Suppose we have $x_3$ distinct $3$-element sets chosen arbitrarily among the $\binom{7}{3}=35 \ge x_3$ subsets of $U=\{1,2,3,4,5,6,7\}$. Consider the $x_4$ distinct $4$-element sets that can be obtained as a union of any two of the above $3$-element sets. I know the probability that a couple of $3$-element sets chosen at random in $U$ have $2$ elements in common and thus their union is a $4$-element set (as computed here). I would like to find a lower bound for $x_4$ as a function of $x_3$, in the worst case:
$$x_4 \ge f(x_3)$$
Ideally, I would like to later extend the result to any $j \lt k \le n$, i.e. find a lower bound:
$$x_k \ge g(x_j, j, k, n)$$
instead of the above case $j=3, k=4, n=7$.
Any hint?
You can solve the problem via integer linear programming as follows. For each $3$-subset $S \subset U$, let binary decision variable $y_S$ indicate whether $S$ is chosen. For a given $x_3$, the problem is to minimize $$\sum_{T \subset U: |T|=4} \max_{(R,S): R \cup S = T} y_R y_S \tag1$$ subject to $$\sum_S y_S = x_3. \tag2$$ The objective function $(1)$, which can be linearized, represents the number of $4$-subsets that are covered exactly by some pair of chosen $3$-subsets. The constraint $(2)$ enforces selection of exactly $x_3$ of the $3$-subsets.
The resulting optimal objective values are $$f(0)=\dots=f(7)=0, f(8)=2, f(9)=4, f(10)=f(11)=5, f(12)=6, f(13)=8, f(14)=11, f(15)=12, f(16)=f(17)=14, f(18)=f(19)=f(20)=15, f(21)=19, f(22)=22, f(23)=24, f(24)=f(25)=25, f(26)=28, f(27)=30, f(28)=f(29)=31, f(30)=33, f(31)=f(32)=34, f(33)=f(34)=f(35)=35.$$
To linearize $(1)$, introduce a binary (or nonnegative) variable $z_T$ for each $4$-subset $T \subset U$ and minimize $$\sum_{T \subset U: |T|=4} z_T \tag3$$ subject to $(2)$ and $$z_T \ge y_R + y_S - 1 \quad \text{for all $T$ and $(R,S): R \cup S = T$}. \tag4$$