I have found this question on a textbook of high school (first year - 14 years old) where there is the Cartesian product topic.
This question there is in the "prove Invalsi" tests for students of 13 years ("terza media") old where they didn't study the Cartesian product in an insiemistic sense. Look the question $D20$ to the pag. 17.
Luisa can't quite remember the combination of the lock of her bicycle. The combination is obtained by turning four small wheels, each bearing all the digits from 0 to 9. Luisa does not remember the second digit of the combination at all but knows that:
- the first digit is 6;
- the third digit is 3 or 4;
- the last digit is 1. How many combinations at most will Luisa have to try to succeed in opening the lock on her bicycle?
I have done this: The combinations with the third digit 3 are ten. \begin{array}{|l|l|l|} \hline 6 & 0 & 3 &1\\ \hline 6 & 1 & 3 & 1\\ \hline 6 & 2 & 3 & 1\\ \hline 6 & 3 & 3 & 1\\ \hline 6 & 4 & 3 & 1\\ \hline 6 & 5 & 3 & 1\\ \hline 6 & 6 & 3 & 1\\ \hline 6 & 7 & 3 & 1\\ \hline 6 & 8 & 3 & 1\\ \hline 6 & 9 & 3 & 1\\ \hline \end{array} With the same mechanism the combinations with the third digit 4 are ten.
\begin{array}{|l|l|l|} \hline 6 & 0 & 4 &1\\ \hline 6 & 1 & 4 & 1\\ \hline 6 & 2 & 4 & 1\\ \hline 6 & 3 & 4 & 1\\ \hline 6 & 4 & 4 & 1\\ \hline 6 & 5 & 4 & 1\\ \hline 6 & 6 & 4 & 1\\ \hline 6 & 7 & 4 & 1\\ \hline 6 & 8 & 4 & 1\\ \hline 6 & 9 & 4 & 1\\ \hline \end{array}
In total I have 20 combinations.
How I can find the $20$ combinations using combinatorial calculus and especially using the Cartesian product? I don't know how to put the digits in the $x-$ and $y-$axis.
Because there are $4$ digits, the number of possibilities is actually a $4$-fold cartesian product $$ D_1 \times D_2 \times D_3 \times D_4, $$ where $D_n$ is the set of possible digits in position $n$, i.e. \begin{align} D_1 &= \{6\} \\ D_2 &= \{0, 1, \dots, 9\} \\ D_3 &= \{3, 4\} \\ D_4 &= \{1\} \end{align} so the number of possibilities is $$ |D_1 \times D_2 \times D_3 \times D_4| = |D_1| \cdot |D_2| \cdot |D_3| \cdot |D_4| = 1 \cdot 10 \cdot 2 \cdot 1 = 20. $$
Of course it's difficult to visualize a $4$-dimensional array, but fortunately, when one or more of the sets has only a single element, we can essentially ignore it. In this exercise, $|D_1| = |D_4| = 1$. In fancy set theory terms, there's a bijection $$ D_2 \times D_3 \;\to\; D_1 \times D_2 \times D_3 \times D_4 $$ mapping $(d_2, d_3) \mapsto (6, d_2, d_3, 1)$. So we can enumerate this $2$-fold cartesian product instead, and this array can be drawn on the plane.