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

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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?

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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 $2^4 =16$

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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 $|S\times T| = |S| \cdot |T|$.

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