# Is this the correct answer for the cardinality of this set?

This is a question from a practice quiz at my university.

Is the question asking for the cardinality of Σ1 = {a,b} to the power of four?

if that's the case, then the set would still have a cardinality of 2 since elements in a set are unique. It wouldn't be {a,b,a,b,a,b,a,b} right?

This is how you should solve it. Let $$\Sigma_1=\Sigma$$

We are asked to find: $$|\Sigma^ 4|$$ Now

$$\Sigma^4 = \Sigma . \Sigma . \Sigma . \Sigma = \{a,b\} . \{a,b\}. \{a,b\} . \{a,b\}$$

$$\text{ Here . means the concatenation operator}$$

So this is equivalent to number of ways of forming strings of length $$4$$ with symbols from $$\Sigma$$.

We have $$4$$ places,

Each of the places have two choices to get filled : either $$a$$ or $$b$$. So,

_   _  _   _
2 . 2. 2. 2


So answer is $$2^4 =16$$

A set "multiplied" with another set is called the cartesian product and is formed of tuples of elements, one from each set you are multiplying.

If you have $$S = \{a, b\}$$ and $$T = \{x, y\}$$, then $$S \times T = \{ (a,x), (a,y), (b, x), (b, y)\},$$ and its cardinality is actually $$|S\times T| = |S| \cdot |T|$$.