What is conditional: must contain a particular item?

If a chosen object is forced in, put it in first, then choose the remaining places from the remaining objects. A team of 5 from 11 that must include the captain C has C fixed, so the other 4 come from the remaining 10: ^10C_4 = 210.

What is conditional: must exclude a particular item?

If an object is banned, simply remove it and choose from what is left. A team of 5 from 11 that must exclude an injured player X chooses all 5 from the remaining 10: ^10C_5 = 252. Combine the two: include C and exclude X leaves 4 to choose from 9, namely ^9C_4 = 126.

What are split a constrained selection into independent groups?

When a committee must contain set numbers from disjoint pools, choose from each pool separately and multiply. A committee of 5 from 6 men and 8 women with exactly 2 men chooses 2 of the 6 men and 3 of the 8 women: ^6C_2 × ^8C_3 = 15 × 56 = 840.

What is at least / at most: count the complement?

"At least k" usually fragments into several size cases, so count the total and subtract the few unwanted cases. A committee of 4 from 5 women and 4 men with at least 2 women is the total ^9C_4 = 126 minus the committees with 0 or 1 woman, ^4C_4 + ^5C_1\,^4C_3 = 1 + 20 = 21, giving 126 - 21 = 105.

NSWMaths Extension 1Syllabus dot point

When the order of a selection does not matter, how do you count the choices, and how do the restriction moves you learned for arrangements carry across to unordered selections?

Count unordered selections (combinations) using ^{n}C_{r} = ^{n}P_{r}/r! = n!/(r!(n-r)!), recognise that an n-element set has 2^n subsets and that the r-element subsets number ^{n}C_{r} with the symmetry ^{n}C_{r} = ^{n}C_{n-r} and row sum 2^n, and apply the at-least, complement, conditional and geometric counting techniques to selection problems

Counting unordered selections with combinations. Why nCr = nPr/r! = n!/(r!(n-r)!), why an n-set has 2^n subsets, the symmetry nCr = nC(n-r), and the row sum 2^n. Restriction technique with combinations: at least k by complement, must contain or exclude an item, committees with constraints, and geometry counts.

Generated by Claude Opus 4.818 min answerUpdated 2026-06-21

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

A combination counts unordered selections: $^{n}C_{r} = \binom{n}{r} = \dfrac{^{n}P_{r}}{r!} = \dfrac{n!}{r!\,(n-r)!}$ , the number of $r$ -element subsets of an $n$ -set, with the division by $r!$ removing the $r!$ orderings each subset is counted with. For example $^{8}C_{5} = \frac{8 \times 7 \times 6}{3!} = 56$ . An $n$ -set has $2^n$ subsets in all (each element in or out), so $^{n}C_{0} + {}^{n}C_{1} + \cdots + {}^{n}C_{n} = 2^n$ , and there is the symmetry $^{n}C_{r} = {}^{n}C_{n-r}$ (choosing the keepers is choosing the leavers). Restriction technique: force an item in with $^{n-1}C_{r-1}$ , ban one with $^{n-1}C_{r}$ , take fixed numbers from separate pools by multiplying ( $^{6}C_{2}\,^{8}C_{3} = 840$ ), and handle "at least" by the complement (at least $2$ women in a committee of $4$ from $5$ women and $4$ men is $126 - 21 = 105$ ). Geometry counts use combinations directly: from $8$ points with no three collinear there are $^{8}C_{2} = 28$ lines and $^{8}C_{3} = 56$ triangles.

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What this dot point is asking
The answer
How exam questions ask about this dot point

What this dot point is asking

The multiplication principle and ordered selections page counted selections where order matters: filling distinct positions, forming words, building numbers. The count of ordered selections of $r$ distinct objects from $n$ is the permutation $^{n}P_{r} = \frac{n!}{(n-r)!}$ . This page is about the other half of selection counting: when order does not matter. Choosing $3$ books to take on holiday, picking a committee of $4$ , drawing $5$ cards, joining $2$ points with a line, these are all unordered selections, and an unordered selection of distinct objects is just a subset of the set you are choosing from.

