☰Starting simple: picking teams

Imagine you have 3 friends: Ali, Bo, and Cara. You want to pick 2 of them to help with a project. How many different pairs can you make? Let's list them all:

List them out

1. Ali + Bo
2. Ali + Cara
3. Bo + Cara

That's 3 different pairs. This is called a combination — picking a group where the order doesn't matter. Ali + Bo is the same pair as Bo + Ali.

What if you had 4 friends and wanted to pick 2? You'd get 6 pairs. With 5 friends picking 2, you'd get 10 pairs. Listing them all works for small numbers but gets tedious fast — we need a shortcut.

♂The shortcut formula (with a simple example first)

From 5 friends, pick 2: the answer is 10. Here's the formula that gives you that answer without listing:

Number of combinations — what it means

In plain English: $\binom{n}{r}$ means "how many ways can I pick $r$ things from $n$ things?"

$$\binom{n}{r} = \frac{n!}{r!\,(n-r)!}$$

The ! means factorial — multiply all whole numbers down to 1.

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

So $\binom{5}{2} = \frac{120}{2 \times 6} = \frac{120}{12} = 10$. ✓

Pascal's triangle — a number pattern

Build a triangle: each row starts and ends with 1. Each number in the middle is the two numbers above it added together.

Row 0: 1
Row 1: 1 1
Row 2: 1 2 1
Row 3: 1 3 3 1
Row 4: 1 4 6 4 1

These numbers are the combination counts! Row 4 gives: $\binom{4}{0}=1,\; \binom{4}{1}=4,\; \binom{4}{2}=6,\; \binom{4}{3}=4,\; \binom{4}{4}=1$.

Binomial expansion — multiplying out brackets

Multiply out $(x + y)^2$ by hand:

$(x+y)(x+y) = x^2 + xy + yx + y^2 = x^2 + 2xy + y^2$

The coefficients are 1, 2, 1 — that's Row 2 of Pascal's triangle!

$(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$ — coefficients 1, 3, 3, 1 — that's Row 3!

So Pascal's triangle tells you the coefficients when you expand $(x+y)^n$:

$$(x+y)^n = \binom{n}{0}x^n + \binom{n}{1}x^{n-1}y + \cdots + \binom{n}{n}y^n$$

✏Worked example

Q: From 6 students, pick a committee of 3. How many ways?

Step by step

Step 1: Identify $n$ and $r$. $n = 6$ (students), $r = 3$ (committee size).

Step 2: Plug into the formula: $\binom{6}{3} = \frac{6!}{3! \times 3!}$

Step 3: Compute: $\frac{720}{6 \times 6} = \frac{720}{36} = 20$.

Answer: 20 different committees.

Check with Pascal's triangle: Row 6 of Pascal's triangle is 1, 6, 15, 20, 15, 6, 1. The middle number is 20. ✓

⚙Teaching notes

The order question

The test for combination vs permutation: does reordering the chosen items give a different result? Picking a committee = no (Ali+Bo+Cara = Cara+Bo+Ali) = combination. Lining up for a photo = yes = permutation.

Why Pascal's triangle matters

It connects counting to algebra. The same numbers that count committees also appear as coefficients when you expand brackets. This is the bridge students will cross again in Year 12 with binomial probability.

Year 11 Methods · Teacher Brush-Up