Yes, I misread your problem. I only consider the case where $$$A$$$ and $$$B$$$ are continuous. The answer should be

$$$2(n-2k)!k\sum\limits_{t=k}^{n-k} \frac{(t-1)!(n-t)!}{(t-k)!(n-t-k)!}$$$.

Suppose $$$A$$$ is in front of $$$B$$$. You should multiply $$$2$$$ to the final answer.

Suppose the last place of $$$A$$$ is $$$t$$$, where $$$k \le t \le n-k$$$. Then, we can choose $$$k-1$$$ places from the first $$$t-1$$$ places to settle $$$A$$$, which is $$$\binom{t-1}{k-1}$$$. And there are $$$k!$$$ ways to place $$$A$$$. For the first $$$t$$$ places, there are $$$t-k$$$ places that are not $$$A$$$. We can choose from $$$n-2k$$$ numbers to fill in these $$$t-k$$$ places because we cannot choose elements from $$$A$$$ as they have already been fixed, and we could not choose elements from $$$B$$$ because they should be placed after $$$A$$$. So there are $$$\binom{n-2k}{t-k}$$$ choices. The $$$t-k$$$ spare places in the range $$$[1, t]$$$ can be randomly placed, so you should multiply $$$(t-k)!$$$. Now, the first $$$t$$$ places are settled, so the set of elements in the last $$$n-t$$$ places are also settled, and they can be placed randomly, so there are $$$(n-t)!$$$ ways.

In all, for each $$$t$$$ such that $$$k \leq t \leq n-k$$$ and $$$A$$$ is in front of $$$B$$$, there are

$$$\binom{t-1}{k-1}\cdot k! \cdot \binom{n-2k}{t-k} \cdot {(t-k)!} \cdot {(n-t)!} = \frac{(n-2k)!k(t-1)!(n-t)!}{(t-k)!(n-t-k)!}$$$ ways.

Check Code:

#include <bits/stdc++.h>
using namespace std;
#define ll long long
int main(void){
    vector<int> x = {1, 2, 3, 4, 5, 6, 7};
    set<int> a = {2, 3}, b = {4, 5};
    ll n = (ll)x.size(), k = 2, ans = 0, num = 0;
    do{
        set<int> pa, pb;
        for(int i = 0; i < n; ++i){
            if(a.count(x[i])){
                pa.insert(i);
            }else if(b.count(x[i])){
                pb.insert(i);
            }
        }
        if(*pa.begin() > *pb.rbegin() || *pb.begin() > *pa.rbegin()){
            ++ans;
        }
        ++num;
    }while(next_permutation(x.begin(), x.end()));
    cout << ans << " " << num << "\n";
    ll fac[9], ans2 = 0;
    fac[0] = 1;
    for(int i = 1; i <= n; ++i){
        fac[i] = i*fac[i-1];
    }
    for(int t = k; t <= n-k; ++t){
        ans2 += (fac[t-1]*fac[n-2*k]*k*fac[n-t])/(fac[t-k]*fac[n-t-k]);
    }
    cout << 2*ans2 << "\n";
}