Your weak spots

Insights load after your first practice round.

Module 4 · L4 of 15 ~40 min ⚡ +95 XP available

Permutations with Restrictions

Real counting problems rarely give you total freedom. In a photo, best friends always stand together. At a dinner table, rivals must not sit adjacent. Someone always needs to be first. Restrictions shrink the count — but they require a systematic strategy. This lesson gives you three: fix the object, glue objects together, or use the gap method.

Today's hook — Six people line up for a photo: Alice, Bob, Carlos, Diana, Eve, Frank. Without any restrictions, there are $6! = 720$ arrangements. But Alice and Bob are inseparable — they must always stand together. Before you read on, estimate what fraction of the 720 arrangements satisfy this condition.

0/5QUESTS

Recall — your gut answer first

+5 XP warm-up

Six people line up for a photo. Without restrictions, $6! = 720$ arrangements exist. Alice and Bob insist on standing next to each other (in either order). Without calculating — what fraction of 720 do you think satisfy this condition?

auto-saved

The three strategies

+5 XP to read

Every restriction problem uses one of three moves. Identify the restriction type, pick the matching strategy, execute.

Fix — a specific object must occupy a specific position. Fix it there, arrange the rest normally.
Glue — objects must be together. Glue them into a single unit, arrange units, then multiply by internal arrangements.
Gap — objects must not be together. Arrange the other objects first, then place the restricted objects in the gaps.

Fix · Glue · Gap

Fix: the simplest restriction

Pin one (or more) objects to their required positions, then apply $^nP_r$ or $n!$ to the remaining objects and positions.

Glue: count internal order

When $k$ objects are glued together, the unit can be internally arranged in $k!$ ways. Always multiply by $k!$ at the end.

Gap: count the spaces

$n$ objects in a line create $n+1$ gaps (including both ends). Place restricted objects into chosen gaps using $^{n+1}P_k$.

What you'll master

Know

Key facts

$n$ objects in a row create $n+1$ gaps (including both ends)
Glue method: treat grouped objects as one unit, then multiply by $k!$ internal arrangements
"Not together" methods: gap method or (total) $-$ (together)

Understand

Concepts

Why fixing an object reduces the problem size by one
Why the glue method gives $k!$ times more arrangements than forgetting the internal order
When to use the complement approach versus the direct gap method

Can do

Skills

Solve "must be together" problems using the glue method
Solve "must not be together" problems using the gap or complement method
Apply fixed-position restrictions to reduce arrangement counts

Key terms

Fixed positionA constraint requiring a specific object to occupy a specific slot. Fix it there and arrange the remaining objects.

Glue methodTreat objects that must be together as a single unit. Arrange units, then multiply by the internal arrangements of the glued objects.

Gap methodArrange unrestricted objects first ($n$ of them), creating $n+1$ gaps. Place restricted objects in chosen gaps.

Complement methodTotal arrangements minus the number of arrangements where the restriction is violated. Useful when "not together" is complex.

UnitThe combined block formed when objects are glued. If 3 objects are glued, they form 1 unit that can be arranged internally in $3!$ ways.

GapA space between or beside objects in a row. $n$ objects create $n+1$ gaps: one before the first, one after each object.

Types of restrictions and strategies

core concept

When a permutation problem includes conditions on where objects can appear, we apply one of the three strategies: Fix, Glue, or Gap.

Strategy 1 — Fix (object in a specific position):

Place the required object in its fixed position first.
Arrange the remaining $n-1$ objects in the remaining $n-1$ positions: $(n-1)!$ ways.
Example: A must be first in a row of 5 $\Rightarrow$ fix A, arrange B,C,D,E in the remaining 4 spots: $4! = 24$ ways.

Strategy 2 — Glue (objects must be adjacent):

Treat the $k$ objects that must be together as a single unit.
Count the arrangements of the units: $(n-k+1)!$ ways if all $n$ objects are being arranged.
Multiply by the internal arrangements of the glued group: $k!$ ways.
Total = $(n-k+1)! \times k!$

Strategy 3 — Gap (objects must not be adjacent):

Arrange the unrestricted objects first: say $m$ objects gives $m!$ arrangements.
This creates $m+1$ gaps (including both ends): _ O _ O _ O _ ...
Choose gaps for the restricted objects: $^{m+1}P_k$ ways (if $k$ restricted objects are all distinct).
Total = $m! \times {^{m+1}P_k}$

Check your hook answer. Alice and Bob must be together in a group of 6. Glue A+B into one unit: now 5 units, $5! = 120$ arrangements. A and B can swap within the unit: $2! = 2$. Total = $120 \times 2 = 240$. That is $\dfrac{240}{720} = \dfrac{1}{3}$ of all arrangements.

