In the last lesson, we counted arrangements. Changing the order usually created a new possibility.
But sometimes order does not matter at all.
A team is not an arrangement
Suppose Alice, Ben, Charlie and Dom are available, and we want to choose two people for a team.
If we use the choices-in-stages idea, there are 4 choices for the first person and then 3 choices for the second:
4 × 3 = 12
But that cannot be the number of different teams. The choice Alice then Ben gives the same team as Ben then Alice.
When we choose a group, the same group may appear several times if we count it as an arrangement. We need to correct for that overcounting.
Let us list the six different two-person teams:
Each team appeared twice in the count 4 × 3: once in each order. So we divide by 2:
(4 × 3) ÷ 2 = 6
Six children are available for a two-person quiz team.
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Reveal answer
There are 6 choices for the first person and 5 for the second, so there are 30 ordered selections.
Each team appears twice, once in each order.
(6 × 5) ÷ 2 = 15
There are 15 different teams.
Choosing more than two
Now suppose we choose three people from six.
If order mattered, there would be:
6 × 5 × 4 = 120
But each group of three people can be arranged in
3! = 3 × 2 × 1 = 6
different orders. So every three-person group has been counted six times.
120 ÷ 6 = 20
To choose r objects from n different objects, we may count the ordered selections and then divide by r!, because each chosen group can be arranged in r! different ways.
Eight people are available. A committee of three is chosen, with no special roles.
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Reveal answer
There are 8 × 7 × 6 ordered selections.
Each group of three appears 3! = 6 times.
(8 × 7 × 6) ÷ 6 = 56
There are 56 committees.
A useful shorthand
Choosing r objects from n objects happens so often that mathematicians give it a special notation.
We write “n choose r”, sometimes written as C(n,r).
C(n,r) = n! ÷ (r!(n-r)!)
For example:
C(6,3) = 6! ÷ (3!3!) = 20
The formula is useful, but the important idea is still the same: count the arrangements, then correct for the orders that do not matter.
A pizza shop has 7 possible toppings. You choose exactly 4 different toppings.
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Reveal answer
C(7,4) = 35
There are 35 different pizzas.
Choosing what stays is the same as choosing what goes
Suppose five people are available and we choose three for a team.
Choosing the three people who are on the team automatically tells us which two people are not on the team.
So:
C(5,3) = C(5,2)
For any n and r, C(n,r) = C(n,n-r). Sometimes choosing the smaller group is much easier to calculate.
Ten books are on a table. You choose eight of them.
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Reveal answer
C(10,8) = C(10,2) = 45
There are 45 different selections.
Seven people are available. You choose a team of three, and then choose one of those three people to be captain.
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Reveal answer
There are C(7,3) = 35 possible teams.
For each team there are 3 possible captains.
35 × 3 = 105
There are 105 outcomes.
What have we learned?
When order does not matter, arrangement counting usually counts each choice too many times.
- Ask whether changing the order creates a new possibility.
- If it does not, count the arrangements and divide by the number of irrelevant orders.
- Choosing r objects from n objects is written C(n,r).
- C(n,r) = C(n,n-r).
The key question is still: what exactly counts as a different outcome?
Next: Pascal’s triangle
The numbers C(n,r) form a remarkable pattern.
In the next lesson we will build that pattern row by row and discover why the same numbers appear in Pascal’s triangle.
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