Some students are confused by the difference between combinations and permutations, so here is a neat example where they can derive the rules themselves, and see the difference in meaning of these concepts.
This is based on an investigation I have given to student teachers in a statistics and probability content and pedagogy course for several years.
Please feel free to make use of it, or any aspects of it for your own classes or children.
Counting Investigation: Ice Cream!!
An ice cream shop has 20 different flavours for sale.
Their specialty is the Triple Ripple Cone, which contains 3 big scoops of ice cream.
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How many different types of “Triple Ripple” ice creams can be made from the 20 flavours? (Allowing repeats?)
Teacher note: This explores basic counting with replacement. Students see that each scoop is independent, so they multiply the number of choices for each scoop:
$20 \times 20 \times 20 = 8000$
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How many different types of “Triple Ripple” ice cream can be made with three different flavours?
Teacher note: Here, students explore counting without replacement. They realize that the number of choices decreases as flavours are used, and they begin to see how order affects the total:
$20 \times 19 \times 18 = 6840$
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Explain the difference between (1) and (2).
Teacher note: Students reflect on how allowing repeats vs requiring distinct choices changes the counting method. They also articulate that “order matters” in cones here.
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Suppose an extremely dexterous ice cream scooper joins the shop, and starts selling a cone with 4 scoops. How would this change the answers to (1) and (2)? What if they served 5 scoops? Or $k$ scoops?
Teacher note: Students can see the pattern by counting choices step by step. For 4 scoops with all distinct flavours:
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First scoop: 20 choices
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Second scoop: 19 choices
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Third scoop: 18 choices
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Fourth scoop: 17 choices
Multiply together: $20 \times 19 \times 18 \times 17$
Students can see the pattern more concretely with additional examples:
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4 scoops: $20 \times 19 \times 18 \times 17 = 116{,}280$
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5 scoops: $20 \times 19 \times 18 \times 17 \times 16 = 1{,}860{,}480$
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6 scoops: $20 \times 19 \times 18 \times 17 \times 16 \times 15 = 27{,}907{,}200$
Notice a pattern: for $k$ scoops from 20 flavours, the total number of ordered arrangements is:
$20 \times 19 \times 18 \times \cdots \times (20-k+1)$
Students can see that this product can be written as a ratio of two factorials:
$\frac{20!}{(20-k)!}$
Why does this work?
$20! = 20 \times 19 \times 18 \times \cdots \times 2 \times 1$
$(20-k)! = (20-k) \times (20-k-1) \times \cdots \times 2 \times 1$Dividing $20! / (20-k)!$ cancels all the terms after $20-k+1$, leaving exactly the product we need. This gives a compact formula for any number of scoops, without introducing permutation notation yet.
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Tragically, the shop runs out of cones! They begin serving the “Triple Ripple” ice cream in bowls. Jenny and Yvonne come in to get their usual ice creams. Despite having very similar tastes, they are bizarrely fussy about the order they eat their ice cream in:
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Jenny: chocolate on vanilla on strawberry
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Yvonne: vanilla on chocolate on strawberry
They are horrified to find that when served in a bowl, the ice creams look identical.
Question: What other cone combinations of these flavours would look identical served in a bowl?
Teacher note: Students explore that each set of 3 flavours has 6 possible arrangements ($3! = 6$) in a cone, but all are identical in a bowl. This is their first insight into combinations vs permutations.
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How many different combinations of 3 distinct flavours can they make serving the ice cream in bowls?
Teacher note: Students now apply the logic from the previous question: divide the total number of ordered cones (6840) by $3! = 6$ to get the number of unique sets:
$\frac{6840 }{6} = 1140$
They see that order no longer matters in bowls, introducing the combination concept.
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How many different combinations of 4 distinct flavours could they make serving their ice cream in bowls?
Teacher note: Students generalise the idea: for 4 scoops, there are $4! = 24$ ways to arrange each group in a cone, but all look the same in a bowl. This reinforces the concept of dividing by the factorial of the number of items to remove duplicates:
$\frac{(20 \times 19 \times 18 \times 17)}{24} = 4845$
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How many different combinations of $k$ distinct flavours could they make serving their ice cream in bowls?
Teacher note: Students see the formula for combinations emerges naturally:
$\frac{20!}{k!(20-k)!}$
They understand the conceptual difference between ordered and unordered counting and why the division by $k!$ is needed. The final step, of course, is to generalise 20 to $n$.
Takeaway
Using ice cream to explore counting allows students to:
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Derive the formula for ordered arrangements (permutations) themselves by counting step by step
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Discover that combinations = ordered arrangements ÷ arrangements of the chosen items
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Clearly understand when order matters (cones) and when it doesn’t (bowls)
This activity gives students a concrete, tasty way to see abstract counting principles in action.
I usually then get students to use these ideas to work on arrangements of letters in a ‘word’, look at the case where a word is made up of two letters repeated some number of times each, and link this to the binomial expansion and why the binomial coefficients are the entries of Pascal’s triangle. A number of prospective maths teachers have said to me that this investigation helped them to a real understanding of these ideas, which they had never had before.