The central object is the combination $^{n}C_{r}$ , the number of $r$ -element subsets of an $n$ -element set, spoken " $n$ choose $r$ " and also written $\binom{n}{r}$ . There are three facts to lock in. First, the formula $^{n}C_{r} = \dfrac{^{n}P_{r}}{r!} = \dfrac{n!}{r!\,(n-r)!}$ , where the division by $r!$ is exactly the step that turns "ordered" into "unordered". Second, the total number of subsets of an $n$ -element set is $2^n$ , which gives the row sum $^{n}C_{0} + {}^{n}C_{1} + \cdots + {}^{n}C_{n} = 2^n$ . Third, the symmetry $^{n}C_{r} = {}^{n}C_{n-r}$ , because choosing the $r$ you keep is the same as choosing the $n-r$ you leave behind. With those three in hand, the rest of the dot point is restriction technique: "at least", the complement, "must contain or exclude" a particular item, committees with gender or role constraints, and geometric counts such as triangles from points on a circle.

The answer

From ordered to unordered: why you divide by $r!$

Start from something you can already count. The number of ordered selections of $3$ distinct letters from the five-letter set $\{A, B, C, D, E\}$ is

^{5}P_{3} = 5 \times 4 \times 3 = 60.

Now ask the unordered question: how many subsets of size $3$ are there? The subset $\{B, C, E\}$ is a single object, but it shows up among those $60$ ordered words as six different words,

BCE,\ BEC,\ CBE,\ CEB,\ EBC,\ ECB \longleftrightarrow \{B, C, E\},

because the three chosen letters can be ordered in $3! = 6$ ways. The same is true of every subset: each $3$ -element subset corresponds to exactly $3!$ ordered words. So the $60$ ordered words count each subset $6$ times, and the number of subsets is

\frac{60}{6} = 10.

The picture is the whole proof in one frame: the six ordered words on the left all funnel into the single subset on the right, so to count subsets you take the ordered count and divide by the $r!$ orderings that each subset is counted with. Doing this in general, there are $^{n}P_{r} = \frac{n!}{(n-r)!}$ ordered selections, each subset is ordered in $r! = {}^{r}P_{r}$ ways, so the number of unordered selections is

^{n}C_{r} = \frac{^{n}P_{r}}{r!} = \frac{n!}{(n-r)!} \div r! = \frac{n!}{r!\,(n-r)!}.

That single division by $r!$ is the only difference between a permutation and a combination. If you ever cannot remember whether to divide, ask "does reordering my chosen objects make a new outcome?" If yes, it is a permutation; if no, divide by $r!$ and it is a combination.

Computing $^{n}C_{r}$ by hand, and the $^{n}P_{r}$ contrast

For actual numbers, do not expand both full factorials. Cancel the largest factorial, $(n-r)!$ , against the top straight away, leaving $r$ falling factors over $r!$ . For $^{8}C_{5}$ ,

^{8}C_{5} = \frac{8!}{5!\,3!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56.

Notice the shortcut: cancel the larger of $5!$ and $3!$ (here $5!$ ), so only $8 \times 7 \times 6$ survives on top and $3!$ on the bottom. That is the same as using the symmetry first, $^{8}C_{5} = {}^{8}C_{3} = \frac{8 \times 7 \times 6}{3!} = 56$ , which is why $^{n}C_{r}$ is easiest to compute when you make the bottom number $\min(r,\,n-r)$ .

The contrast with the permutation is worth seeing side by side. Choosing $5$ objects from $8$ in order gives

^{8}P_{5} = 8 \times 7 \times 6 \times 5 \times 4 = 6720,

and dividing by $5! = 120$ recovers the combination, $6720 \div 120 = 56$ . The ordered count is $120$ times larger because each unordered choice of $5$ can be arranged in $5!$ orders.

