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. $\endgroup$ – астон вілла олоф мэллбэрг 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

We break it up, starting with the easy:

1) If the same flavour is used three times, there are six ways to choose this flavour. Hence the answer is six in this case.

2) Suppose there are two different flavours in your sundae. Split it:

i)There are $\binom 62 = 15$ ways to choose the two flavours in the sundae.

ii) There are two ways to decide which of these two is to be doubled in the sundae! (Just to make things absolutely clear).

hence, the answer on this step is $15 \times 2 = 30$.

3) There are three different flavours in the sundae. These are chosen in $\binom 63 = 20$ ways.

So your answer of $30+20+6 = 56$ is correct.

In the general case of: $n$ flavours, $k$ scoop sundae, with each flavour available unlimited , you can rephrase the problem as follows:

For each of $n$ flavors, let $x_m = $ number of times that flavour $m$ is used in the sundae. Then, note that $x_i \geq 0$ and $\sum_{i=1}^m x_i = k$. So all we have to do, is to find the number of possible solutions of $x_i$.

There is a known method called stars and bars. Without touching too much on it, I will tell you that the solution is given as $\binom {n+k-1}k$. So, in the case $n=6,k=3$ (six flavors, triple sundae!), this gives $\binom 83 = 56$, which is the right answer.

Hence, we can give an answer for general $n$ and $k$. Having said that, some of these $56$ sundaes are boring. I mean, the same flavour thrice?


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