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An ice cream store offers 14 different flavors. Customers can purchase a single scoop or a double scoop ice cream. The double scoop portion DOES NOT allow two scoops of the same flavor. How many different ice creams can be purchased?

I am not too sure how to do this problem. Do I just do $_{14}\complement_1$ (Single scoop) + $_{13}\complement_1$ (Double scoop)? Or is there a different way to tackle this problem?

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3 Answers 3

up vote 3 down vote accepted

You have the single scoops correct, but the double scoops would be $14\choose2$. Thus you have $14+{14\choose2}$ possibilities.

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From an educational point of view, I think it's wise to drop the Pr Cr, and look at just the numbers for a moment. Of course, there are 14 different single scoops.

A double scoop can be made by adding a different scoop to a single scoop. For every single scoop, we have 13 different flavours, so we end up with $14 + 14 * 13 = 14^2$ different scoops in total.

Now a subtlety is that a cone with for example vanilla on the bottom and chocolate on top is counted as different from a cone with vanilla on top and chocolate on the bottom. The cones are what we (in math) call a permutation of each other: they contain the same things and the same amount of them, just in a different order.

The question if cones with the same flavours but different stacking should be treated as the same is not addressed in this exercise. We now counted them as if they were different (I hope you can understand why, this is difficult for me to explain in words: a picture of a mathematical tree might help here, but it's a lot of effort to make one right now). All we have to do to correct for this, is divide the permutations for the double scoops by two (since every double scoop combination has two permutations). So we have $14 + \frac{14 * 13}{2} = 14 + 7 * 13 = 105$ different combinations.

Of course, you can also use nCr nPr on your calculator, but I just wanted to show that the math behind it is pretty easy to understand and you don't really need a calculator for small examples like this.

(The following paragraph is not necessary to understand my answer, but hopefully it shines some light on combinations and permutations).
Typically, if you have n different things, there are $n! = n * (n - 1) * (n - 2) * ... * 2 * 1$ different orders (so $n!$ different permutations). We can derive this quite easily. We can make every order by first choosing the first element, then the second, and so on, until we have picked all elements. Obviously, for the first element, we have n choices, for the second $n - 1$ (because we already picked one element that we thus can't use again), for the third $n - 2$ (for the same reason), for the third $n - 3$, and so on. With the same logic, we don't have to necessarily stop until there are no more elements left, we can also stop earlier.

Using this method, if we define a nPr b as "How much different permutations are there if we choose a sequence of b things from a set of a different things, without picking one twice?", we find $a \text{ nPr } b = a * (a - 1) * (a - 2) * ... * (a - (b - 1)) = a! / (a - b)!$. Notice that there are $b$ factors, which correspond to the $b$ times we have to choose the next element in the order. Now if we want to count only the combinations and thus regard different permutations as the same if they have the same things, we just need to divide by the number of permutations of $b$ things. This means dividing by $b!$. So we have $a \text{ nCr } b = a! / (b! * (a - b)!)$

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You can pair the different flavours in ${14 \choose 2}$ ways.

How many single-scoop ice creams are there?

Combine them and you'll have the number of different ice creams.

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