How Do You Calculate Combinations for Ice Cream Flavors with Constraints?

[roadrunner]

Homework Statement

i think i might have it right but i want a 2nd opinion (mainly part c)

31 flavors of icecream. we are picking 12 cones(1 scoop).

a) cannot have same flavor twice
b) can have any flavor up to 12 times
c) can have any flavor up to 11 times but not 12

The Attempt at a Solution

a) no idea where to even start...
b)(31+12-1)c(31)?
c)(31+12-1)c(31)-31*30...since there are 31 flavors, you will have 31 cases of 11 of one flavor and 1 of another (of the 30 remaining flavors)

 

[roadrunner]

thinking a is maybe 31C12?

 

[dynamicsolo]

thinking a is maybe 31C12?

(a) Think about how the cones would get filled. You have 31 choices for the first, 30 for the second, etc. Are we producing all 12 cones in one go, or are you going to B-R on 12 separate occasions? I'm asking whether order matters.

(b) Since you can have *any* flavor up to 12 times, does it matter what is in each cone? How many ways could you fill 12 cones?

(c) Make the twelfth cone a "set-aside". How many ways could you fill 11 cones? Now, in these cases where all eleven are the same flavor, you must choose the twelfth to be different from all the rest. Otherwise, it doesn't matter what's in the twelfth cone.

 

[roadrunner]

a) order does NOT matter which is why i don't understand it lol...if order mattered its 31!/20!
(whats B-R)?

b) your not impoying 31^12 are you? (its in the chapter on combinations with repettitions...and 31^12 woold be if order matters)
c)

 

[dynamicsolo]

a) order does NOT matter which is why i don't understand it lol...if order mattered its 31!/20!
(whats B-R)?

B-R is Baskin-Robbins, home of the 31 flavors. I asked about order because I wasn't sure about the context of the problem. So order doesn't matter because all twelve are being made at once. So the answer should be 31C12, since it is essentially 31 choices taken from a bin without replacement ( 31·30·...·20 = 31!/(31-12)! = 31P12 ), divided by 12! possible arrangements of the cones that we are counting as only a single possibility.

b) your not impoying 31^12 are you? (its in the chapter on combinations with repettitions...and 31^12 woold be if order matters)

Now that I know that order doesn't matter, I guess this gets modified a little. What I was suggesting, though, is that if any amount of repetition is allowed, it doesn't really matter what choices you make for the twelve. So we have 31 choices taken from the bin with replacement. That would be 31^12, but, as you say, we don't want to count the different orders of removal as distinct, so we would again divide by 12! (Even if all twelve cones got the same flavor, we want to count it as one outcome, but there are 12! arrangements in which individual scoops could end up in cones. I'm still thinking about this, though...)

This is a much larger number than for part (a), which makes sense.

c)

The last one's a bit more complicated, since we have to make sure we don't allow more than 11 duplicates. I think we can start with the result for part (b), as you proposed, and subtract outcomes. But I think there are only 31 cases to be removed: those where all 12 cones have the same flavor. You don't have to deal with the cases of 11 identical cones and what to do with the twelfth, because those already exist in the set we've counted in part (b). So I believe (at the moment, anyway) that this would be
[(31^12)/12!] - 31.

I'll think about this some more, but I don't think the nCr combinatorial function comes into the last two parts, since you have the equivalent of selection from an urn with replacement (have you looked at lottery problems?). But feel free to argue with this -- combinatorics is one of the harder parts of probability theory to evaluate clearly.

 

[roadrunner]

ok part a makes sense
for part b now I am thining combination with repetition
so (31+12-1)C(12) posssibly?
and so taht would make C (31+12-1)C(12)-31?

 

[dynamicsolo]

ok part a makes sense
for part b now I am thining combination with repetition
so (31+12-1)C(12) posssibly?
and so taht would make C (31+12-1)C(12)-31?

Why would you have 42C12? Where would the 42 come from?

These combinatorial formulas have to be applied carefully. They do not work for all sorts of counting problems; you need to be able to justify why the factors in the product would appear in the situation you are considering.

The flavors of ice cream can be treated like balls in a bin that can be withdrawn one by one. In the last two parts, since you can re-use any of them, these problems can be treated like "drawing with replacement". So you have all 31 to choose from each time.