The I Scream Shop specializes in triple scoop sundaes. How many different triple scoop sundaes could be made using these flavors: vanilla, chocolate, strawberry, cherry, lime, and lemon?

I attempted to solve this problem by breaking it down into three cases: The first case involves the same flavor 3 times in a row. The second is 2 of a kind and then another flavor. The third is for ice cream containing three different flavors.

EDIT: The order of the scoops does not matter for each ice cream served.

The main problem I have is that I feel my work is not only incorrect, but also for the cases that I did manage to find an answer, I wish to find a more concrete answer instead of counting.

Case $1$: For three different ice creams, we have $\binom63= 20$

Case $2$: This case we can choose two different flavors. There are $\binom62$ ways of doing this. Now we must consider which one of these flavors doubles. At this point, I had a gut feeling that it is twice as much, but I'm not to sure how to do this for $n$ scoops.

Anyways, the total here is $\binom62\times{2}= 30$

Case $3$: We choose one flavor, so there is $\binom61= 6$ ways.

Therefore, there is a total of $20+30+6=56$ possible ice cream scoops. Again, my main concern is if there is a more concrete and general way to do the second case, or in the general, the cases between the two extremes- choosing one and choosing all different.

  • $\begingroup$ I don't know much about ice creams (haven't had one in over a year), but does it matter what order the scoops are in i.e. if chocolate goes under vanilla, is it different from vanilla going before chocolate? If it does,then you have to take that into account in your answer. $\endgroup$ – Teresa Lisbon Dec 4 '16 at 5:11
  • $\begingroup$ @астонвіллаолофмэллбэрг I forgot to mention that order does not matter. $\endgroup$ – Ian Limarta Dec 4 '16 at 5:13
  • $\begingroup$ Alternative reasoning for case $2$: Select a flavour which doubles in $\binom61$ ways. Then choose the flavour which occurs once in $5$ ways (exclude the twice occurring flavour because otherwise this will overlap with your case $3$) $\endgroup$ – Shraddheya Shendre Dec 4 '16 at 5:15

The simplest solution to this problem is to use the stars and bars method. It states that if you have $k$ non-negative integers that have a sum of $n$ , then there are ${n + k - 1}\choose{k-1}$ or ${n + k - 1}\choose{n}$ ways to form this sum. Thus, in your question, there are $${{3 + 6 - 1}\choose{3}} = {{8}\choose{3}} = 56$$ ways to choose the flavors.

Hope that helps.

  • $\begingroup$ Is the linear equation f1+f2+f3+f4+f5+f6 = 3 where fn are the flavors? $\endgroup$ – Ian Limarta Dec 4 '16 at 6:21
  • $\begingroup$ That is absolutely correct! Though remember the bounds on $f_n$ is that it has to be a nonnegative integer. $\endgroup$ – Rajat Mittal Dec 4 '16 at 14:37

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