The name of the Knewton tech blog is “N choose K”. If it’s been a long time since you set foot in a math class, this may not have any meaning to you. However, it’s an interesting concept, and really not all that difficult. In this post, I’ll go over what the term means in simple English and discuss how we use it here at Knewton.
Imagine you’re visiting the newest, trendiest lounge in New York. Like many such lounges, this one specializes in producing excellent cocktails from before the Prohibition era. You look over the menu and see that they offer one drink for each type of alcohol: champagne, gin, rum, tequila, and vodka. You decide that you want to try all of them, in pairs rather than one at a time, to see how well they go with each other. You want to know how many different pairs‚ (one for each hand) you can make with these 5 drinks.
One way of doing this would be to count each of them. To do this, we can label each of the drinks with their main ingredient:
and systematically count each of the pairs (note that order of ingredients is not important):
for a total of 10 different pairs of drinks. Mathematically, this is known as n choose k, and is written as with n being the total number of items, and k being the number to choose each time. You’re asking “how many ways can I choose 2 items out of 5,” which we know from our graph is:
It’s not too difficult to count each of these by hand. But now imagine that you’re an octopus at a different bar, which has 20 drinks, and you want to find out how many ways you can choose 8 drinks out of 20, i.e.
Clearly, counting is not an option here.
The great thing about n choose k is that there’s a simple formula to figure it out: you just plug in your values for n and k, and out comes the answer. Here’s the formula:
which is a lot less complicated than it seems. The exclamation point is known as a factorial, and you can find any factorial by multiplying together all the numbers from that number to 1. So, 3! is simply: Let’s try it on our original problem, how many ways can we choose 2 drinks out of 5?
Note how we can make the calculations easier by canceling the top and bottom of the fraction when they’re exactly the same.
It’s a good thing octopi have a strong tolerance for liquor.
Of course, “n choose k” pertains to more than just mixing drinks at bars. In fact, you can see this in all sorts of applications, and often in exactly the sort of work we do here at Knewton. For example, let’s say we have a graph that has only paths to adjacent nodes:
The question we need to answer is how many shortest paths are there from A to Z? (These are called lattice paths). There are algorithms that allow you to determine this exactly, but they require polynomial time to complete. Instead, let’s solve an easier problem: What is the upper bound of the number of shortest paths from A to Z? So, what would the graph would look like where the upper bound is the same as the actual number?
Well, a fully-connected graph would fit this scenario:
We can count the length of the path from A to Z by counting the length of the path:
Clearly we cannot have a shorter path than this, and so all shortest paths must have exactly 6 steps. The only independent variable then is the choice of which steps we take to the right; the dependent variable is which steps we take down (it’s dependent because it is completely determined by the other variables). So in the example above:
Independent: steps 1, 2, 3, 4 are to the right
Dependent: steps 5, 6 are to the bottom (whatever is left)
Starting to sound familiar? It’s the same as “n choose k”! Given the total number of steps, how many ways can we choose to go right? And so we get:
More generally, for an m x n lattice grid, we can use the formula:
where j is one of the dimensions of the grid and k is the other; which one you pick doesn’t matter, as you’ll see below.
Be aware, you might see this formula as . This is correct if you count the starting point A as (0, 0) and ending point Z as (m, n), which subtracts one in each direction.
You might be interested in why we chose which to the right and which down; in other words, can we switch the independent variable? The answer is yes, and it doesn’t matter, and it’s pretty neat, because as you could calculate:
In fact, if you calculate a number of these and group them into a triangle shape, it results in a pattern known as Pascal’s triangle:
To find for example, count to the fifth row and the second column, and you’ll find that . As you can see, the left side of the triangle has the exact same values as the right side, which is why , and more generally:
These values are also known as binomial coefficients; if you’re interested in why, you can easily find more information on the web.
There are a couple neat patterns that Pascal’s triangle has. First, each number in the triangle is the sum of the two numbers above (you can think of the entire triangle surrounded by zeros if that helps). This is pretty miraculous if you recall our original formula for determining n choose k.
The second pattern has to do with its shape: if you take Pascal’s triangle and color only the odd numbers, you get a Sierpinski triangle, which you may be familiar with: it’s our n choose k blog logo!
Our work at Knewton allows us to play with ideas all day and turn those ideas into a reality. We chose the binomial coefficient to be the title of our tech blog because it reflects many of our core values as Knewton developers: our passion for mathematics and the language of ideas, and our commitment to intellectual endeavor.
In this entry we discussed what n choose k means and how you can use it to limit your alcohol consumption find the number of ways groups of things can be chosen. Then we found a simple formula to calculate it. We saw a practical use of formula in finding an upper bound of the number of shortest paths in a particular graph. Finally we looked at Pascal’s Triangle and the relationships between the binomial coefficients, aka the results of multiple n choose k computations. I hope you had some fun, learned a bit about math, and have something new to discuss at your next cocktail party.
Thanks to Kevin Wilson and Solon Gordon for their help with this post.