Exam technique

Markers reward a clean, checkable layout. Lock in the marks like this:

Cancel the larger factorial first. Write $^{8}C_{5} = \frac{8 \times 7 \times 6}{3!}$ , not two full factorials, so the arithmetic is short and visibly correct.
Use the symmetry to make the bottom small. $^{n}C_{r} = {}^{n}C_{n-r}$ , so compute $^{20}C_{18}$ as $^{20}C_{2} = \frac{20 \times 19}{2} = 190$ , never as a $20$ -term factorial.
State "order does not matter" once when you decide a count is a combination. That one line often secures the method mark even if a number slips.
For a multi-stage choice, write the product of combinations ( $^{6}C_{2} \times {}^{8}C_{3}$ ), then evaluate, so each stage is auditable.
Keep one source of truth. If a count such as $^{9}C_{4} = 126$ appears in the complement and again in a check, compute it once and reuse it.
Read "at least", "at most", "not" and "excluding" as complement warnings. They almost always signal "total minus unwanted" is the shorter route.

Every subset at once: an $n$ -set has $2^n$ subsets

Step back from a fixed size and count all subsets of an $n$ -element set together. Building a subset means going through the $n$ elements one at a time and deciding, independently, whether each is in or out, two choices per element. By the multiplication principle that is

\underbrace{2 \times 2 \times \cdots \times 2}_{n \text{ factors}} = 2^n

subsets in total, where the all-out choice gives the empty set and the all-in choice gives the whole set. For the five-element set $\{A, B, C, D, E\}$ this is $2^5 = 32$ subsets.

The display sorts those $32$ subsets into columns by size. The column heights are $^{5}C_{0}, {}^{5}C_{1}, \dots, {}^{5}C_{5}$ , which read $1, 5, 10, 10, 5, 1$ , and they add to $32$ . That gives a second way to count all subsets: count them size by size and add. Equating the two counts is the row sum

^{n}C_{0} + {}^{n}C_{1} + {}^{n}C_{2} + \cdots + {}^{n}C_{n} = 2^n,

true because both sides count every subset of an $n$ -set exactly once, the left by sorting on size and the right all together. For $n = 5$ this is $1 + 5 + 10 + 10 + 5 + 1 = 32 = 2^5$ , exactly the columns in the figure.

The symmetry $^{n}C_{r} = {}^{n}C_{n-r}$

Look at the figure again and notice the columns read the same left to right as right to left: $1, 5, 10, 10, 5, 1$ . That is the symmetry

^{n}C_{r} = {}^{n}C_{n-r}.

The reason is a perfect pairing, not a coincidence. Every time you choose an $r$ -element subset to keep, you simultaneously leave behind an $(n-r)$ -element subset, its complement. Choosing the keepers and choosing the leavers are the same act described two ways, so the number of $r$ -subsets equals the number of $(n-r)$ -subsets. Choosing $2$ people from $5$ to make the tea is the same as choosing the $3$ who will not, so $^{5}C_{2} = {}^{5}C_{3} = 10$ . In practice the symmetry is a labour-saver: compute $^{n}C_{r}$ using whichever of $r$ and $n-r$ is smaller.

Restriction technique with combinations

The reshaping moves from the grouping, complement and cases page carry straight across to selections, because a combination is itself a count you can subtract from or split. Four moves cover almost every restricted-selection question.

Conditional: must contain a particular item: If a chosen object is forced in, put it in first, then choose the remaining places from the remaining objects. A team of $5$ from $11$ that must include the captain $C$ has $C$ fixed, so the other $4$ come from the remaining $10$ : $^{10}C_{4} = 210$ .
Conditional: must exclude a particular item: If an object is banned, simply remove it and choose from what is left. A team of $5$ from $11$ that must exclude an injured player $X$ chooses all $5$ from the remaining $10$ : $^{10}C_{5} = 252$ . Combine the two: include $C$ and exclude $X$ leaves $4$ to choose from $9$ , namely $^{9}C_{4} = 126$ .
Split a constrained selection into independent groups: When a committee must contain set numbers from disjoint pools, choose from each pool separately and multiply. A committee of $5$ from $6$ men and $8$ women with exactly $2$ men chooses $2$ of the $6$ men and $3$ of the $8$ women: $^{6}C_{2} \times {}^{8}C_{3} = 15 \times 56 = 840$ .
At least / at most: count the complement: "At least $k$ " usually fragments into several size cases, so count the total and subtract the few unwanted cases. A committee of $4$ from $5$ women and $4$ men with at least $2$ women is the total $^{9}C_{4} = 126$ minus the committees with $0$ or $1$ woman, $^{4}C_{4} + {}^{5}C_{1}\,^{4}C_{3} = 1 + 20 = 21$ , giving $126 - 21 = 105$ .

