How many anagrams, even meaningless ones, are there of the word MONTE?
The answer is very simple with combinatorial calculus. The word MONTE has 5 letters all distinct from each other; its anagrams are $5!=120$.
How is it solved with the Cartesian product?
PS: The exercise was taken from a first-year high school science book Cartesian product paragraph.
You could interpret your solutions as $5$-ary Cartesian product
$$ \{(x_1, \ldots, x_5) | x_i \in \{\text{M,O,N,T,E}\} \text{ for }i \in \{1,...,5\} \}$$
Rule: In general is the cardinality of the Cartesian product the product of the cardinality of each set.
In this example $5 \cdot 5 \cdot 5 \cdot 5 \cdot 5=5^5$.
For your solution you need all elements with different letters. So the Cartesian product will consist of 5 sets with 5, 4, 3, 2 and 1 element. Using the rule above we have $5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$.