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We reviewed the binary factorization, and then explored "Yanghui's triangle" (and you should look yours over carefully):
We attempted to verify these rules, and managed to find that they are consistent; we couldn't contradict the rules -- although Becca noticed that there's a problem in the next to last row.
Upon closer examination, we discovered that there is an error in the table! Whoever wrote this, in 1200 AD or so, must have made a little mistake, and lost a stroke. See if you can find it.
We'll talk more about symmetry as we get further into the course, but here's one place where it really shows up. Because we expect the table to be a mirror reflection of itself, left and right, we can spot errors -- where the symmetry breaks down.
But there are still some "open questions", which we haven't figured out. I asked you to try to answer some questions (mathematicians look for patterns):
There were still some open questions about the bamboo counting rods way of writing numbers. Let's nail those down first. How does this writing work?
One can also think of this as a connection to "Plinko" (if you've ever heard of that); and to something called a Galton Board. (I'll show off my "Hexstat Probability Generator" -- which shows you that Pascal's triangle is also very closely tied to probability calculations, too.)
Imagine flipping coins for each ball: heads, go left; tails, go right. Let's draw the decision tree.
Pascal's triangle simply doesn't care if you go right then left, or left then right: you end up in the same plinko spot. So Pascal says "2 ways to do that". "1 way to get HH", "1 way to get TT", but two ways to get a head and a tail (HT and TH). Pascal's triangle just does the accounting: \[ 1 \hspace{1in} 2 \hspace{1in} 1 \] The next row is the ways of getting three heads, two heads, one head, and no heads: \[ 1 \hspace{1in} 3 \hspace{1in} 3 \hspace{1in} 1 \]
And, by counting the leaves of the binary tree, we see that it's always a power of 2! So the sum of the rows of Pascal's triangle has to be a power of 2.... Now we understand Emilee's "power of 2" rule;
but also the rule about where adding the "two above" to get the value in the next row. That's the plinko rule!:) Or the Hexstat rule. Where do balls come from at any point in the triangle? From the two spots right above.
Now, if we don't care about "who's who", then we can see all the different types of Facebooks we can have:
Now how many different Facebook configurations are there for, say
For example, there is only one way to choose all; and only one way to choose none! Those are the easy cases. It should be obvious that there are $n$ ways to choose 1 from among $n$; or to leave one out of $n$ (note the symmetry).