Geometry counts: lines and triangles from points

A clean family of combination questions hides inside geometry. Given points with no three on a straight line, every unordered pair of points determines exactly one line (chord), and every unordered triple determines exactly one triangle, so the counts are simply combinations.

For the $8$ points shown, the chords number $^{8}C_{2} = 28$ , the triangles number $^{8}C_{3} = 56$ , and the quadrilaterals number $^{8}C_{4} = 70$ . The highlighted triangle $\{P_1, P_4, P_6\}$ is one of those $56$ : notice that choosing the three vertices in any order gives the same triangle, which is exactly why it is a combination and not a permutation. The "no three collinear" condition matters, because three collinear points would give a degenerate "triangle" (a straight line) and the count $^{n}C_{3}$ would overcount; on a circle no three points are ever collinear, so the count is exact.

Verifying a small case by listing

Whenever a count is small, the safest check is to list and tally. Take the triangles from $6$ points $\{1,2,3,4,5,6\}$ on a circle. The formula gives $^{6}C_{3} = \frac{6 \times 5 \times 4}{6} = 20$ . Listing every $3$ -element subset in increasing order gives

123,\,124,\,125,\,126,\,134,\,135,\,136,\,145,\,146,\,156,\,234,\,235,\,236,\,245,\,246,\,256,\,345,\,346,\,356,\,456,

which is $20$ triangles, matching the formula. Listing in a fixed order (here, smallest digit first) guarantees you write each subset once and never repeat, which is the discipline that makes a by-hand check trustworthy.

How exam questions ask about this dot point

The wording tells you which move to reach for. Map the phrase to the method:

"...select / choose / pick a group / committee / team / hand", "in how many ways can $r$ be chosen from $n$ ". A plain combination: $^{n}C_{r}$ , because order does not matter.
"...how many subsets", "how many possible selections of any size", "any number of toppings". All subsets: $2^n$ .
"...show that $^{n}C_{r} = {}^{n}C_{n-r}$ ", "why are the answers the same". Symmetry by complement: choosing the $r$ kept is the same as choosing the $n-r$ left out.
"...must include / contains a particular ...". Fix that item, choose the rest from the rest: $^{n-1}C_{r-1}$ .
"...must not include / excludes / without a particular ...". Delete that item, choose normally: $^{n-1}C_{r}$ .
"...exactly $k$ of one type and the rest another", "two men and two women". Multiply per-pool combinations: $^{a}C_{i} \times {}^{b}C_{j}$ .
"...at least / at most / a majority of ...". Complement: total minus the unwanted size cases.
"...how many lines / chords / triangles / diagonals from these points", with no three collinear. A geometry combination: $^{n}C_{2}$ for lines, $^{n}C_{3}$ for triangles.

Worked examples

Evaluate a combination and contrast it with the permutation

Evaluate $^{8}C_{5}$ from the formula, and explain why the number of ways to choose $5$ books from $8$ to put on a shelf in order is much larger.

Choose the formula and cancel the larger factorial. Order does not matter for "which $5$ books", so this is a combination. Cancel $5!$ against $8!$ :

^{8}C_{5} = \frac{8!}{5!\,3!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56.

Confirm with the symmetry. Since $^{8}C_{5} = {}^{8}C_{3}$ , the same arithmetic with the smaller bottom gives $\frac{8 \times 7 \times 6}{3!} = 56$ , a useful cross-check.

Compute the ordered count. Putting $5$ chosen books on a shelf in order is a permutation:

^{8}P_{5} = 8 \times 7 \times 6 \times 5 \times 4 = 6720.

