perhaps an easier way (or when you don't want to solve for a general case n, or recurrence doesnt strike immediately) to approach this (could be a bit wrong though)
the total number of pairings would indeed be (10P5)*(2^-5)
assuming left can be paired with left and right can be paired with right
say that we have gloves as 1L, 2L, 3L, 4L, 5L and 1R, 2R, 3R, 4R, 5R
Possible pairings
for 1L: (1R, 2L, 2R)
2L: (1L, 1R, 2R, 3L, 3R)
...
5L: (4L, 4R, 5R)
so basically for 1L we have 3C1, 2L: 5C1, 3L: 5C1, 4L: 5C1, 5L: 3C1
How the total number of pairings would be (10P5)*(2^-5)?
perhaps an easier way (or when you don't want to solve for a general case n, or recurrence doesnt strike immediately) to approach this (could be a bit wrong though)
the total number of pairings would indeed be (10P5)*(2^-5)
assuming left can be paired with left and right can be paired with right
say that we have gloves as 1L, 2L, 3L, 4L, 5L and 1R, 2R, 3R, 4R, 5R
Possible pairings
for 1L: (1R, 2L, 2R)
2L: (1L, 1R, 2R, 3L, 3R)
...
5L: (4L, 4R, 5R)
so basically for 1L we have 3C1, 2L: 5C1, 3L: 5C1, 4L: 5C1, 5L: 3C1
3 + 5 + 5 + 5 + 3 = 21
(considered only for L since order doesnt matter)
21/total_no_of_pairings = 1/45
How can you add the choices like that to make 21?
This assumes that glvoes don't have handedness, if the gloves have handedness, right and left then the recurrence would be fn=fn-1+fn-2
yes, it assumes left-left can also be paired