This question is motivated by a real-world application related to an art project that involves displaying images, but my search hit a dead end after finding the wikipage about [**Kirkman systems**](https://en.wikipedia.org/wiki/Kirkman%27s_schoolgirl_problem) (other related terms include [**Steiner systems**](https://en.wikipedia.org/wiki/Steiner_system) and the [**Social golfer problem**](https://en.wikipedia.org/wiki/Social_golfer_problem)) and looking over references linked there. A few people have written programs for this specific question that established lower bounds (so: more than $2100$) but none has found an exact answer.

**The question is:**

> Given a set of $40$ elements, what is the maximum number of subsets that can be created such that no triple appears more than once?

(For example, if the set includes $A,B,C,D,E$ as elements, then one cannot include in the collection of subsets *both* $\{A,B,C,D\}$ *and* $\{A,B,C,E\}$ since, in this scenario, we would have the triple $A,B,C$ appearing more than once.)

As an excerpt from the [**History**](https://en.wikipedia.org/wiki/Kirkman%27s_schoolgirl_problem#History) section of the aforelinked wikipage, there is under the first bullet point a question attributed to Wesley Woolhouse (1844):

>"Determine the number of combinations that can be made out of $n$ symbols, $p$ symbols in each; with this limitation, that no combination of $q$ symbols, which may appear in any one of them shall be repeated in any other." 

followed by the formula:

$$\frac{n!}{q!(n-q)!)} \div \frac{p!}{q!(p-q)!}$$

Unfortunately, one finds that this formula is *false* already for small examples (e.g. $n=5, p=4, q=3$) and, indeed, reading further on that page indicates that the formula only holds in certain scenarios.

Rephrased, I am looking for an answer to Woolhouse's question for the case of $n=40, p=4, q=3$ either by a counting argument, an effective program, or a reference. Please tag/retag as appropriate; thanks!