The Yanghui Triangle, Part II

Welcome back for round two!  If you missed Part I, you can catch up here.

An Alternative Explanation

In the first part of this post, we explored the following specific consequence of the triangle that we generalized:

\displaystyle \binom{6}{2} = \binom{5}{1}+\binom{5}{2}.

I recognized that I lacked an example of why this might be true from a counting perspective.  Using the triangle below, we see that the equation above is saying that 15=5+10.

Here is Yanghui’s Triangle for convenience.

So, why is this true from a counting perspective?

Let’s say we want to select 2 of these 6 letters: ABCDEF.  There should be 15 ways of doing this according to the triangle.  Let’s list them below.


Now, let’s try it another way.  Let’s first make a decision on the F.  We can either select it, or not select it.  Let’s explore both cases.

We Select The F. Now, our task is to select 1 more of the 5 letters ABCDE.  We can do this in \binom{5}{1}=5 different ways.  We are left with AF, BF, CF, DF, and EF.

We Do Not Select F. Now, our task is to select 2 of the 5 letters ABCDE.  We can do this in \binom{5}{2} = 10 ways. Here they are: AB, AC, AD, AE, BC, BD, BE, CD, CE, DE.

Now, combine these two together for the total number of ways to select 2 of 6 letters.

OK, now onto some more fun properties.

Powers of 2

If we sum each row of Yanghui’s Triangle, you will notice a pattern.

  1. Sum of 0th row: 1
  2. Sum of 1st row: 2
  3. Sum of 2nd row: 4
  4. Sum of 3rd row: 8
  5. Sum of 4th row: 16

And so on.  From the title of this section, you probably have already guessed the pattern.  If we think of the first row as the zeroth row instead, then we simply take 2 to the power of the row to get the sum.

Why does this work? You may have to review the first post to recall that the numbers in each row can be used as the coefficients when expanding expressions such as (x+y)^6.  From the previous post,

\displaystyle (x+y)^6 = x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6.

Let’s use this result, and plug in 1 for both x and y. The left part of the equation says that this is just 2^6.  The right part of the equation says that this is the sum of the numbers in the 6th row of Yanghui’s Triangle.

Fibonacci Sequence

By using hexagons to construct the triangle, the lines that create Fibonacci are easy to construct

The picture above says it all.  By creating lines that run parallel to the top left side of the hexagons and summing up the numbers that it passes through, we get Fibonacci’s Sequence.

Why is this?

Let’s explore the line that goes through 1, 4, and 3, which sums to 8.  The 1, 4, and 3 correspond with the following counting principles: choosing 0 of 5 things, 1 of 4 things, and 2 of 3 things.  To see why this must sum up to the 5th Fibonacci number (again, thinking that the first 1 in Fibonacci’s sequence is the null, or zeroth number), review my explanations in Approach 1 and Approach 2 in Jumping Lily Pads. You can also watch the quick video there.

For a brief explanation of why \binom{5}{0}+\binom{4}{1}+\binom{3}{2} sums up to the fifth Fibonacci number, it is like counting the number of ways that a frog can jump from shore to the fifth lily pad if the frog can jump one or two lily pads at a time.

He can make 5 total jumps, 0 of which are skipping any lily pads: \binom{5}{0}=1.

He can make 4 total jumps, 1 of which must be a double: \binom{4}{1} = 4.

He can make 3 total jumps, 2 of which must be doubles: \binom{3}{2} = 3.

As was described in that post, this is equivalent to summing up the number of ways you can jump to lily pad 3 (3) and the number of ways you can jump to lily pad 4 (5).

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