Halloween Candy and Counting

Happy Halloween! It’s that time of year in which we celebrate ghosts, pumpkins, and fear itself. So, what better time to discuss a very common fear among adults these days: Mathematics!

If you’re reading this blog, I’m guessing you’re probably already not too afraid of mathematics. But I hope you share this post with people who are somewhat spooked by it but like to face their fears now and then. And let’s face it, even for math lovers, every difficult-sounding math problem is always a little scary at first… until you work it out and realize that there’s only beauty behind the mask.

I recently made up the following problem for a friend teaching a discrete mathematics class:

Five kids, dressed as a Ghost, a Witch, a Monster, a Skeleton, and a Black Cat, knock at your door. You open it and welcome them in, but you realize you only have $3$ Snickers bars and $3$ Kit Kats left in your candy stash!

Since you have $6$ pieces of candy and there are only $5$ kids, you decide to give both a Kit Kat and a Snickers bar to the scariest costume, and then give the remaining four kids one piece each. How many different ways can you choose who to give what candy to?

Eeek! Ghosts and witches! And combinatorics!

Well, let’s consider. There are $5$ ways you can decide on the scariest costume, since there are $5$ kids to choose from. But for each of those choices, you also have to pick which two of the remaining four get the Kit Kat and which two get the Snickers.

To make this a little easier, suppose we picked the Ghost as the scariest, and so we have to choose which two of the Witch, Monster, Skeleton, and Cat get the Snickers bars (and then the other two will get the Kit Kat). Well, we could pick one of the following options:
1. Witch and Monster
2. Witch and Skeleton
3. Witch and Cat
4. Monster and Skeleton
5. Monster and Cat
6. Skeleton and Cat.
(Notice that I ordered these by first choosing all those that can pair with the Witch, and then all those without the Witch, to make sure I didn’t miss any.)

So there are $6$ ways to choose who gets the Snickers and who gets the Kit Kats, assuming we chose Ghost as Scariest. But this is the same computation no matter who of the $5$ we chose as Scariest. If we chose the Witch as scariest there would still be $6$ possibilities, and same for the other three. Therefore, there are $5\cdot 6=30$ total possibilities.

The nice thing here is that there’s a known formula for computing the $6$ possibilities for the Kit Kats and Snickers – the number of ways of choosing $2$ things out of $4$ things is written $\binom{4}{2}$, pronounced “four choose two”. The formula for $\binom{a}{b}$, the number of ways to choose $b$ things from $a$ things, is known to be $\frac{a!}{b!\cdot (a-b)!}$ where $n!$ means the product of all the numbers from $1$ to $n$. So to compute $\binom{4}{2}$, we just compute $$\frac{4!}{2!\cdot 2!}=\frac{1\cdot 2 \cdot 3 \cdot 4}{1\cdot 2 \cdot 1\cdot 2}=6.$$ This is a shortcut for counting the possibilities without listing them all out.

Can you see why the formula for computing $\binom{a}{b}$ is true?