I'm not learning much from my earlier note on making a plan before executing. So 10a was done before writing this.
Basically, I just sorted the list, since you had to use all of them, just calculate the difference between the element and the next one, making sure to add a 0 to the head, and val[length] + 3 to the last item. Then count 1's and threes.
Given the list of deduced differences;
[1,1,1,1,3,1,1,1,1,3,1,1,1,1,3,1,1,1,1,3,1,1,3,1,1,3,1,1,1,1,3,3,3,1,1,1,1,3,3,1,1,1,1,3,1,1,3,1,3,3,1,1,1,1,3,3,1,1,3,1,1,1,1,3,1,1,1,3,1,1,1,1,3,1,1,1,3,1,1,3,1,1,1,1,3,1,1,1,1,3,1,3,1,1,1,1,3]
We can state the following;
[1] = 1 permutation (2 ^ 1-1) = 1
[1, 1] == [2] == 1 + 1 steps (2 ^ 2-1) = 2
[1a, 1b, 1c] == [2, 1] == [2, 1] == [1, 2] 3 extra permutations (2 ^ 3-1) = 4
[1a, 1b, 1c, 1d] == [2, 1c, 1d] == [1a, 2(a), 1d] == [1a, 2(b), 1d]== [3, 1] == [3, 1] == [1, 3] (we can't go to 4)
I.e. Any sequence of 1s = 2^(n-1). So if we take the diff of 1s, we should be able to calculate the total permutations.