All possible permutations of joining/merging two lists - Circular vs. Non-Circular I have two lists List A = [a,b,c,d] and List B = [1,2,3,4] and would like to combine/interweave/shuffle them to make a list so that all elements of both lists are in it.
Here  it is given that this would be
$$(a+b)!\over  a!\cdot b!$$ where a and b refer to the number of elements in A and B.
Which in my case would be $8!/(4!\cdot 4!)$.
I presume that this maintains the relative order of elements in BOTH lists, A and B.
Elsewhere (I cannot find the link), I had read that all permutations of A and B  preserving the order of ONLY A but not B, is
(a+b)!/ a!
My core question is this, if List A and B are circular, then applying the notion of minus 1 for circular permutations, would the answers above change to

*

*$(a+b-1)!\over  (a-1)!(b-1)!$ <--- maintains the relative order of both A and B

*$(a+b-1)!\over (a-1)!$ <--- maintains the relative order of only A but not B (I think this is also called the rising factorial from the twelvefold way).

In my example, maintaining the order of both lists
Not assuming circular 8!/(4!*4!) = 70
Assuming circular 7!/(3!*3!) = 140
It just seems anti-intuitive that circular yields a larger number of permutations.

Question: Am I correct that the number of ways to interleave two circular lists of lengths $a$ and $b$ is equal to $\frac{(a+b-1)!}{(a-1)!(b-1)!}$? If so, how can it be that is this more than the number of ways to interleave two linear lists? If not, what is the correct formula?

 A: Surprisingly, I could not find a straightforward answer to this question of combining two lists with all unique elements such that all elements are included. However, thinking through circular permutations, the numerator is correct (a+b-1)! but the denominator is probably incorrect and should only be minus 1 on the larger set (a).
So, the answer is (a+b−1)!/(a−1)!(b)!
This would be all possible placements of 1234 into abcd, maintaining the order: such as 1234abcd, 123a4bcd,123ab4cd.....etc.
This is extra, not raised in the original question.
If we would further like to include all outcomes that go through all circularly equivalent combinations of 1234 (i.e., 4123, 3412, 2341 as well as the rotation 4321, 3214 etc.), we can find the equivalent combinations by using b!/((b-1)!/2)
which in this case is 24/3 = 8. Then multiply this with the (a+b−1)!(a−1)!(b)!
This then simplifies to (a+b−1)!/(a−1)!*((b-1)!/2)
Note - If flipping is not desired, skip the division by 2. Also, b should be more than 2.