Explain the factor. Each unordered choice of $5$ books can be shelved in $5! = 120$ orders, so the ordered count is $120$ times the unordered count: $56 \times 120 = 6720$ , and dividing back, $6720 \div 120 = 56$ .

Final answer. $^{8}C_{5} = 56$ ; the ordered shelving gives $^{8}P_{5} = 6720$ , which is $5! = 120$ times larger because order multiplies by $5!$ .

Count selections of every size, then a fixed size

A music streaming app lets you build a playlist from $5$ saved songs, choosing any number of them (including the empty playlist). (a) How many different playlists are possible? (b) How many playlists contain exactly $3$ songs? (c) Verify that the playlists, grouped by number of songs, add up to the answer in part (a).

Part (a): count all subsets. A playlist is a subset of the $5$ songs (order aside), and each song is in or out:

2^5 = 32 \text{ playlists.}

Part (b): count one size with a combination. Choosing $3$ of the $5$ songs is

^{5}C_{3} = \frac{5 \times 4 \times 3}{3!} = \frac{60}{6} = 10.

Part (c): add the sizes and match. The playlists of size $0, 1, 2, 3, 4, 5$ number

^{5}C_{0} + {}^{5}C_{1} + {}^{5}C_{2} + {}^{5}C_{3} + {}^{5}C_{4} + {}^{5}C_{5} = 1 + 5 + 10 + 10 + 5 + 1 = 32,

which equals the $2^5$ from part (a), as the row-sum identity predicts.

Final answer. (a) $32$ ; (b) $10$ ; (c) the size totals $1 + 5 + 10 + 10 + 5 + 1 = 32 = 2^5$ agree.

Choose a committee with a conditional and a complement

A committee of $4$ is chosen from $5$ women and $4$ men. (a) In how many ways with no restriction? (b) In how many ways if a particular woman, $W$ , must be on the committee? (c) In how many ways if there must be at least $2$ women?

Part (a): a plain combination. Choose any $4$ of the $9$ people:

^{9}C_{4} = \frac{9 \times 8 \times 7 \times 6}{4!} = \frac{3024}{24} = 126.

Part (b): force $W$ in, choose the rest from the rest. With $W$ fixed, choose the other $3$ members from the remaining $8$ people:

^{8}C_{3} = \frac{8 \times 7 \times 6}{3!} = 56.

Part (c): use the complement. "At least $2$ women" is the total minus the committees with $0$ or $1$ woman. With $0$ women all $4$ are men ( $^{4}C_{4} = 1$ ); with $1$ woman, choose her and $3$ men ( $^{5}C_{1}\,^{4}C_{3} = 5 \times 4 = 20$ ):

126 - (1 + 20) = 126 - 21 = 105.

Check part (c) directly. $^{5}C_{2}\,^{4}C_{2} + {}^{5}C_{3}\,^{4}C_{1} + {}^{5}C_{4}\,^{4}C_{0} = 60 + 40 + 5 = 105$ , the same value.

Final answer. (a) $126$ ; (b) $56$ ; (c) $105$ .

Count triangles and lines from points on a circle

Ten points are spaced around a circle so that no three lie on a straight line. (a) How many straight lines join pairs of the points? (b) How many triangles have their vertices among the points? (c) How many of those triangles use a particular point, $A$ , as a vertex?

Part (a): a line is an unordered pair. The two endpoints in either order give one line, so count pairs:

^{10}C_{2} = \frac{10 \times 9}{2} = 45.

Part (b): a triangle is an unordered triple. No three points are collinear, so every $3$ points form a genuine triangle:

^{10}C_{3} = \frac{10 \times 9 \times 8}{3!} = \frac{720}{6} = 120.

Part (c): fix $A$ , then choose the other two vertices. With $A$ a vertex, the remaining $2$ vertices are any $2$ of the other $9$ points:

^{9}C_{2} = \frac{9 \times 8}{2} = 36.

Sanity check on (c). Each triangle has $3$ vertices, so the triangles-through-a-fixed-point should be $\frac{3 \times 120}{10} = 36$ , which agrees.

