In the last lesson, we learned to count choices that happen in stages by multiplying the number of choices at each stage.
Now we will use exactly the same idea to count arrangements — different orders of the same objects.
Three books on a shelf
Suppose you have three different books, labelled A, B and C.
How many different ways can you arrange the three books in a row?
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Reveal answer
If A is first, the arrangements are ABC and ACB.
If B is first, they are BAC and BCA.
If C is first, they are CAB and CBA.
So there are 6 arrangements.
But we do not actually have to list them.
There are 3 choices for the first position. Once one book has been used, there are 2 choices for the second position. Then only 1 choice remains.
3 × 2 × 1 = 6
For n different objects, there are n choices for the first position, then n−1 choices, then n−2, and so on until only one object remains.
Alice, Ben, Charlie and Dom stand in a line.
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Reveal answer
There are 4 choices for the first place, then 3, then 2, then 1.
4 × 3 × 2 × 1 = 24
There are 24 arrangements.
A useful shorthand: factorials
Products like
5 × 4 × 3 × 2 × 1
appear so often in counting that mathematicians use a shorthand:
5! = 5 × 4 × 3 × 2 × 1
We read 5! as “five factorial”.
| Expression | Value |
|---|---|
| 1! | 1 |
| 2! | 2 |
| 3! | 6 |
| 4! | 24 |
| 5! | 120 |
The factorial symbol is just a compact way to write the descending product that comes from the multiplication principle.
Five runners reach the finish line. There are no ties.
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Reveal answer
5! = 120
There are 120 possible finishing orders.
What if one position is fixed?
Restrictions often reduce the number of choices.
Suppose five different books are arranged on a shelf, but the dictionary must be first.
The dictionary’s position is already decided. We only need to arrange the other four books:
4! = 24
Five children stand in a row for a photograph. The captain must stand in the middle.
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Reveal answer
The captain uses the middle position, leaving four children to arrange in the remaining four places.
4! = 24
There are 24 arrangements.
When one object has several allowed positions
Six runners finish a race. Alice must finish either first or second.
There are 2 choices for Alice’s position. After that, the other five runners can be arranged freely.
2 × 5! = 240
How many different arrangements can be made from the letters M, A, T, H, using each letter exactly once?
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Reveal answer
We are arranging four different objects.
4! = 24
There are 24 arrangements.
What counts as a different arrangement?
In an arrangement, changing the order creates a new result.
These are different because the objects occupy different positions.
Five children — Alice, Ben, Charlie, Dom and Eva — stand in a row.
Alice and Ben want to stand next to each other.
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Reveal answer
Treat Alice and Ben as one block. Then we are arranging four objects: the AB block, Charlie, Dom and Eva.
4! = 24
But inside the block, Alice and Ben can appear as AB or BA.
2 × 4! = 48
There are 48 arrangements.
What have we learned?
- Arrangements can be counted using choices in stages.
- For n different objects, the number of arrangements is n!.
- Fixed positions reduce the number of free choices.
- Sometimes a restriction suggests treating several objects as one block.
n! = n × (n−1) × (n−2) × ··· × 2 × 1
Next: Choosing without ordering
Suppose we choose Alice, Ben and Charlie for a team.
ABC, ACB, BAC, BCA, CAB and CBA all describe the same three-person team.
Our arrangement method would count the same choice six times. In the next lesson, we will learn how to correct that.
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