1. ## Quick explanation

Suppose the set A has m elements and the set B has n elements. There are 2^(m*n) relations from A to B and n^m functions from A to B.

Can some one just give me combinatorial explanation. I don't really need to see anything worked out.

Thanks,
--Dan

2. Originally Posted by Danneedshelp
Suppose the set A has m elements and the set B has n elements. There are 2^(m*n) relations from A to B and n^m functions from A to B.
You need to recall that any relation from A to B is a subset of $A \times B$.
If $\left| A \right| = m~\& ~\left| B \right| = n \Rightarrow \quad \left| {A \times B} \right| = m \cdot n\quad$.
There are $2^{\left| {A \times B} \right|} = 2^{mn}$ subsets of $A \times B$.

There it is!

3. ## Second Part

The second part is because there are n choices to send the first element in A to, n choices to send the second element in A to, n choices to send the ...., n choices to send the mth element to. That is $n^m$ distinct choices.