Final answer. (a) $45$ lines; (b) $120$ triangles; (c) $36$ triangles through $A$ .

Build a team with role and gender constraints

A mixed sports squad has $6$ men and $8$ women. (a) A team of $5$ is chosen with exactly $2$ men; in how many ways? (b) From the chosen team of $5$ , one player is then named captain; in how many ways can the team-and-captain be chosen, still with exactly $2$ men in the team?

Part (a): choose each gender separately and multiply. The men and women are chosen from separate pools, so multiply the two combinations. Choose $2$ of the $6$ men and $3$ of the $8$ women:

^{6}C_{2} \times {}^{8}C_{3} = 15 \times 56 = 840.

Part (b): add the role as a separate ordered choice. First form the team as in part (a), then choose the captain from its $5$ members:

840 \times 5 = 4200.

Confirm part (b) another way. Equivalently, the captain can be a man or a woman. Captain a man: choose the captain ( $6$ ), the other man ( $^{5}C_{1} = 5$ ) and the $3$ women ( $^{8}C_{3} = 56$ ), giving $6 \times 5 \times 56 = 1680$ . Captain a woman: choose the captain ( $8$ ), the $2$ men ( $^{6}C_{2} = 15$ ) and the other $2$ women ( $^{7}C_{2} = 21$ ), giving $8 \times 15 \times 21 = 2520$ . Together $1680 + 2520 = 4200$ , matching.

Final answer. (a) $840$ ; (b) $4200$ .

Verify $^{n}C_{r} = {}^{n}C_{n-r}$ and list a small case

(a) Show by the formula that $^{7}C_{2} = {}^{7}C_{5}$ . (b) List all the $2$ -element subsets of $\{a, b, c, d\}$ and confirm there are $^{4}C_{2}$ of them, then state how many ordered pairs the same letters give.

Part (a): write both in factorial form. The two combinations are

^{7}C_{2} = \frac{7!}{2!\,5!}, \qquad ^{7}C_{5} = \frac{7!}{5!\,2!}.

These are identical because multiplication is commutative ( $2!\,5! = 5!\,2!$ ), so $^{7}C_{2} = {}^{7}C_{5} = \frac{7 \times 6}{2} = 21$ . The reason is the complement pairing: each $2$ -subset of a $7$ -set is matched with the $5$ -subset of elements left out.

Part (b): list the unordered pairs. The $2$ -element subsets of $\{a, b, c, d\}$ , written in fixed order, are

\{a,b\},\ \{a,c\},\ \{a,d\},\ \{b,c\},\ \{b,d\},\ \{c,d\},

which is $6$ subsets, agreeing with $^{4}C_{2} = \frac{4 \times 3}{2} = 6$ .

Contrast with ordered pairs. The same letters give $^{4}P_{2} = 4 \times 3 = 12$ ordered pairs ( $ab$ and $ba$ both appear), exactly $2! = 2$ times the unordered count, since each pair has $2!$ orders.

Final answer. (a) $^{7}C_{2} = {}^{7}C_{5} = 21$ ; (b) there are $^{4}C_{2} = 6$ unordered pairs and $^{4}P_{2} = 12$ ordered pairs, a factor of $2! = 2$ apart.

