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Flavor Combinations!

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Flavor Combinations!


Post by ZardoZ » Mon Jan 18, 2016 4:12 am

An ice cream store has 20 different flavours. In how many ways can we order a dozen different ice cream cones, if each cone has 2 different flavours?
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Re: Flavor Combinations!


Post by Demetres » Mon Jan 18, 2016 4:12 am

I am assuming that the order in which we order the cones does not matter. I am also assuming that the order in which the flavours are placed in the cone also does not matter. (Although it did matter to me when I was a kid. :)) Furthermore, since it says "2 different flavours" I am assuming that you cannot order two scoops of the same flavour.

EDIT: A further assumption that I have made in my solution is that in our ordering we are allowed to order the same cones. (I am thinking of a group of twelve people wanting ice cream cones, so some of them might want exactly the same flavours in their cones.)

We have \(\binom{20}{2} = 190\) different possible cones that we can order. We need to decide how many of these we want for a total of 12 cones. I.e. we need to find the number of non-negative integer solutions to the equation \[x_1 + \cdots + x_{190} = 12.\] This is equal to \(\binom{201}{12}\). To see this consider a sequence consisting of 12 \(\ast\)'s and 189 \(\dagger\)'s in some order. This sequence gives a solution to the above equation as follows: \(x_1\) is going to be the number of \(\ast\)'s before the first \(\dagger\), \(x_2\) is going to be the number of \(\ast\)'s after the first \(\dagger\) but before the second \(\dagger\) and so on. It is immediate that this correspondence is bijective and so it is enough to count the number of such sequences. But this is clearly given by \(\binom{201}{12}\).
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