Combinations: Selections — notes, question types & practice · Permutation, Combination & Probability

01

Selection, not arrangement

A combination is a selection. Order does not matter. AB and BA are the same team.

Committees, teams, handshakes and groups of friends are all selections.

02

The nCr formula

$^{n}C_r$ counts the selections of $r$ items from $n$ distinct items.

^{n}C_r = \frac{n!}{r!\,(n-r)!}

Every selection of $r$ items can be lined up in $r!$ orders. So arrangements $=$ selections $\times\ r!$ , which gives $^{n}C_r = \dfrac{^{n}P_r}{r!}$ .

$^{5}C_2 = 10$ : two friends from five make 10 pairs.

03

Committees and teams

Split the choice into independent groups and multiply.

From 5 men and 3 women, choose 2 men and 1 woman: $^{5}C_2 \times ^{3}C_1 = 10 \times 3 = 30$ committees.

If a group may send any number of members (including none), its share is $2^m$ for $m$ members — the sum of all its nCr terms.

Rule: "2 men and 1 woman" means two selections joined by "and", so multiply.

04

At least one

"At least one woman" means one, two, three or more women. Counting every case is slow. The fast way:

\text{at least one} = \text{total selections} - \text{selections with none}

From 5 men and 4 women, a committee of 4 with at least one woman: total $^{9}C_4 = 126$ . All-men committees: $^{5}C_4 = 5$ . Answer $126 - 5 = 121$ .

Tip: "At least one" always flips to "none". Subtract and finish in two lines.

05

Handshakes, matches and diagonals

Any question about pairs is $^{n}C_2$ .

10 friends shake hands once each: $^{10}C_2 = 45$ handshakes. 12 teams in a league: $^{12}C_2 = 66$ matches.

A polygon with $n$ sides has $^{n}C_2 - n$ diagonals: join any two vertices, then remove the $n$ sides.

A knockout tournament with $n$ players has $n - 1$ matches, because every match removes exactly one player.

06

Useful nCr facts

These facts save real time:

$^{n}C_r = ^{n}C_{n-r}$ , so $^{15}C_{12} = ^{15}C_3 = 455$ .
$^{n}C_0 = ^{n}C_n = 1$ .
$^{n}C_x = ^{n}C_y$ means $x = y$ or $x + y = n$ .
$^{n}C_0 + ^{n}C_1 + \cdots + ^{n}C_n = 2^n$ .
$^{n+1}C_{r+1} = ^{n}C_r + ^{n}C_{r-1}$ .

Quick use: $^{10}C_8 = ^{10}C_2 = 45$ — flip first, compute second.

Note: The sum fact answers subset questions too: a set of $n$ elements has $2^n$ subsets, or $2^n - 1$ non-empty ones.

07

Practice the wording

Read each question and ask one thing: does order matter?

A president and a secretary: order matters, use $^{n}P_r$ . A committee of two: order does not, use $^{n}C_r$ . Numbers and words: order matters. Teams and hands: order does not. When in doubt, ask if swapping two members creates a new outcome — if not, it is a combination.

08

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

Evaluate nCr

How to spot it:

A direct ask: 'Find the value of 12C4', or an expression built from nCr terms.

^{n}C_r = \frac{n!}{r!\,(n-r)!}

Method

Use $^{n}C_r = ^{n}C_{n-r}$ to shrink the lower index.
Cancel before multiplying.
Simplify step by step.

Why it works:

Factorials cancel heavily, so dividing first keeps every number small.

Try this

Find the value of $^{8}C_3$ .

Show solution

$\dfrac{8 \times 7 \times 6}{3 \times 2 \times 1}$ .
$= 56$ .

Answer

56

Type 2very common2 practice Q

Committee or team from two groups

How to spot it:

'From 6 men and 4 women choose…' — a fixed number from each group is required.

Method

Choose from each group separately.
Multiply the group counts.
Check the joining word: 'and' multiplies, 'or' adds.

Why it works:

Each group's selection is independent, so the counts multiply.

Try this

From 6 men and 4 women, a committee of 3 men and 2 women is formed. In how many ways?

Show solution

Men: $^{6}C_3 = 20$ .
Women: $^{4}C_2 = 6$ .
$20 \times 6 = 120$ .

Answer

120

Type 3very common2 practice Q

At least one (or at least r)

How to spot it:

'At least one woman', 'at least one defective', 'at least 3 girls' in the team.

Method

One or two cases: count each and add.
Many cases: take total minus none.
'At least r' with few people: add the cases $r, r+1, \ldots$

Why it works:

The complement of 'at least one' is 'none', and 'none' is a single clean count.

Try this

From 6 boys and 4 girls, a team of 4 with at least one girl is chosen. In how many ways?

Show solution

Total: $^{10}C_4 = 210$ .
No girl: $^{6}C_4 = 15$ .
$210 - 15 = 195$ .

