Math Chat: Rearranging Letters And Spinning Sprinklers

Old challenge

A recent National Public Radio Weekend Edition puzzle asked to replace one letter in HOMEGAME by another letter to get the name of a famous philosopher. Bruce Cobi asks how many different letter sequences you can obtain this way. What if you are also allowed to rearrange the letters? (The NPR answer was none of these, but to replace the Greek letter OMEGA with U to get HUME.)

Answer

Laura Sabel and Michael Stern correctly figure that each of the 8 letters of HOMEGAME may be replaced of any of the other 25 letters of the alphabet for a total of 8 x 25 = 200 sequences.

Erik Randolph, Ari Turner, and Derek Smith compute the number of rearrangements. If all of the letters of a sequence were different, the number of arrangements would be eight choices for which letter to use first, times seven choices for which letter to use second, times six choices for which letter to use third, and so on, for a total of 8x7x6x5x4x3x2x1 = 40,320 arrangements. But if for example there are two E's and two M's, then switching the E's or switching the M's or switching both pairs makes no difference, and we have counted each arrangement four times, so that the number of arrangements is really 40,320/4 = 10,080.

To get a collection of letters with two E's and two M's from HOMEGAME, you have to replace one of the four other letters by one of the 20 letters not in use, which can be done in 4 x 20 = 80 ways. Therefore the total number of arrangements for this case is 80 x 10,080 = 806,400. There are lots of other cases: three E's and one M, three E's and two M's, and so on. Each case must be computed separately.

The grand total is 1,794,240 possible arrangements. Ironically, trying every single one of these would not have produced NPR's tricky answer!

Mathematicians stumped

Martin Hildebrand sent in the notorious "3n+1" problem that has stumped mathematicians for decades. Start with any natural number. If it is even, divide by 2. If it is odd, triple it and add 1. Keep going. For example, if you start with 13, you go 13, 40, 20, 10, 5, 16, 8, 4, 2, 1. Do you always get down to 1, no matter what number you start with?

In an excellent survey article in a 1985 American Mathematical Monthly (Vol. 92, pp. 3-23), Jeffrey Lagarias reports that the result had been checked for all numbers up through 12 digits by Nabuo Yoneda at the University of Tokyo.

The quaternions

There are some fascinating and useful systems of arithmetic beyond the real numbers and imaginary numbers. One of the most interesting and important is the "quaternions."

Among real numbers, x2 = -1 has no solution (because any number squared is nonnegative) and (-1) is undefined. The complex numbers come from adding i = (-1) to the real numbers. Among the complex numbers, x2 = -1 has two solutions, i and -i. The quaternionic numbers come from adding to the real numbers three square roots of -1: i, j, k, with ij = k unequal to ji = -k, so that multiplication quaternions is not commutative, the order matters. Among the quaternions, x2 = -1 has infinitely many solutions, not just i, -i, j, -j, k, and -k, but also for example (3/5)i + (4/5)j, because ((3/5)i + (4/5)j)x((3/5)i + (4/5)j) = (9/25)i2+ (12/25)ij + (12/25)ji +(16/25)j2 = (9/25)(-1) + (12/25)k + (12/25)(-k) + (16/25)(-1) = -9/25 + 0 - 16/25 = -1. The quaternions can be used to describe all possible rotations of space.

New sprinkler challenge

An old-fashioned water sprinkler with s-shaped spiral arms is propelled counterclockwise as the water spurts out. Which way will it turn if you place it upside down under water and have it suck water in?

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