Question

How many different element does \({{\bf{A}}^{\bf{n}}}\), have when A has m elements and n is a positive integer.

Short answer

\({A^n}\) contains \({m^n}\) elements.

Step-by-step solution

Cartesian product

The cartesian product of A and B denoted by\({\bf{A \times B}}\), is the set of all ordered pairs (a,b).

Therefore, \(A \times B = \left\{ {\left( {a,b} \right)\left| {a \in A\Lambda b \in B} \right.} \right\}\)

To find number of elements

\(\)

Here,

A has m elements

\(\left| A \right| = m\)

Now, \({A^n}\) represent the Cartesian product of A and A and A ........and so on till last A.

Therefore, the number of elements in Cartesian product is the product of elements in each of these sets as in the Cartesian product.

\(\begin{aligned}{c}\left| {{A^n}} \right| = \left| A \right| \cdot \left| A \right| \cdot \left| A \right|........\left| A \right| = {\left( {\left| A \right|} \right)^n}\\ = {m^n}\end{aligned}\)

Therefore, \({A^n}\) contains \({m^n}\) elements.