Answer

195

Type 4common2 practice Q

Particular people included or excluded

How to spot it:

'Two particular players must be included', 'one particular student must be excluded'.

Method

Included people are fixed: take them out of the pool.
Excluded people are banned: remove them from the pool.
Select the rest from the reduced pool.

Why it works:

Fixing or banning people just shrinks the pool that the remaining choice works on.

Try this

A team of 4 is chosen from 10 players, and 2 particular players must be included. In how many ways?

Show solution

2 are fixed; choose 2 of the remaining 8.
$^{8}C_2 = 28$ .

Answer

28

Type 5common2 practice Q

nCr identities and equations

How to spot it:

An equation like $^nC_4 = ^nC_6$ , or a big sum of nCr terms.

^{n}C_0 + \cdots + ^{n}C_n = 2^n

Method

For $^{n}C_x = ^{n}C_y$ : the answer is $x = y$ or $x + y = n$ .
For the full sum, the answer is $2^n$ .
For a missing term, use $2^n$ minus the known terms.

Why it works:

The symmetry of nCr and the subset count behind $2^n$ settle these at sight.

Try this

If $^{n}C_3 = ^{n}C_5$ , find $n$ .

Show solution

The indices differ, so they must add to $n$ .
$n = 3 + 5 = 8$ .

Answer

8

Type 6very common2 practice Q

Handshakes, matches and diagonals

How to spot it:

'Every pair shakes hands / plays once', or the diagonals of a polygon are asked.

\text{pairs} = ^{n}C_2 = \frac{n(n-1)}{2}

Method

Pairs of any kind are $^{n}C_2$ .
League matches: $^{n}C_2$ ; knockout: $n - 1$ .
Diagonals: $^{n}C_2$ minus the $n$ sides.

Why it works:

A handshake, a match and a diagonal are all just a pair chosen from $n$ things.

Try this

12 teams play a league in which every pair meets once. How many matches are played?

Show solution

$^{12}C_2 = \dfrac{12 \times 11}{2}$ .
$= 66$ .

Answer

66

09

Formula sheet

Select r of n

^{n}C_r = \frac{n!}{r!\,(n-r)!}

Link with nPr

^{n}C_r = \frac{^{n}P_r}{r!}

Divide arrangements by $r!$ for selections.

Symmetry

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

Flip a large lower index before computing.

Sum of all nCr

\sum_{r=0}^{n} {}^{n}C_r = 2^n

The number of subsets of an $n$-element set.

Pairs

^{n}C_2 = \frac{n(n-1)}{2}

Handshakes, league matches, diagonals.

10

Shortcuts that save time

⚡ Total minus none for 'at least one'

Instead of counting the one, two, three… cases, count everything and remove the all-avoid case.

Example

From 7 men and 3 women, a committee of 3 with at least one woman is formed. How many committees are possible?

Show solution

Total: $^{10}C_3 = 120$ .
No woman: $^{7}C_3 = 35$ .
$120 - 35 = 85$ .

Answer

85

⚡ Flip nCr when the lower index is large

$^{n}C_r = ^{n}C_{n-r}$ . Turn $^{18}C_{16}$ into $^{18}C_2$ before computing.

Example

Find $^{18}C_{16}$ .

Show solution

$= ^{18}C_2$ .
$\dfrac{18 \times 17}{2} = 153$ .

Answer

153

11

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Using $^{n}P_r$ where a selection is asked — a president and a secretary is $^{n}P_r$ , a committee is $^{n}C_r$ .

Mistake 02

Adding group selections joined by 'and' — 2 men of 5 and 1 woman of 3 is $10 \times 3$ , not $13$ .

Mistake 03

Forgetting the subtraction in 'at least one' — the answer is total minus the all-none case, never the total itself.

Mistake 04

Missing that $^{n}C_x = ^{n}C_y$ forces $x + y = n$ — $^{n}C_3 = ^{n}C_5$ gives $n = 8$ , not $n = 3$ or $n = 5$ .

Mistake 05

Counting knockout matches as $^{n}C_2$ — a knockout has $n - 1$ matches; only a league has $^{n}C_2$ .

12

Quick revision

Read this the night before the exam.

Order irrelevant, so it is a combination.
$^{n}C_r = \dfrac{n!}{r!(n-r)!} = \dfrac{^{n}P_r}{r!}$ .
$^{n}C_r = ^{n}C_{n-r}$ — flip before computing.
Groups joined by 'and' multiply; joined by 'or' add.
At least one $=$ total $-$ none.
Pairs, handshakes, league matches: $^{n}C_2$ ; knockout: $n-1$ .
$\sum_{r=0}^{n} {}^{n}C_r = 2^n$ .

13

Practice: 12 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 12 questions

Suggested time 8 min · wrong answers go to your mistake notebook automatically.