Zoë's Math

[oreophile]

Recently I was at the Dayton book author event, and heard John read Chapter 9 of Zoë's Tale. I bought a copy of the book, had John autograph it, and just finished reading it last night. During the reading, something bothered me, and after I read the book it was clear to me that John had done some *bad math*.

On page 81 Zoë and Grethchen break up a fight between two clots of boys from different home worlds. Zoë says: "And there are colonists from ten separate worlds. That's a hundred different possible idiotic teenage boy fight situations." Gretchen notes about the Colonial Mennonites: "They're pacifists. So it's only eighty-one possible idiotic teenage boy fight combinations."

Their calculation is obviously ten squared and nine squared combinations. But that's not how combinations work. The combination of ten items taken two at a time is 10*9/2 or 45, and nine items taken two at a time are 9*8/2 or 36.

Admittedly, that's still a lot of potential fights, but less than half of what the girls' calculate.

Maybe this error can be corrected in future editions of the book.

Oreophile

[Lauren]

They're girls! Hyperbolic speaking is well within their capabilities.
Alternately, they have yet to be taught combinatorics.

[txjak]

I have not read the book, but if you count the situation where group A starts a fight with group B and the situation where group B starts a fight with group A as two situations and also count a fight breaking out within a group as a situation, you have n * (n-1) + n possibilities, or n squared possible situations.

For example, if there were three groups, A, B, and C, then the 9 situations would be A-A, A-B, A-C, B-A, B-B, B-C, C-A, C-B, and C-C.

[oreophile]

I have not read the book, but if you count the situation where group A starts a fight with group B and the situation where group B starts a fight with group A as two situations and also count a fight breaking out within a group as a situation, you have n * (n-1) + n possibilities, or n squared possible situations.

For example, if there were three groups, A, B, and C, then the 9 situations would be A-A, A-B, A-C, B-A, B-B, B-C, C-A, C-B, and C-C.

First, I highly recommend you read Zoë's Tale.

Secondly, the girls are talking about combinations of groups, not permutations or within group arguments. So in your example, the only combinations are A-C, B-A, nad B-C. It doesn't make any difference who starts the confrontation. Combinations of three items, two at a time are 3*2/2 = 3.

[txjak]

First, I highly recommend you read Zoë's Tale.

Oh, I fully intend to.

My comment is only based on what you wrote. It wasn't obvious to me that Zoë was talking about combinations as you have assumed, since she didn't use that word in the quote you provided.

[Nightsky]

For example, if there were three groups, A, B, and C, then the 9 situations would be A-A, A-B, A-C, B-A, B-B, B-C, C-A, C-B, and C-C.

The girls assume that boys won't fight with boys from their own homeworld, knocking out A-A, B-B, and C-C.

I forget when I first was formally exposed to combinatorics (Algebra II?) but I first ran across the concept in PBS's old "Square One" show, as a mere slip of a girl.