Ten pin bowling and Combinatorics

This is just an idea I had tonight while on camp with Year 7 and 8. It doesn't relate to anything thing I'm doing in class at the moment, but I wanted to get it written down somewhere before I forgot.

So we went ten pin bowling tonight, and I noticed that after each bowl the scoreboard showed a quick video of the pins getting knocked over (I wish I'd gotten a photo of it). Maybe this is pretty standard and you all know exactly what I'm talking about, but we don't get to see much bowling in a town as small as ours.

Anyway, initially I couldn't tell if the video was really recorded live with a camera I couldn't see, or if it was faked by showing a prerecorded video with the appropriate pins being knocked over. (I'm pretty sure the videos were real, but that's beside the point). It got me thinking - how many videos would need to be prerecorded in order to be able to show every combination of pins falling over?

The simple solution is this: there are ten pins, and each pin has two possible states, standing or knocked over. So the total number of combinations is 210, or 1024 as any addict of "2048" would be able to tell you. Or if we imagine a more general sport called n-pin bowling, then the number of videos needed is 2n.

The thing is, this is not the first solution I tried. Instead, I tried to use combinatorics:

  • There is C(10,0) = 1 combination with no pins knocked over.
  • There are C(10,1) = 10 combinations with 1 pin knocked over.
  • There are C(10,2) = 45 combinations with 2 pins knocked over.

etc. Then we add them together. This is much more work than the other method (particularly when trying to do it in you head while supervising a bunch of excited 12-14 year olds), but should give us our result. We can make this simpler by remember that combinations are contained in Pascal's triangle, so we can just add the numbers in row 10 to get our result:

1 + 10 + 45 + 120 + 210 + 252 + 210 + 120 + 45 + 10 + 1 = 1024

But this should always work for our hypothetical game of n-pin as well, by adding the entries in row n. Because the number of videos should be the same regardless of which method we use, this leads to the following result:

The sum of the entries in row n of Pascal's Triangle is 2n.

So that's kind of cool. A more typical proof of that statement is to consider the binomial expansion of (a + b)n with both a and b set to 1. So now we can relate the problem to algebra as well!

I know this isn't a lesson yet, but I wanted to get it down while it was still fresh in my mind. Hopefully I'll remember to look up this post later in the year, when I actually have to teach this stuff! There's also few possible extensions to this problem I thought of:

  • Are there any combinations we can eliminate and not make videos for because they are impossible to occur?
  • My 2048 reference was really just a joke. But thinking now, 2048 is all about powers of 2. Pascal's triangle is (in a rather sneaky way) also about powers of 2. Is there some hidden connection between 2048 and Pascal's Triangle?

For the record, I scored 133 and 126 in our two games of ten pin. Not great scores, but pretty good for me. I did start with two strikes, so now some of the kids think I'm some sort of bowling genius.

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Shaun used to be maths, IT and ocassional physics teacher at a small P-12 school (primary and secondary) in rural Victoria, Australia. He is currently in the process of starting his career again in the United States.

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