OK, so, given your more rigorous wording, here's my amended working:
[Edit: as I acknowledged in a later post, this working was wrong. I will not delete it from the historical record, but I have corrected it further below]
1-set-combos, both complete and incomplete (individual sets of four, three, two and one):
4C4 + 4C3 + 4C2 + 4C1
2-set-combos, complete (individual sets of three-and-one, two-and-two, one-and-three):
4C3*1C1 + 4C2*2C2 + 4C1*3C1
2-set-combos, incomplete (individual sets of two-and-one, one-and-two, one-and-one):
4C2*2C1 + 4C1*3C2 + 4C1*3C1
3-set-combos, complete (individual sets of two-and-one-and-one, one-and-two-and-one, and one-and-one-and-two):
4C2*2C1*1C1 + 4C1*3C2*1C1 + 4C1*3C1*2C2
3-set-combos, incomplete (one-and-one-and-one):
4C1*3C1*2C1
4-set-combos, complete (one-and-one-and-one-and-one):
4C1*3C1*2C1*1C1
Now combining into equivalent sets of combinations, and then dividing by the number of possible permutations of each set of combinations:
4C4 + 4C3 + 4C2 + 4C1 + (4C3*1C1 + 4C1*3C1) / 2 + 4C2*2C2 + (4C2*2C1 + 4C1*3C2) / 2 + 4C1*3C1 + (4C2*2C1*1C1 + 4C1*3C2*1C1 + 4C1*3C1*2C2) / (3*2) + 4C1*3C1*2C1 + 4C1*3C1*2C1*1C1
= 1 + 4*3*2/(3*2*1) + 4*3/(2*1) + 4 + (4*3*2/(3*2*1)*1 + 4*3) / 2 + 4*3/(2*1)*1 + (4*3/(2)*1 + 4*3*2/(2*1)) / 2 + 4*3 + (4*3/(2*1)*2*1 + 4*(3*2)/(2*1)*1 + 4*3*1) / (3*2) + 4*3*2 + 4*3*2*1
= 104
So, which of us is wrong, and why?
[Edit: here is the corrected version. Aside from a few minor slip-ups, the major thing I missed was dividing by the factorial of each number of subsets which shared a common size, as noted in
post #238. I've marked all changes to the above working in bold green - each bolded, green item is either an addition or an amendment]
1-set-combos, both complete and incomplete (individual sets of four, three, two and one):
4C4 + 4C3 + 4C2 + 4C1
2-set-combos, complete (individual sets of three-and-one, two-and-two, one-and-three):
4C3*1C1 + 4C2*2C2
/2! + 4C1*3C
3
2-set-combos, incomplete (individual sets of two-and-one, one-and-two, one-and-one):
4C2*2C1 + 4C1*3C2 + 4C1*3C1
/2!
3-set-combos, complete (individual sets of two-and-one-and-one, one-and-two-and-one, and one-and-one-and-two):
4C2*2C1*1C1
/2! + 4C1*3C2*1C1
/2! + 4C1*3C1*2C2
/2!
3-set-combos, incomplete (one-and-one-and-one):
4C1*3C1*2C1
/3!
4-set-combos, complete (one-and-one-and-one-and-one):
4C1*3C1*2C1*1C1
/4!
Now combining into
sets of equivalent combinations, and then dividing by the number of
repetitions of equivalent combinations:
4C4 + 4C3 + 4C2 + 4C1 + (4C3*1C1 + 4C1*3C
3) / 2 + 4C2*2C2
/2 + (4C2*2C1 + 4C1*3C2) / 2 + 4C1*3C1
/ 2 + (4C2*2C1*1C1 + 4C1*3C2*1C1 + 4C1*3C1*2C2)
/ (3*2) + 4C1*3C1*2C1 / (3*2) + 4C1*3C1*2C1*1C1
/ (4*3*2)
= 1 + 4*3*2/(3*2*1) + 4*3/(2*1) + 4 + (4*3*2/(3*2*1)*1 + 4*
1) / 2 + 4*3/(2*1)
/2 + (4*3/(2)*
2 + 4*3*2/(2*1)) / 2 + 4*3
/ 2 + (4*3/(2*1)*2*1 + 4*(3*2)/(2*1)*1 + 4*3*1) / (3*2) + 4*3*2
/ (3*2) + 4*3*2*1
/ (4*3*2)
=
51