Problem of the Week 1163

Walk a Mile in My Shoes

Solution

Solved by Alan Lipp and Joseph DeVincentis. One incorrect solution was received.

DeVincentis writes:

The number of ways for each person to select a left shoe is 5! = 120. The number of ways for each person to select a right shoe which doesn't match the left shoe is the number of derangements of 5, the number of permutations of the numbers 1 through 5 where the nth element is never n. Such derangements must either be a cycle of 2 and a cycle of 3, or a single cycle of 5.

For the first case, there are Binomial[5,2] = 10 ways the cycle of 2 can appear, and one way to assign the shoes in that cycle, and two ways to assign the shoes in the cycle of 3, making 20.

For the second case, the first guy can have any of four shoes other than his matching one. The guy with that match can have any of three shoes other than his matching one or the first guy's matching one. The guy with his match can have any of two shoes, and the last two shoes have only one choice that completes the cycle, so there are 24 of these, or 44 total derangements.

So the answer is 44 * 120 = 5280 — the number of feet in a mile.

This problem is not too difficult, but I could not resist the 5280 connection. More challenging versions are in the cited article: Sally Cockburn and Joshua Lesperance, "Deranged socks," Math Mag. 86 (2013) 96-109.

[Back to Problem 1163]

© Copyright 2013 Stan Wagon. Reproduced with permission.


17 September 2013