Choose Your Own Mathventure

Hello everyone!

Today, I'd like to talk about one of the most fundamental concepts in combinatorics: binomial coefficients.  We define (read " choose ") to be the number of ways to choose of things.  So, for example, there are ways to choose of blocks to color red, as shown below.simple_choose

 

Now, this is combinatorics, so most of what we're going to do today is about counting the same thing in two ways.  We'll start with the simplest first.  Let's show .  That is, there are exactly as many ways to choose of items as there are to choose of them.  Why?  Well, there are ways we could color of blocks red, as we saw above.  But we could just as well choose the other blocks to color blue, as below.  So .  By the same logic, we get that .choose_inverse

 

Let's look at another interesting result: .  Why is it true?  Let's count all the ways to pick any number of blocks from a row of length to color red.  One way we could see this is to break up the possibilities by the number of red blocks.  If we want red blocks from our total blocks, there are to choose them.  Summing over all valid , we get total possibilities.

Another way to think of it is that for each block, there are two possibilities: red or white.  The colors for each are independent, so we can multiply the possibilities together.  There are blocks, each with possible colorings, for a total of possibilities.  We've counted the same colorings with these two methods, so the counts must be equal.  Thus, .

 

It turns out binomial coefficients are also related to each other by the recurrence .  We already know that counts the ways to choose of tiles to color red.  So let's find a way to count the same thing to get the other side of the equation.  In this case, we can divide the possibilities into those where the last tile is red and those where it isn't.  If the last tile is red, then we have more tiles to choose from, and we want to be red.  If it isn't, we want to choose of the remaining tiles.  This gives us , as we wanted.choose_by_lastThis last relationship gives us a nice way to construct the binomial coefficients.  We can place them in a triangle structure, as below.  You can see down in the sixth row.  From the recurrence above, we know that each number in this triangle is the sum of the two above it.  You might have seen this pattern of numbers before.  It's known as Pascal's triangle.  There are a lot of pretty combinatorial and number theoretic results that come out of it.  For today, I'll just mention that this is one way to derive it.pascal

 

One last thing before I sign off for the day.  I've been calling the binomial coefficients.  Why do they get that name?  It turns out that is the coefficient of in .  To see why it works, imagine writing out the copies of we want to multiply, as below.  To get a copy of , we  must choose of the terms to give us an .  Equivalently, we can imagine a block for each term which is colored red if we choose the and blue if we choose the .  This can happen in ways, so we get a total of copies of , which we then add together.  Thus, the coefficient of is .binomial

Facebooktwitterredditpinterestlinkedinmailby feather

Leave a Reply

Your email address will not be published. Required fields are marked *