Distribution & Grouping — notes, question types & practice · Permutation, Combination & Probability

01

Splitting into groups

To split $n$ different people into groups of sizes $a$ , $b$ and $c$ :

\text{ways} = \frac{n!}{a!\,b!\,c!}

Divide by the factorial of each group size, because order inside a group does not matter.

9 students into groups of 4, 3 and 2: $\dfrac{9!}{4!\,3!\,2!} = 1260$ .

Two groups of the same size can swap with each other, so divide by $2!$ as well. 8 people into two teams of 4: $\dfrac{1}{2!} \times \dfrac{8!}{4!\,4!} = 35$ .

Watch: Groups of the same size are interchangeable. Divide by the number of equal groups.

02

Sharing with everyone at least one

Distribute $n$ identical items among $r$ people, each getting at least one.

\text{ways} = \,^{n-1}C_{r-1}

8 identical chocolates, 3 children, each at least one: $^{7}C_2 = 21$ .

Why it works: lay the 8 chocolates in a row and cut twice. There are 7 gaps between them and 2 cuts, so $^{7}C_2$ .

Test the rule on a tiny case: 3 chocolates, 2 children, each at least one gives $^{2}C_1 = 2$ — the splits (1, 2) and (2, 1). Listing confirms it.

03

Sharing with zero allowed

Now empty hands are allowed — a child may get nothing.

\text{ways} = \,^{n+r-1}C_{r-1}

5 identical sweets, 3 children, zeros allowed: $^{7}C_2 = 21$ .

Rule: "Each at least one" gives $^{n-1}C_{r-1}$ . "Zero allowed" gives $^{n+r-1}C_{r-1}$ . One changed word changes the formula.

04

Different things to different people

When the items are different, each item chooses its person.

3 different letters into 2 letterboxes: each letter has 2 choices, so $2^3 = 8$ ways.

Two letters, two boxes: $2^2 = 4$ — both letters in one box (two ways) or split between the boxes (two ways).

$r$ different items to $n$ people give $n^r$ ways. If each person may hold at most one item, the count becomes $^{n}P_r$ instead.

05

Identical things to different people

Identical items use the cut formulas of the last two sections. Watch the wording closely.

7 identical pens among 4 shops, zeros allowed: $^{10}C_3 = 120$ .

The algebra version is the same idea. Positive solutions of $x + y + z = 12$ : $^{11}C_2 = 55$ . Non-negative: $^{14}C_2 = 91$ .

Example: "Each basket at least 2"? Put 2 into each basket first, then share what remains by the usual formula.

06

Making pairs

$2n$ different people into $n$ unordered pairs:

\text{ways} = \frac{(2n)!}{2^n\,n!}

8 people into 4 pairs: $\dfrac{8!}{2^4 \times 4!} = 105$ .

The $2^4$ removes the order inside each pair; the $4!$ removes the order of the pairs themselves.

07

Question types you will see

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

Type 1very common3 practice Q

Identical items, everyone at least one

How to spot it:

'Distribute n identical items among r people, each gets at least one' — or 'positive integer solutions'.

^{n-1}C_{r-1}

Method

Lay the $n$ items in a row: there are $n-1$ gaps.
Choose $r-1$ gaps to cut.
Answer $^{n-1}C_{r-1}$ .

Why it works:

Each set of cuts splits the row into $r$ non-empty heaps, one per person.

Try this

9 identical toffees are shared by 3 children, each getting at least one. In how many ways?

Show solution

Gaps: 8; cuts: 2.
$^{8}C_2 = 28$ .

Answer

28

Type 2very common2 practice Q

Identical items, zero allowed

How to spot it:

'…a person may get none' or 'non-negative integer solutions of x + y + z = n'.

^{n+r-1}C_{r-1}

Method

Note that empty hands are allowed now.
Answer $^{n+r-1}C_{r-1}$ .
Check: 'at least one' would give $^{n-1}C_{r-1}$ instead.

Why it works:

Allowing empty heaps adds the extra spots where cuts may fall.

Try this

5 identical sweets are given to 3 children; a child may get none. In how many ways?

Show solution

$^{5+3-1}C_{3-1} = ^{7}C_2$ .
$= 21$ .

Answer

21

Type 3common2 practice Q

Split n different people into groups

How to spot it:

'Divide 9 students into groups of 4, 3 and 2', or two teams of equal size.

\frac{n!}{a!\,b!\,c!}

Method

Write $n!$ over $a!\,b!\,c!$ for the group sizes.
Equal groups: divide further by the number of such groups ( $2!$ for a pair).
Compute with cancellation.

