Year 11 Mathematical Methods — Selections (Combinations)

Cambridge Methods 1&2 — Counting Methods (10C)  •  Sun 14 June 2026
📚 Also for this topic: 📄 Printable worksheet ✅ Solutions (answer key) → Next: 10D Probability

Today's lesson

A selection (or combination) is a choice of objects where order does not matter — a team, a hand of cards, a committee. Compare with yesterday's arrangements, where order did matter. Today's tool is $^nC_r$.

Learning intentions

Part 1 — Order doesn't matter: $^nC_r$

From $4$ numbers $1,2,3,4$, the selections of $3$ are $\{123\},\{124\},\{134\},\{234\}$ — just $4$. But each set of $3$ can be arranged in $3!=6$ orders, so there are $4\times6=24=\,^4P_3$ arrangements. So selections $=$ arrangements $\div\,r!$:

Combinations. The number of ways to choose $r$ objects from $n$ when order does not matter is $$^nC_r=\binom{n}{r}=\frac{^nP_r}{r!}=\frac{n!}{r!\,(n-r)!}.$$ Here $n$ is how many you choose from, and $r$ is how many you choose.

A useful symmetry: $^nC_r=\,^nC_{n-r}$ (choosing $r$ to keep is the same as choosing $n-r$ to leave out). For example $^{13}C_7=\,^{13}C_6=1716$.

On your CAS. Use the built-in combination function nCr (in the probability/maths menu): nCr(13, 7) gives $1716$. Selections grow huge fast — $^{52}C_7$ (seven-card hands) $=133\,784\,560$.

📺 Walkthrough: why we divide $^nP_r$ by $r!$ — each unordered selection corresponds to $r!$ ordered arrangements, so $^nC_r=\,^nP_r\div r!$.

EXAMPLE 1 — basic selections
(a) How many netball teams of $7$ can be chosen from $13$ players? (b) How many three-scoop sundaes (different flavours, order irrelevant) from $20$ flavours?
  1. (a) Order of a team doesn't matter: $^{13}C_7=\dfrac{13!}{7!\,6!}=\boxed{1716}$.
  2. (b) A sundae is a set of $3$ flavours: $^{20}C_3=\dfrac{20!}{3!\,17!}=\boxed{1140}$.

Now you try: How many ways to choose $4$ books from $10$?   Answer: $^{10}C_4=210$.

Part 2 — "Some guaranteed" & "at least"

When certain members must be included, place them first and choose the rest. For "at least", it is often quicker to use the complement (total $-$ unwanted).

EXAMPLE 2 — some guaranteed
A team of $3$ is chosen from a squad of $8$. The captain must be in the team. How many teams are possible?
  1. The captain is guaranteed, so $1$ place is filled. Choose the other $2$ from the remaining $7$ players.
  2. $1\times\,^7C_2 = \boxed{21}$ teams.
  3. Check by complement: all teams $^8C_3=56$; teams without the captain $^7C_3=35$; so $56-35=21$. ✓

Now you try: A committee of $4$ is chosen from $10$ people; one particular person must be on it.   Answer: $^9C_3=84$.

EXAMPLE 3 — exactly / at least
A team of $5$ is chosen from $5$ boys and $5$ girls. (a) How many teams have exactly $2$ boys? (b) How many have at least $1$ boy?
  1. (a) Exactly $2$ boys means $2$ boys and $3$ girls: $^5C_2\times\,^5C_3 = 10\times10 = \boxed{100}$.
  2. (b) At least $1$ boy — use the complement. All teams $^{10}C_5=252$; teams with no boys (all girls) $^5C_5=1$. So $252-1=\boxed{251}$.

Tip: "exactly" → split into groups and multiply; "at least one" → total $-$ none.

Part 3 — Selection of any size: $2^n$

What if you may choose any number of items — none, some, or all? For each of the $n$ items you make an independent yes/no decision (take it or not), so there are $2\times2\times\cdots\times2 = 2^n$ possible selections.

Selection of any size. $^nC_0+\,^nC_1+\cdots+\,^nC_n = 2^n.$ This counts every subset, including the empty one (choosing nothing). If at least one item must be chosen, the count is $2^n-1$.
EXAMPLE 4 — any size
Joanne is offered $5$ different sweets. She may take any number of them — including all, or none. In how many ways can she make her choice?
  1. For each of the $5$ sweets she decides take / don't take: $2^5 = \boxed{32}$ ways (this includes taking none).
  2. If she must take at least one sweet, exclude the empty choice: $2^5-1=31$.

Check: $^5C_0+\,^5C_1+\,^5C_2+\,^5C_3+\,^5C_4+\,^5C_5 = 1+5+10+10+5+1 = 32 = 2^5$. ✓

Practice 3.1 — selection or arrangement? Then compute.

  1. Choose $3$ pizza toppings from $8$ (order irrelevant)
  2. A hand of $7$ cards from $52$
  3. A team of $11$ from $15$ players
  4. A committee of $4$ from $10$ that must include one named person
  5. Choose any number of items (incl. none) from $6$ items
  6. From $5$ boys, $5$ girls choose a team of $5$ with the girls outnumbering the boys
a) $^8C_3=56$    b) $^{52}C_7=133\,784\,560$    c) $^{15}C_{11}=1365$    d) $^9C_3=84$    e) $2^6=64$    f) $3,4$ or $5$ girls: $^5C_3{}^5C_2+{}^5C_4{}^5C_1+{}^5C_5{}^5C_0=100+25+1=126$

Part 4 — Quick quiz (5 min)

Pick the correct answer for each, then click Mark.

Q1. For a problem where order does not matter, you should use:

Q2. $^8C_3$ equals:

Q3. A team of $3$ from $8$ must include the captain. Number of teams?

Q4. Choosing any number of items (including none) from $5$ items gives:

Q5. Which equals $^{13}C_7$?

Working program — Cambridge — Counting Methods Ex 10C

After the quiz, open Cambridge Methods 1&2 (Counting Methods, Chapter 10) and complete the set work:

Set workExtension
Ex 10C — Q1–5, 8, 10Harder "at least / exactly" selections and group-selection problems.

First ask: does order matter? If no, it's a selection ($^nC_r$).

Exit ticket — write in your book

Before you pack up, write one sentence each:
  1. How do you decide between $^nP_r$ and $^nC_r$?
  2. Write the formula for $^nC_r$ and explain why we divide by $r!$.
  3. How many selections of any size (including none) can be made from $n$ items?

— Mr Wong