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  1. If there are 16 different ice-cream flavours, how many combinations are there for a two scoop?

  2. If there are still 16 different ice-cream flavours, how many combinations are there for a three scoop?

How would you work out the above combinations? I found it just sitting in my notes app and I don't recall I ever found an answer.

My thoughts at the moment are 16 * (no. of scoops) but I am still lost since of course the flavours can appear in any order. (As my SE profile will demonstrate, I'm not much of a mathematician!)

So what would an equation be for the following as applicable to both Problem 1 and Problem 2?

A. Working out the number of combinations including duplicate scoops (e.g. chocolate-chocolate-vanilla)

B. Working out the number of combinations where a flavour only appears once in each possible combination (e.g. chocolate-vanilla-strawberry and then vanilla-chocolate-strawberry).

Any help would be much appreciated!

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1) working out the number of combinations including duplicate scoops (e.g. chocolate-chocolate-vanilla)

Consider the case where there is only one scoop of ice cream. There are 16 flavors (choices), and thus 16 "combinations." The next case is 2 scoops. One way to think about this problem is to consider how many choices you have per scoop. There are 16 choices for the first scoop and 16 choices for the second scoop since duplicates are allowed. This works out to $16^2 = 256$. It should be more clear as to how you can expand this to more scoops.

2) working out the number of combinations where a flavour only appears once in each possible combination (e.g. chocolate-vanilla-strawberry and then vanilla-chocolate-strawberry)

Now try the case of 3 scoops. There are 16 choices for the first scoop, but 15 choices for the second scoop since duplicates are not allowed. For the third scoop there are 14 choices, which works out to $16*15*14=3,360$. Note that with this way of combining flavors, the order is important since chocolate-vanilla-strawberry and vanilla-chocolate-strawberry are both counted.

EDIT (for cases where duplicates are allowed):

1.If there are 16 different ice-cream flavours, how many combinations are there for a two scoop?

There's a decent explanation from another post. The formula is indeed $$ \binom{n+k-1}{k} $$

where $n$ is the number of flavors and $k$ is the number of scoops. This is called a combination. Note that using this formula gives a different answer than I originally provided, since combinations do not consider the order in which objects are counted $$ \binom{16+2-1}{2}=\binom {17}2=\frac{17\cdot16}{1\cdot2}=136 $$

  1. If there are still 16 different ice-cream flavours, how many combinations are there for a three scoop?

Using the formula above it can be worked out similarly $$ \binom{16+3-1}{3}=\binom {18}3=\frac{18\cdot17\cdot16}{1\cdot2\cdot3}=816 $$

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  • $\begingroup$ Thank you. Though I mustn't have been as clear as possible. 1) and 2) do not correlate to the first and second predicaments; they are meant as sub points for each problem if that makes sense. I have edited the question to reflect this. $\endgroup$ – Dog Lover Jul 13 '16 at 2:26
  • $\begingroup$ Would you be able to edit your answer accordingly, as so far your answer is excellent. $\endgroup$ – Dog Lover Jul 13 '16 at 2:32
  • $\begingroup$ @DogLover I added a little more to my answer including the more likely formula that you're looking for. I know I'm not as comprehensive as I could be, but I hope this helps your intuition as to how this problem is worked out. $\endgroup$ – Jonathan So Jul 13 '16 at 3:01
  • $\begingroup$ Thank you! Just what I was after, plus I learnt something new: mathematical combinations. Although is there a reason why 1 is subtracted from the number of scoops? Is it something to do with zero-indexing? $\endgroup$ – Dog Lover Jul 13 '16 at 3:10
  • $\begingroup$ @DogLover You're welcome! Glad I could help. And to answer your question, kind of. This problem is very related to what is known as the Stars and Bars technique. A little bit of it was shown in the post I linked, and a more detailed description and usage can be found here. Without additional reading I can't honestly say I remember why 1 is subtracted, but in stars and bars there will always be 1 less bar separating the $k$ objects. $\endgroup$ – Jonathan So Jul 13 '16 at 3:17
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1: This problem is equivalent to putting two identical balls into 16 boxes (with repetition allowed).
$\binom{2+16-1}{2}$

Where 2 is the number of balls, and $16-1$ is the number of separators needed to divided a line segment into 16 smaller segments,each segment represents a box.

Each way of putting the balls (each combination of flavors) will be corresponds to a arrangement of $2$ balls and $16-1$ seperators (where balls and seperators are indistinguishable among themselves).

2.$\binom{3+16-1}{3}$.

A. As shown above, where we assume repetition is allowed.

B. If repetition is not allow,

  1. $\binom{16}{2}$

  2. $\binom{16}{3}$

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