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 Post subject: Flavor Combinations!Posted: Mon Jan 18, 2016 4:12 am

Joined: Wed Nov 11, 2015 12:47 pm
Posts: 13
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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 Post subject: Re: Flavor Combinations!Posted: Mon Jan 18, 2016 4:12 am

Joined: Mon Nov 09, 2015 11:52 am
Posts: 76
Location: Limassol/Pyla Cyprus
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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