Question 394474: Spiffy the Spider has 8 legs, each of which is a different color. Spiffy has a sock and a shoe for each of his eight legs (so that he has one sock and one shoe in each of the eight colors). In how many ways can Spiffy put on his socks and shoes, if each sock and shoe must go on with the matching color, and the sock must go on before the shoe?
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! There are 16 "objects" (socks and shoes) to be put on in some order. Hence, there are 16! ways to put on the socks and shoes, if we include any order. However, for each color sock/shoe, we have two cases:
1: ...shoe1....sock1...
2: ...sock1....shoe1...
We divide by 2 for each color to account for over-counting, so we divide by 2^8 since there are eight colors. This leaves us  ways to distribute the socks and shoes.
The problem is, the placement of the socks and shoes matters (i.e. the spider's legs are distinguishable). For example, if ROYGBIV + X were the colors, then in counterclockwise order, the positions
ROYGBIVX and
BGIORVXY are distinct. Since there are 8! ways to distribute the colors to each of the legs, we multiply by 8! to obtain  .
(I hope it's correct...)
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