If there are 71 algs for 6CO, and 47 possible 6CP cases for each 6CO state, that leads to a total of 3337 algs, not including 6CO states that will lead to a lucky skip. Granted, there is probably more calcating needed than just 71 x 47, but ~3337 algs is the base answer. One-looking L6C (Last 6 Corners) could be, to some extent, possible, but pure 1LL6C is probably not practical nor efficient. I would rather suggest something more intuitive for metha, such as FB + FR/RB edges -> 6CO etc., which would be much more feasible. Hope this helps!