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Seeking assistance with the following question please:

A store sells $11$ different flavours of ice cream. In how many ways can a customer choose $6$ ice cream cones, not necessarily of different flavours?

I'm thinking this is a pigeon hole type question where you want to allocate $6$ similar objects (cones) to $11$ pigeon holes (flavours), without restrictions on the number to go in to each pigeon hole. In which case there would be $\binom{6+11-1}6 = \cfrac {15!}{6!\ 9!} = 5005$ ways.

Is this correct?

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    The term *pigeonhole* usually refers to a different kind of problem, but your analysis is correct: in effect you’re counting the ways to distribute $6$ indistinguishable objects amongst $11$ distinguishable boxes, and the answer is indeed $\dbinom{6+11-1}{11-1}=\dbinom{6+11-1}6$. However, this is $\dbinom{16}6=8008$.2012-11-20
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    @TryinHard As pointed out by Scott, People usually call this a [*ball-and-urn problem*](http://www.artofproblemsolving.com/Wiki/index.php/Ball-and-urn) or a *bin-and-ball problem*. The term *pigeonhole* usually refers to [something else](http://en.wikipedia.org/wiki/Pigeonhole_principle).2012-11-20
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    Your answer is right, BTW. See ["stars and bars"](http://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)). Another way of seeing this is that you're counting nonnegative integer solutions to $x_1 + x_2 + \dots + x_{11} = 6$.2012-11-20

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