Cardinality of Cartesian Product where (a,b) are elements of A x B I'm having issues with a question regarding the cardinality of a cartesian product.

Question:
Let $A=\{0, 1, 2, 3, 4, 5, 7\}$ and $B=\{0, 2, 4, −1, 12\}.$

How many elements are in $\{(a, b) ∈ A × B \; |  \; a &lt; 7 \text{ and } b &lt; 4\}$?

I want this to be $6 \cdot 3 = 18$. but that seems to be wrong.
You may begin by omitting all $a \in A$ such that $a\ge 7$ and all $b \in B$ such that $b \ge 4$ because we know ordered pairs with said elements will not be included in the Cartesian product we're looking for. So, after omission, we have $A = \{0,1,2,3,4,5\}$ and $B=\{0,2,-1\}$. Now, the cardinality of the Cartesian product between sets $A$ and $B$, denoted as $|A$ x $B|$, is given by $|A$ x $B|$ = $|A| \cdot |B| = 6 \cdot 3 = 18$. So you are correct! If your book says differently, then I highly suspect your book contains a mistake.
If I've understood your question correctly, you'll begin by taking the appropriate subsets of A and B. 

Since A = {0,1,2,3,4,5,7} and B = {0,2,4,−1,12}, and you're restricting your subsets to a &lt; 7, and b &lt; 4, this leaves the sets C = {0,1,2,3,4,5} and D = {0,2,-1}. 

If you allow for repetition, which your question seems to assume, then there are 6 ways to select the first element, and 3 ways to select the second, implying a total of 18 ways to construct ordered pairs on the subsets.

Note that your textbook might be looking for un-ordered subsets of size 2, which is a different question.