A counting question about pairs of three-digit numbers the question asks how many pairs of $3$ digit numbers exist such that when one is added to the other, we never have to use a "carry-over". For clarification, a carry-over occurs  when you add $75 + 68$, because $5 + 8=13$, which contributes the "$1$" to the next round of addition "$7+6$". Note that you need to have another carry-over because $7 + 6 + 1=14$. Any thoughts would be appreciated.

Assuming I am understanding the question correctly, we are trying to find the number of non-ordered pairs $\{\overline{abc}, \overline{def}\}$ such that $\overline{abc} + \overline{def}$ does not require a "carry-over."
I will not give away the specifics of the solution so you can try things out on your own. My general recommendation is to look at the digits separately and count the number of combination possible for $\{c,f\}, \{b,e\}, \{a,d\}$. Note that $\{c,f\}$ and $\{b,e\}$ should have the same number of possible combinations (whereas $\{a,d\}$ is slightly different given that none of them can be $0$). Also note that to not have a "carry-over" when adding two digits means the sum of the two digits should be less than 10 (i.e. less than or equal to 9).
Multiplying the number of combinations for each digit gives you the total number of combinations. In addition, we need to handle the double-counted pairs. In this case, dividing by $2$ should suffice since addition is symmetrical.
Edit: oops sorry yah the comment below is right haha, do pick out those when doubled also don’t cause any carry-overs, also fixing typo on $\leq 9$
