Is my reasoning here correct or is there a better way to solve this problem?
Given a set of items $I$ and a set of baskets $B$, where each basket contains $k$ items, what is the maximum number of pairs of items that show up at least once among the baskets? A pair is defined as $(i_1, i_2); i_1 \neq i_2; i_1, i_2 \in B_j$ i.e. pairs do not cross basket boundaries.
(I think) I solved the problem by doing the following:
$$ \text{max number of pairs} = \begin{cases} \binom{\lvert I \rvert}{2}, & k\lvert B \rvert \lt \binom{\lvert I \rvert}{2} \\ \binom{\lvert I \rvert}{k\lvert B \rvert}\binom{k}{2}\lvert B \rvert, & \text{otherwise} \end{cases} $$
My reasoning was that I could maximize the number of pairs if I could get the maximum distinct number of pairs in each of the $\lvert B \rvert$ baskets. This should work when $k\lvert B \rvert \lt \binom{\lvert I \rvert}{2}$. Otherwise, there will be at least one pair repeated, but it is possible to create an instance of the problem where all $\binom{\lvert I \rvert}{2}$ possible pairs from $I$ are observed among the $\lvert B \rvert$ baskets.
Is this correct? Was there a better (or just different) way to do it?
There are only ${|I| \choose 2}=\frac 12|I|(|I|-1)$ pairs, so if there are lots of baskets you will have that many. Your first inequality is in the wrong direction because you want lots of baskets to get the maximum number of pairs.
Each basket can contain at most $\frac 12k(k-1)$ pairs, so if there are lots of items and all the pairs are distinct there will be $\frac 12k(k-1)|B|$ pairs.
It gets hard in the middle range. Suppose there are two baskets, each of which can hold six items, and eight kinds of items. There are $28$ pairs and each basket can hold $15$ pairs, but you can't get them all. If an item is in only one basket it can only participate in $5$ pairs of the $7$ it could participate in. You only have room to put four types of item in each basket, so your baskets have $1,2,3,4,5,6$ and $1,2,3,4,7,8$ or something like it. This means you are missing four pairs-$57,58,67,68$ so have only $24$ pairs. I don't know of an easy formula here.