Maximizing and minimizing the number of positive product $k$-subsets of an $n$-set

Question

The question is simple but require some definitions. I came across resolving a certain inequality. If there is no closed answer is there a related sequence describing the situation?

Let

$$S\ :=\ \{X=(x_1,\ldots,x_n)\in(\mathbb{R}\setminus\{0\})^n\,:\ \exists_{(i,j)} \ x_i\cdot x_j<0\}.$$

For $k$, $1\le k\le n$, there is $\binom{n}{k}$, $k$-tuples choices among the entries of $X\in S$. A choice is denoted by $X_k^{(j)}=(x_1^{(j)},\ldots,x_k^{(j)})$, $j=1,\ldots,\binom{n}{k}$. Let $P_k^{(j)}=\prod_{i=1}^kx_i^{(j)}$.

What is the maximum (minimum) number of positive products $P_k^{(j)}$ over the set $S$ for $k$ fixed? over $S$ and $k$ variable?

The cases $k=1,2$ are perhaps direct but for else it may be not simple.

$k=2$ , you should get the maximum number of positive products is when the vector $X$ has a single positive entry and thus equals $\binom{n-1}{2}$. The minimum is for $X$ having exactly $\lfloor n/2\rfloor$ positive entries

Answer 1

Here are some minor remarks. Since the actual numbers do not matter, the question can be rephrased as follows. Let $\sigma: [n] \to \{-, +\}$. Say that a $k$-subset of $[n]$ is $\sigma$-positive if it contains an even number of elements $a$ such that $\sigma(a)=-$. The question is to find the maximum and minimum number of $\sigma$-positive $k$-subsets of $[n]$ over all $\sigma$ with at least one positive and one negative element.

Clearly, the number of $\sigma$-positive $k$-subsets only depends on the size of $\sigma^{-1}(-)$. Therefore, letting $f(k,n)$ and $g(k,n)$ be the corresponding maximum and minimum values, we have that $f(k,n)=\max_{m \in [1,n-1]} \sum_{2i \leq m} \binom{m}{2i} \binom{n-m}{k-2i}$ and $g(k,n)=\min_{m \in [1,n-1]} \sum_{2i \leq m} \binom{m}{2i} \binom{n-m}{k-2i}$, as already observed by Domotorp in the comments above.

For $k$ odd, let $\gamma$ be obtained from $\sigma$ by reversing the sign of every element in $[n]$. Then a $k$-subset of $[n]$ is $\sigma$-positive if and only if it is not $\gamma$-positive. Therefore, for all odd $k$, we have $f(k,n)=\binom{n}{k} - g(k,n)$.

You have already given the answer for small values of $k$. Here are the answers for large values of $k$. We have $f(n,n)=1$ and $g(n,n)=0$ if $n \geq 3$, $f(n-1, n)=n-1$ if $n$ is even (take exactly one element to be positive), and $f(n-1, n)=n-2$ if $n$ is odd (take exactly two elements to be positive).