# How many combinations of $3$-scoop ice cream cones are possible from $6$ flavors, if all flavors on the cone must be different? (Parent seeking help.)

My 5th-grade daughter needs to figure out this problem and it has been years for me.

Two girls go into an ice cream shop. There are $$6$$ flavors of ice cream. How many combinations of a $$3$$-scoop ice cream cone can you make?

All the flavors have to be different on each cone, meaning you can't have strawberry, strawberry, and peach on a cone.

If you can help not only with the answer but the equation I would appreciate it.

• I changed the tag, because this is not abstract-algebra – J. W. Tanner Sep 7 '20 at 1:40
• Would ice cream cones that differ only in the order of the three scoops be considered to be different? I'd be curious to see a picture of the assignment. – littleO Sep 7 '20 at 1:41
• and are you asking about how many combinations are possible for one of the two girls? – J. W. Tanner Sep 7 '20 at 1:42
• The question is a bit ambiguous which is why folks are commenting. There are ${6 \cdot 5 \cdot 4 \over 3 \cdot 2 \cdot 1}$ different $3$ scoop combinations, but the 2 girsl part could be relevant and change the meaning. – copper.hat Sep 7 '20 at 1:45
• I think this would be challenging for a $5$th grader? – copper.hat Sep 7 '20 at 1:45

The simple answer that a fifth-grader should have no trouble with is that for a three-scoop ice cream cone, you have three scoops, and each one needs to be a flavor. For the first scoop, you've got 6 options, because there are six flavors total. So if the question was how many possibilities you can have with one unique scoop the answer would be six.

Now for each of the options for the first scoop, you've got to find all the options for the second scoop, so we want to figure out how many options there are for scoop 2, and multiply by options for scoop 1. Because we already used a flavor, we have 5 possibilities left, so scoop 2 can be any of 5 flavors. Multiply that by 6, and the number of options for two scoops is 30. Except that if we think about the list of 30 options we made, the first scoop can be any of the 6 flavors, strawberry let's say, and the second can be any of the others, let's say peach. But our list of 30 flavors here (it might be useful to have your daughter write up the list of 30 so she can prove this for herself) also includes an option where scoop 1 is peach and scoop 2 is strawberry, which depending on the wording of the question is the same ice cream cone. Up to you to figure out if different orders of the scoops are considered different depending on her question's wording. But, if strawberry+peach is considered the same as peach+strawberry, we've got to eliminate all those reordered cases somehow. So we divide our total number of options by how many repeats there are for each. There are two ways of repeating two scoops (A+B and B+A), so we divide by 2. Note that we didn't have to deal with this for the one scoop case because there's only one way of arranging one scoop! So for two scoops the answer is:

$$\frac{6 \ \textrm{(options for first scoop)} * 5 \ \textrm{(options for second scoop)}}{2 \ \textrm{(ways of rearranging two things)}} = 15$$

Now we're starting to get close to a formula. To finish off the problem, and find the formula, let's consider the third scoop. We used one flavor for the first scoop, and another flavor for the second scoop, so we've only got 4 flavors left for the third. So $$6 * 5 * 4$$ gives us the number of ways to make ice cream cones with 3 unique scoops chosen from 6 flavors. Now to account for rearrangements, there are 6 ways of rearranging 3 things:

$$ABC$$ $$ACB$$ $$BAC$$ $$BCA$$ $$CAB$$ $$CBA$$

So we divide by 6, and get:

$$\frac{6 \ \textrm{(options for first scoop)} * 5 \ \textrm{(options for second scoop)} * 4 \ \textrm{(options for third scoop)}}{6 \ \textrm{(ways of rearranging two things)}} = 20$$

To finish off the formula, I'll tell you that given $$K$$ things, there are K! (factorial K) ways of arranging them. This is best found experimentally by trying to find all the ways of rearranging letters as I did above and looking for patterns. So there is $$1$$ way of rearranging one item, $$1*2$$ ways of rearranging two items, $$1*2*3$$ ways of rearranging 3 items, and so on. We also figured out that this factorial pattern also applies for the scoops, where if there are $$N$$ flavors to choose from, the first scoop has $$N$$ options (not accounting for different orders), the second has $$N-1$$ so first times second is $$N * (N-1)$$, the third has $$N-2$$ so first times second times third is $$N * (N-1) * (N-2)$$, and so on. For $$K$$ total scoops this can be expressed (somewhat confusingly) in factorial form as $$\frac{N!}{(N-K)!}$$ because for example if we have $$8$$ flavors and $$3$$ scoops, then most of the numbers on the top cancel with something on the bottom, and the factorials come out as follows:

$$\frac{6!}{3!} = \frac{8 * 7 * 6 * 5 * 4 * 3 * 2 * 1}{5 * 4 * 3 * 2 * 1} = 8 * 7 * 6$$

So our total formula, after accounting for repeats in different orders, given that there are $$K!$$ possible orderings for $$K$$ things, is:

$$\frac{N!}{K!(N-K)!}$$

Just to verify that the number $$20$$ that we got above is correct, we can plug in our 6 flavors for $$N$$ and our 3 scoops for $$K$$ and we get:

$$\frac{N!}{K!(N-K)!} = \frac{6 * 5 * 4 * 3 * 2 * 1}{3 * 2 * 1 * 3 * 2 * 1} = \frac{6 * 5 * 4}{3 * 2 * 1} = 20$$

And there's your answer, and hopefully a way that a fifth-grader with a very patient mother can understand this problem well enough to solve future similar problems on her own.

There are $$\pmatrix{6\\3}$$ combinations. The notation means "six choose three". Out of six items (flavors) choose three.

$$\pmatrix{n\\k}=\frac{n!}{k!(n-k)!}.$$

$$\pmatrix{6\\3}=\frac{6!}{3!3!}.$$

Think of it this way. There are 6 ways to choose a flavor. Once you choose, there are 5 ways to choose the next. After that, there are 4 flavors left. which is $$6!/3! = \frac{6\cdot 5\cdot 4\cdot 3 \cdot 2 \cdot1}{3\cdot2\cdot 1}=6\cdot 5\cdot 4 = 120$$.

But, you could have chosen $$\{$$chocolate,vanilla,strawberry$$\}$$ and you get the same combination as $$\{$$vanilla, strawberry, chocolate$$\}$$ so we have to divide by $$3!=3\cdot2\cdot1=6$$ to account for the order of choosing.

So the number of combinations of flavors is $$\pmatrix{6\\3}=\frac{120}{6}=20.$$