this when |B| = 2, such as when B = {0,1}, so that the functions f:A→{0,1} can be put in a 1-1 correspondence with the subset of A:

given a subset S of A, we define:

f(a) = 1, if a is in S
f(a) = 0, if a is not in S.

thus the number of functions f:A→{0,1} is the same number as 2|A|, the cardinality of the power set of A.

so, yes, the correct answer is n(n2).