Common traps

Using a permutation where order does not matter: A committee, team, hand or subset is unordered, so it is $^{n}C_{r}$ , not $^{n}P_{r}$ . Forgetting to divide by $r!$ inflates the answer by a factor of $r!$ . Ask: "does reordering the chosen objects make a new outcome?" If no, divide by $r!$ .
Expanding both full factorials: Cancel the larger factorial straight away, and use $^{n}C_{r} = {}^{n}C_{n-r}$ to keep the bottom number small: compute $^{20}C_{18}$ as $^{20}C_{2} = 190$ , not a $20$ -term product.
Adding when you should multiply (or vice versa): Choosing from separate pools (men and women) multiplies the combinations. Splitting into mutually exclusive cases (exactly $2$ , then $3$ , then $4$ ) adds them. Multiply within a single selection, add across alternative cases.
Counting "at least" directly when the complement is shorter: "At least $2$ women" out of a committee is faster as total minus ( $0$ or $1$ woman). Reading "at least", "at most" and "not" as complement cues saves work and avoids missed cases.
Mishandling a forced or banned item: "Must contain $X$ " means choose the rest from the rest, $^{n-1}C_{r-1}$ , not $^{n}C_{r}$ with $X$ left free. "Must exclude $X$ " means delete $X$ first, $^{n-1}C_{r}$ .
Forgetting the no-three-collinear condition in geometry counts: $^{n}C_{3}$ counts triangles only when no three points are collinear; collinear triples give degenerate triangles. Points on a circle are always safe, but points on a grid are not.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marks(i) Evaluate

^{9}C_{4}

. (ii) Evaluate

^{9}P_{4}

. (iii) Hence state, in one sentence, why selecting

4

people from

9

to form a committee gives a smaller answer than choosing

4

people from

9

to fill four distinct offices (chair, secretary, treasurer, member).

Show worked solution →

Part (i): use the formula for a combination. Cancel the larger factorial against the numerator:

^{9}C_{4} = \frac{9!}{4!\,5!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = \frac{3024}{24} = 126.

Part (ii): use the formula for a permutation. Here order matters, so do not divide by $4!$ :

^{9}P_{4} = 9 \times 8 \times 7 \times 6 = 3024.

Part (iii): explain the gap. Filling four distinct offices is an ordered selection, so it counts $^{9}P_{4} = 3024$ . A committee is unordered, so each committee of $4$ is counted $4! = 24$ times among those ordered lists, and dividing removes that overcount: $3024 \div 24 = 126$ . Order multiplies the answer by $4!$ .

Answer. (i) $126$ ; (ii) $3024$ ; (iii) the committee is unordered, so it equals $^{9}P_{4} \div 4! = 126$ , a factor of $4!$ smaller.

foundation2 marksA pizza shop offers

7

toppings. A customer may choose any number of toppings, from none at all up to all

7

. How many different pizzas (counted by their set of toppings) are possible?

Show worked solution →

Recognise this as counting subsets. A pizza is just a subset of the $7$ toppings, because the order you list the toppings in does not change the pizza.

Count every subset at once. Each topping is independently in or out, giving two choices per topping across $7$ toppings:

2^{7} = 128.

Confirm by adding the subset sizes. The pizzas with $0, 1, 2, \dots, 7$ toppings number $^{7}C_{0} + ^{7}C_{1} + \cdots + ^{7}C_{7} = 1 + 7 + 21 + 35 + 35 + 21 + 7 + 1 = 128$ , the same total.

Answer. There are $128$ possible pizzas (including the plain pizza with no toppings).

core3 marksA committee of

4

is chosen from

5

women and

4

men. In how many ways can the committee be chosen so that it contains at least

2

women?

Show worked solution →

Read 'at least' as a cue for the complement. Counting at least $2$ women directly means adding the cases of exactly $2$ , $3$ and $4$ women. The complement (fewer than $2$ women, that is $0$ or $1$ ) is shorter, so subtract it from the total.

Count all committees. Choose any $4$ of the $9$ people:

^{9}C_{4} = 126.

Count the unwanted committees ( $0$ or $1$ woman). With $0$ women all $4$ come from the $4$ men, and with $1$ woman the other $3$ come from the men:

^{4}C_{4} + {}^{5}C_{1} \times {}^{4}C_{3} = 1 + 5 \times 4 = 1 + 20 = 21.

Subtract.

126 - 21 = 105.

Check directly. Summing the wanted cases, $^{5}C_{2}\,^{4}C_{2} + {}^{5}C_{3}\,^{4}C_{1} + {}^{5}C_{4}\,^{4}C_{0} = 60 + 40 + 5 = 105$ , which agrees.

Answer. There are $105$ committees with at least $2$ women.

core3 marksEight points lie on a circle, with no three of them on a straight line. (i) How many straight lines (chords) join pairs of these points? (ii) How many triangles have all three vertices among these points? (iii) How many of those triangles have a particular point,

P

, as a vertex?

