The License Plate Problem (Combinations/Permutations)

The idea for this post came to me from my good friend Johnny Aceti.

Suppose you’re on a road trip, someone else is driving, and you’re trying to while away the hours by staring out the window. Now after a while, this is bound to become boring, so you devise a game to play with yourself to stay entertained. You start looking at the cars as they pass and more importantly, their license plates. Each one seems to be different, all just a jumbled collection of characters including letters and/or numbers. You think to yourself, as any good mathematician would, “I wonder how many ways there are to create a unique license plate given an alphabet a and a format for the plate?” Let’s go through and figure this out.

First, define the alphabet. That is, the possibilities for each character that we can put on our license plate. For this example, let’s use 0-9 and A-Z. There are 10 digits and 26 letters so that gives us an alphabet of 36 characters.  (a=36)

The next thing we will do is define the format for the plates in question. Assume that a license plate, for simplicity’s sake, can have 6 characters, either numbers or letters. Let’s first figure out how many possibilities for a license plate with only one character there are. There are 36 characters in our alphabet, so there are 36 possibilities for one “column”. Now there are also 36 choices for each of the remaining five spaces. So we have to do 36*36*36*36*36*36. That’s one time for each of the spaces. To write that simpler, we can generalize this to a^c, where a is the alphabet size and c is the number of characters on each “plate”.

Now, this problem was assuming that we could repeat characters on a license plate. For instance, this assumes that 333333 could be a valid license plate. What if we had a rule that we couldn’t repeat characters?

The first column remains at 36 possibilities, but when we get to the second column, we’ve already used one of our characters and we can’t repeat, so we only have 35. Then we have 34 for the next column. So for this we’ll need to use a factorial (!). However, since we have 36 characters and only 6 spots, we will still have 31 possibilities even in the last spot. We don’t need the whole factorial. Thus, we will use only a!/(a-c)! where a is the alphabet size and c is the number of “columns” on the license plate.

Hope this was a good simple combinatorics problem and displays a way that a mathematical principle can be used in a real life problem.


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