Why it works:

Shuffling people inside a group changes nothing, and equal groups can swap with each other.

Try this

In how many ways can 10 people be divided into groups of 5, 3 and 2?

Show solution

$\dfrac{10!}{5!\,3!\,2!}$ .
$= 2520$ .

Answer

2520

Type 4common2 practice Q

Different items to different people

How to spot it:

'3 different letters into letterboxes', '4 different toys among children' — the items are different.

n^r

Method

Each item picks its own person.
With $n$ people and $r$ items the answer is $n^r$ .
At most one item per person? Use $^{n}P_r$ instead.

Why it works:

Independent choices for each item multiply into a power.

Try this

4 different toys are distributed among 3 children. In how many ways can this be done?

Show solution

Each toy: 3 choices.
$3^4 = 81$ .

Answer

81

Type 5common2 practice Q

Divide into pairs

How to spot it:

'8 people form 4 pairs', or doubles teams are made from a group.

\frac{(2n)!}{2^n\,n!}

Method

Arrange all $2n$ people in a row.
Cut into $n$ pairs; divide by $2^n$ for the order inside each pair.
Divide by $n!$ for the order of the pairs themselves.

Why it works:

Swaps inside a pair and swaps of whole pairs do not create new pairings.

Try this

In how many ways can 6 people be divided into 3 pairs?

Show solution

$\dfrac{6!}{2^3 \times 3!}$ .
$= \dfrac{720}{48} = 15$ .

Answer

15

Type 6common2 practice Q

Buy or choose one or more

How to spot it:

'At least one item is bought' from several different designs, shops or menus.

2^n - 1

Method

Each of the $n$ different items is taken or left: $2^n$ .
Remove the empty choice.
Answer $2^n - 1$ .

Why it works:

A take-or-leave decision per item doubles the outcomes, and only 'take nothing' is excluded.

Try this

A shop has 6 different designs. In how many ways can a customer buy one or more designs?

Show solution

$2^6 - 1$ .
$= 63$ .

Answer

63

08

Formula sheet

Groups of sizes a, b, c

\frac{n!}{a!\,b!\,c!}

$n$ different people; divide again by $k!$ for $k$ equal groups.

Identical items, each at least one

^{n-1}C_{r-1}

$n$ items, $r$ people, no one empty.

Identical items, zeros allowed

^{n+r-1}C_{r-1}

Different items to n people

n^r

$r$ different items, each choosing a person.

2n people into n pairs

\frac{(2n)!}{2^n\,n!}

One or more selections

2^n - 1

$n$ different items, at least one taken.

09

Shortcuts that save time

⚡ Give everyone the minimum first

'Each gets at least 2'? Hand out 2 to each person, then share the rest with the usual formula.

Example

6 identical toffees go to 2 children, each getting at least 2. In how many ways?

Show solution

Give 2 to each: 2 left.
Share 2, zeros allowed: $^{3}C_1 = 3$ .

Answer

3

⚡ Equal groups divide further

Two groups of the same size can swap without changing anything, so divide by $2!$ .

Example

6 people are divided into two groups of 3 each. In how many ways?

Show solution

$\dfrac{1}{2} \times \dfrac{6!}{3!\,3!}$ .
$= \dfrac{720}{72} = 10$ .

Answer

10

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Using $^{n-1}C_{r-1}$ when zeros are allowed — that formula is only for 'each gets at least one'.

Mistake 02

Skipping the extra division for equal groups — two teams of 4 need $\frac{1}{2} \times ^{8}C_4 = 35$ , not $70$ .

Mistake 03

Using $n^r$ for identical items — powers count different items choosing their person.

Mistake 04

Dividing $n!$ by the sum of the group sizes instead of the product of their factorials.

Mistake 05

Ignoring the minimum in 'each basket at least 2' — place the minimums first, then share the rest.

11

Quick revision

Read this the night before the exam.

Groups of sizes $a, b, c$ : $\dfrac{n!}{a!\,b!\,c!}$ ; equal groups: divide again.
Identical items, each at least one: $^{n-1}C_{r-1}$ .
Identical items, zeros allowed: $^{n+r-1}C_{r-1}$ .
Different items to $n$ people: $n^r$ ; at most one each: $^{n}P_r$ .
$2n$ people into $n$ pairs: $\dfrac{(2n)!}{2^n n!}$ .
One or more from $n$ different designs: $2^n - 1$ .

12

Practice: 13 questions

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

Topic test · 13 questions

Suggested time 10 min · wrong answers go to your mistake notebook automatically.