4. Choosing without ordering

Counting & Combinatorics overview Lesson 4 of 6 · Choosing without ordering

In the last lesson, we counted arrangements. Changing the order usually created a new possibility.

But sometimes order does not matter at all.

First question: does order matter?

Before calculating anything, decide what counts as a genuinely different outcome.

Start with something familiar
Choose 3 toppings
Vegetarian pizza with several toppings
mushroom olive pepper
=
pepper mushroom olive

Same toppings. Same pizza.
Naming the same choices in a different order does not create a new outcome.

The key question: if I list the same chosen objects in a different order, have I actually made something new?
Quick decision game
Does order matter?

Decide first, then open the answer.

🍕 Choose 3 extra pizza toppings
Order does not matter here. Assume the pizza base, sauce and cheese are already fixed. We are only choosing the extra toppings: mushroom, olive and pepper is the same choice as pepper, mushroom and olive.
🥇 Award gold, silver and bronze medals
Order matters. Gold, silver and bronze are different positions, so changing who gets which medal changes the outcome.
📚 Choose 2 books to borrow
Order does not matter. The same two books are the same selection whichever one you name first.
🔢 Make a 4-digit code
Order matters. 1234 and 4321 are different codes.
👥 Choose 3 people for a committee
Order does not matter. A committee has no first, second or third position.
📸 Arrange 3 people in a photograph
Order matters. Alice–Ben–Charlie is a different line-up from Charlie–Ben–Alice.

A team is not an arrangement

Alice, Ben, Charlie and Dom are available. Choose two people for a team.

Main idea
Reversing the pair does not make a new team
AB = BA

If we count ordered picks, every team appears twice: once in each order.

There are 4 choices for the first person and 3 for the second, giving 12 ordered picks. Now watch them collapse into six genuine teams:

AB↔BA
AB is one team
AC↔CA
AC is one team
AD↔DA
AD is one team
BC↔CB
BC is one team
BD↔DB
BD is one team
CD↔DC
CD is one team

4 × 3 ordered picks ÷ 2 orders per team = 6 teams

Connection
You have just built a graph. Alice, Ben, Charlie and Dom can be vertices; each possible pair becomes an edge. Graph with Alice, Ben, Charlie and Dom as vertices and the six possible two-person teams as edges Explore this idea in Graphs →
Practice puzzle
Choose two from six

Six children are available for a two-person quiz team. How many different teams are possible?

Reveal answer

Count ordered picks first: 6 × 5 = 30.

Every pair appears twice, so divide by 2.

30 ÷ 2 = 15

Choosing more than two

The same idea works when we choose larger groups. Suppose we choose three people from six.

Step 1 · Count ordered selections
6 choices × 5 choices × 4 choices = 120
Step 2 · See how one group is overcounted

If the chosen people are A, B and C, these six ordered selections all describe the same group:

ABCACBBACBCACABCBA

3! = 6 orders of the same three people

Step 3 · Collapse each set of six duplicates

120 ÷ 6 = 20 groups

Practice puzzle
A committee of three

Eight people are available. A committee of three is chosen, with no special roles.

Reveal answer

Ordered selections: 8 × 7 × 6.

Each committee appears in 3! = 6 orders.

(8 × 7 × 6) ÷ 6 = 56

Important rule

Arrange first, then divide by the orders that do not matter

We have already seen this with two-person teams and with groups of three. Now we can write the pattern in general.

Ordered selections

If we choose k objects from n and order matters, the choices are

n n−1 … n−k+1

This looks like n! with a tail cut off.

n n−1 … n−k+1 × n−k n−k−1 … 2 1

The grey tail is (n−k)!.

To turn the shortened product back into a factorial, multiply by the missing tail — and divide by the same thing. The extra fraction is equal to 1, so it changes nothing:

n(n−1)···(n−k+1) × (n−k)! (n−k)!

The ghost fraction

= n! (n−k)!

So n!/(n−k)! is simply a compact way of writing the ordered-selection product.

If order does not matter, each chosen group of k objects has been counted k! times — once for each ordering of those k objects. Therefore:

unordered choices = ordered selections ÷ k!
(nk) = n! k!(n−k)!
A note about notation

We will use (nk) throughout this course. It is read “n choose k”.

Calculators often show the same operation as C(n,k) or nCr. The notation is different, but the meaning is the same: the number of ways to choose k objects from n when order does not matter.

Worked example
Choose 3 from 10

We want (103). Using the formula:

(103) = 10! 3!7!

Look for the ghost tail. Instead of expanding both factorials completely, expand the larger factorial only until the smaller factorial appears as its tail — then cancel that matching tail.

Expand only as far as 7!
10! = 10 9 8 × 7!

The grey block is the ghost tail: 7! = 7 × 6 × 5 × ··· × 2 × 1.

(103) = 10 × 9 × 8 × 7! 3 × 2 × 1 × 7!

Now the 7! on top and bottom cancel.

= 10 × 9 × 8 3 × 2 × 1

You can simplify before multiplying:

9 ÷ 3 = 3    and    8 ÷ 2 = 4
10 × 3 × 4 = 120

So (103) = 120. Cancelling first keeps the arithmetic small and makes mistakes less likely.

Useful habit: when you see a fraction such as 10! ÷ (7! × 3!), expand the larger factorial only until it reaches the smaller one: 10! = 10 × 9 × 8 × 7!. Then cancel the common factorial before doing any large multiplication.

The formula is useful, but the idea underneath is more important: count arrangements first, then correct for the orders that do not matter.

Practice puzzle
Pizza toppings

A pizza shop has 7 possible toppings. You choose exactly 4 different toppings.

Reveal answer

Use (74):

(74) = 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. Every choice of three automatically determines the two people who are left out.

Interactive exploration
Choose exactly 3 for the team

Tap three people. Watch the other two become the people you leave out.

Choose 3 people.
Chosen
3 people
↔
Left out
2 people

(53) = (52)

Main idea
Choose k — or choose the n − k objects to leave out

(nk) = (nn−k). Sometimes the smaller side is much quicker to count.

Practice puzzle
Eight from ten

Ten different balls are on the table. You choose eight of them.

Leave out any two balls
Tap two balls. The faded balls are left out; the eight bright balls are chosen.
Tap a ball to leave it out.
Reveal answer

Choosing 8 is equivalent to choosing the 2 that are left out.

(108) = C(10,2) = 45

Pause here if you want
You have covered the main idea of choosing without ordering

It is a good place to pause and return to the ideas later.

Counting & Combinatorics overview Lesson 4 of 6 · Choosing without ordering