Weighing your options…

…because you pay by the ounce.

One of my best friends from high school, who currently lives in NYC, is in town right now. So a bunch of us met up for pho (which is reportedly awful in New York), then went to a make-your-own-mix fro-yo shop for dessert.

I can’t get fro-yo without thinking two things:

1) All frozen yogurt shops are absolutely adorable, right down to the spoons.

2) I wonder how many different combinations I can make.

On the drive to the shop, I decided that I’d actually calculate out the number of combinations while we enjoyed our dessert. But upon our arrival, when I was reminded of exactly how many flavors and toppings were available, I realized the numbers would be staggering.It wouldn’t be so bad if you only picked one fro-yo flavor and one topping. If you want to figure out how many combinations are possible in a situation like this one, you just multiply together the number of options you have in each category. There were 14 fro-yo flavors and about 20 toppings (and that’s admittedly a very conservative estimate), resulting in 280 possible combinations.

But who actually limits themselves to one flavor and one topping? If you’re choosing more than one item from each category, your calculations become more complex.

Let’s look at flavors first. Based on the calculation we just did, it’s easy to assume that, if you picked two different flavors, you’d have 14 options for the first one and 13 options for the second one; multiply them together and you get 182 possible pairings. However, that calculation does not take into account the fact that (mathematically, at least) it doesn’t matter what order you pick your flavors in. Multiplying 14 by 13 would count red velvet followed by vanilla and vanilla followed by red velvet as different pairings.

There’s a mathematical way to get rid of these duplicates.

If n equals the number of available options, r equals the number of those options that you want to select, and the order of your choices doesn’t matter, you can use this formula:

n!/((n-r)!*r!)

In math, the exclamation point is used as shorthand for the factorial function.

2!=2*1

3!=3*2*1

4!=4*3*2*1

and so on.

To choose 2 flavors from a selection of 14, the calculation looks like this:

14!/((14-2)!*2!) = 14!/(12!*2!) = 91 combinations

You can apply the same method when calculating possible topping combinations, since order doesn’t matter there either. If you pick 2 of the 20 toppings, your calculation looks like this:

20!/((20-2)!*2!) = 20!/(18!*2!) = 190 combinations

Now we’re ready to look at the total number of fro-yo/topping combinations that are possible in this shop. Let’s assume that someone picks 1 or 2 flavors and anywhere from 1 to 4 toppings.

number of ways to pick 1 fro-yo flavor: 14
number of ways to pick 2 fro-yo flavors: 91
total number of fro-yo flavor options: 105
number of ways to pick 1 topping: 20
number of ways to pick 2 toppings: 190
number of ways to pick 3 toppings: 1140
number of ways to pick 4 toppings: 4845
total number of topping options: 6195

Now we can multiply our total number of flavor options by our total number of topping options to get our total number of combinations, which is 650,475.

How many of these combinations actually taste good? That’s a much more difficult question.

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