Show worked solution →

Part (i): a line needs an unordered pair of points. Choosing the two endpoints in either order gives the same line, so count unordered pairs:

^{8}C_{2} = \frac{8 \times 7}{2} = 28.

Part (ii): a triangle needs an unordered triple of points. No three points are collinear, so every choice of $3$ points really does form a triangle:

^{8}C_{3} = \frac{8 \times 7 \times 6}{6} = 56.

Part (iii): fix $P$ , then choose the rest. With $P$ already a vertex, the other $2$ vertices are any $2$ of the remaining $7$ points:

^{7}C_{2} = \frac{7 \times 6}{2} = 21.

Answer. (i) $28$ lines; (ii) $56$ triangles; (iii) $21$ triangles through $P$ .

exam4 marksA netball squad has

12

players, including the captain

C

and two players

J

and

K

who are both injured. A team of

7

is to be chosen. (i) In how many ways can a team of

7

be chosen with no restriction? (ii) In how many ways can the team be chosen if the captain

C

must be included? (iii) In how many ways can the team be chosen if both

J

and

K

must be excluded?

Show worked solution →

Part (i): choose $7$ from $12$ , order irrelevant. A team is an unordered selection:

^{12}C_{7} = \frac{12!}{7!\,5!} = 792.

Part (ii): force $C$ in, then choose the rest. With $C$ already on the team, choose the other $6$ players from the remaining $11$ :

^{11}C_{6} = \frac{11!}{6!\,5!} = 462.

Part (iii): remove $J$ and $K$ , then choose freely. Excluding $J$ and $K$ leaves $10$ available players, from whom all $7$ are chosen:

^{10}C_{7} = \frac{10!}{7!\,3!} = 120.

Sanity check on (iii). By the symmetry $^{10}C_{7} = {}^{10}C_{3} = \frac{10 \times 9 \times 8}{6} = 120$ , which agrees.

Answer. (i) $792$ ; (ii) $462$ ; (iii) $120$ .

exam4 marksA team of

5

is chosen from

6

men and

8

women. (i) In how many ways can the team be chosen with exactly

2

men? (ii) From a separate group of

10

people, a working party of

4

is chosen and then one of those

4

is named as its leader. In how many ways can the working party and its leader be chosen?

Show worked solution →

Part (i): choose the men and the women separately, then multiply. The two genders are chosen independently, so use the multiplication principle on two combinations. Choose $2$ of the $6$ men and $3$ of the $8$ women:

^{6}C_{2} \times {}^{8}C_{3} = 15 \times 56 = 840.

Part (ii): an unordered choice followed by a distinguished role. First choose the unordered working party of $4$ from $10$ , then pick its leader from those $4$ :

^{10}C_{4} \times 4 = 210 \times 4 = 840.

Confirm part (ii) another way. Pick the leader first ( $10$ ways), then choose the other $3$ members from the remaining $9$ : $10 \times {}^{9}C_{3} = 10 \times 84 = 840$ , the same value. The role is a separate, ordered choice on top of the unordered selection.

Answer. (i) $840$ ; (ii) $840$ .

What this dot point is asking

The answer

From ordered to unordered: why you divide by r!r!r!

Computing nCr^{n}C_{r}nCr​ by hand, and the nPr^{n}P_{r}nPr​ contrast

Every subset at once: an nnn-set has 2n2^n2n subsets

The symmetry nCr=nCn−r^{n}C_{r} = {}^{n}C_{n-r}nCr​=nCn−r​

Restriction technique with combinations

Geometry counts: lines and triangles from points

Verifying a small case by listing

How exam questions ask about this dot point

Practice questions

Related dot points

From ordered to unordered: why you divide by $r!$

Computing $^{n}C_{r}$ by hand, and the $^{n}P_{r}$ contrast

Every subset at once: an $n$ -set has $2^n$ subsets

The symmetry $^{n}C_{r} = {}^{n}C_{n-r}$