
Proof by storytelling - CarolineW
http://chalkdustmagazine.com/features/proof-by-storytelling/
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javpaw
You can also learn more about this kind of proof in statistics 110 class from
Harvard:
[https://www.youtube.com/watch?v=KbB0FjPg0mw&list=PL2SOU6wwxB...](https://www.youtube.com/watch?v=KbB0FjPg0mw&list=PL2SOU6wwxB0uwwH80KTQ6ht66KWxbzTIo)

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losteverything
Are all courses in this course avail? How

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drostie
It's worth pointing out that the conventional formula for N choose K has a
"story proof" as well, and it helps introductory students who can't
necessarily keep straight whether, in these groups, order matters.

"There are N people in a room. Place a line of tape across the room dividing
it into two halves; we'll say this tape goes North-South. Now ask K of them
(politely, but also uniformly-at-random) to be on the East side of the tape
and the N-K remainder are on the West side of the tape. We know from the
definition that there are N choose K ways to do this.

"Now we ask the people on the East side to stand along the East wall and the
people on the West side to stand along the West wall; along the way we will
have to choose among the K! and (N-K)! different orderings for each set. So we
have a uniform distribution of these people chosen from K! (N-K)! (N choose K)
different possibilities.

"Now kindly ask them to walk in the order they're in to the North wall, but to
remain on their side of the tape. You suddenly see that this result must be
the same as simply asking the N people to line up uniformly-at-random along
the North wall and you then place a piece of tape that divides them into the
two groups. Every possibility in the one story comes from exactly one
configuration in the other story and vice versa, therefore N! = (N choose K)
K! (N - K)! no matter what N or K were."

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gohrt
That explanation leaves me more confused than when I started -- there's too
much to try to follow and visualize. Maybe it works in person.

~~~
drostie
Yeah. I think it works when you take three fingers from your left hand and
mash them together with all five from your right, representing a set of 8
elements. Now to perform 8-choose-3 and choose a set of 3 out of the 8
elements, you separate your hands and you've "chosen" (albeit in advance by
the topology of your hands) a group of 3 out of the 8, but your fingers are
still bunched up because these are sets and you haven't ordered everyone yet.
Now we ask for both sides to line up, you "order" them by flattening out your
hands facing each other palm-to-palm, as if you were holding a box on its
sides: the 3 and 5 fingers on either side are now lined up. Now if you've been
following the math you have (8 choose 3) * 3! * 5! ways that you could have
done those steps.

Now you take these 2 hands facing each other and turn them palms-down so that
they form one line of 8 fingers, a permutation of the original set-of-8 which
happens to have a piece of tape (or other division) separating the first 3
from the last 5. Ignoring this division you have just the permutation of 8.

With a little reasoning you get the one-to-one and onto properties of this
mapping: if you got to some permutation of the 8 this way, there was only one
choice of the 3 out of the 8, and only one ordering of the left, and only one
ordering of the right, that generates that permutation of the 8. Furthermore
every permutation can be generated by those very steps of choosing the first 3
of the permutation in the 8-choose-3 step and then ordering the two groups
appropriately. Since this process is one-to-one and onto, it must be the case
that (N choose K) K! (N - K)! = N!, the total number of possibilities for each
must be the same.

Does that help?

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ThrustVectoring
These are fun. There's another similar technique, which proves that a function
f(n) maps positive integers to other positive integers by describing the set
that the function counts.

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gohrt
any ideas on the last one? It's totally different structure from all the
others -- it doesn't involve combinations at all. It seems totally out of
place.

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tofupup
neat

