# Sock Probability

• Oct 20th 2009, 03:07 PM
sweeetcaroline
Sock Probability
3 pairs of socks are placed side-by-side of a straight clothes line. The socks in each pair are identical, but the pairs themselves are of different color. If I stand on one side of the line, how many different color arrangements can I see if no sock is next to its mate?
• Oct 20th 2009, 04:21 PM
Soroban
Hello, sweeetcaroline!

I found no elegant approach to this problem.
I was forced to use "brute force" listing . . . *blush*

Quote:

3 pairs of socks are placed side-by-side of a straight clothes line.
The socks in each pair are identical, but the pairs themselves are of different color.
If I stand on one side of the line, how many different color arrangements can I see
if no sock is next to its mate?

Suppose the colors are Red, Blue, and Green.

Then we have: .$\displaystyle \{R,R,B,B,G,G\}$

We have 3 choices for the color of the first sock (#1).
Suppose it is Red.
The other Red can be in positions #3, 4, 5 or 6.

Reds in (1,3): .$\displaystyle R\:\_\:R\:\_\:\_\:\_$
Then there are two cases:
. . #2 is $\displaystyle B$, and the last three must be $\displaystyle GBG.$
. . #2 is $\displaystyle G$, and the last three must be $\displaystyle BGB.$

Reds in (1,4): .$\displaystyle R\:\_\:\_\:R\:\_\:\_$
Then there are four cases:
. . #2, 3 are $\displaystyle BG$, and the last two are $\displaystyle BG$ or $\displaystyle GB.$
. . #2, 3 are $\displaystyle GB$, and the last two are $\displaystyle BG$ or $\displaystyle GB.$

Reds in (1,5): .$\displaystyle R\:\_\:\_\:\_\:R\:\_$
Then there are two cases:
. . #2, 3, 4 are $\displaystyle BGB$, and the last is $\displaystyle G.$
. . #2, 3, 4 are $\displaystyle GBG$, and the last is $\displaystyle B.$

Reds in (1,6): .$\displaystyle R\:\_\:\_\:\_\:\_\:R$
Then there are two cases:
. . #2, 3, 4, 5 are $\displaystyle BGBG.$
. . #2, 3, 4, 5 are $\displaystyle GBGB.$

Hence, there are: .$\displaystyle 2 + 4 + 2 + 2 \:=\:10$ arrangements that begin with Red.

Since there are 3 choices for the color of sock #1,
. . there are: .$\displaystyle 3 \times 10 \:=\:\boxed{30}$ arrangements with no adjacent mates.