Fix: pin object to position, arrange rest — reduces problem by 1 object and 1 slot; Glue: k objects together treat as 1 unit, multiply by k! internal arrangements

Pause — copy the restriction strategies into your book: Fix — pin object to a position, arrange the rest; Glue — treat $k$ adjacent objects as one unit ($k!$ internal arrangements), then arrange the combined group.

Quick check: Three books (A, B, C) must be kept together on a shelf of 6 books. How many units must be arranged after gluing?

The gap method in detail

core concept

We just saw that "must be adjacent" problems use the glue technique (treat glued objects as one unit, then multiply by internal arrangements) and "must be in specific positions" uses the fix technique. That raises a question: what about "must NOT be adjacent" — gluing doesn't work here, so what does? This card answers it → the gap method: arrange the free objects first, then count the gaps they create, and place the restricted objects into those gaps.

The gap method is the most powerful technique for "not adjacent" problems. The key insight is to arrange the unconstrained objects first, then count ways to insert the constrained objects into the gaps.

Five boys create six gaps. Choosing four of these six gaps for girls guarantees no two girls are adjacent.

Key formula: if $m$ unrestricted objects are arranged first and $k$ restricted objects must not be adjacent (and $k \leq m+1$):

$$\text{Total} = m! \times {^{m+1}P_k}$$

Gap method steps: (1) arrange m free objects in m! ways; (2) count m+1 gaps; (3) place k restricted objects: ^{m+1}P_k; Total = m! {^{m+1}P_k}

Pause — copy the gap method steps into your book: (1) arrange $m$ free objects: $m!$ ways; (2) count $m+1$ gaps; (3) place $k$ restricted objects in gaps: $^{m+1}P_k$; total $= m! \times\ ^{m+1}P_k$.

Did you get this? True or false: four objects placed in a row create five gaps (including both ends).

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · GLUE METHOD

In how many ways can 6 people be arranged in a row if two particular people (Alice and Bob) must sit together?

Glue Alice and Bob into a single unit [AB]. Now arrange 5 units: $5! = 120$

The 6 people become 5 units: {AB}, C, D, E, F. They can be arranged in any order.

PROBLEM 2 · FIXED POSITION

How many arrangements of the letters A, B, C, D, E have A at the beginning?

Fix A in position 1. The remaining positions are 2, 3, 4, 5.

A is pinned. The restriction is fully satisfied — no need to count arrangements where A is elsewhere.

PROBLEM 3 · GAP METHOD

In how many ways can 5 boys and 4 girls be arranged in a row so that no two girls are adjacent?

Arrange 5 boys in a row: $5! = 120$ ways. This creates $5+1=6$ gaps.

_ B _ B _ B _ B _ B _ — one gap before, one after, one between each adjacent pair.

Fill the gap: If 4 boys are arranged first, they create gaps for girls to occupy.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Forgetting internal arrangements in the glue method

When you glue $k$ objects together, the unit can be internally arranged in $k!$ ways. Students who write $(n-k+1)!$ and stop lose marks every time. Always multiply by $k!$ to account for the order within the glued group.

Trap 02

Miscounting gaps

$n$ objects create $n+1$ gaps, not $n-1$. The end positions (before the first object and after the last) are valid gap positions. Forgetting the ends underestimates the count. Always draw the diagram: _ O _ O _ O _ to count gaps explicitly.

Trap 03

Using subtraction incorrectly for "not together"

The complement (total $-$ together) works when only one pair must be separated. For larger groups, the complement gets messy — use the gap method directly instead. But if you use the complement, remember to multiply the "together" count by $k!$ for the internal arrangements of the glued group.

Did you get this? True or false: for a group of 6 arranged in a row with 2 specific people not together, the answer equals $6! - 5! \times 2!$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

In how many ways can 4 boys and 3 girls be arranged in a row if the 3 girls must sit together?

How many arrangements of the letters of the word LEADER have the two E's together?

In how many ways can 6 people be arranged in a row if two particular people must not sit together? (Use the complement method.)

Seven people line up. Two specific people (X and Y) must not be next to each other. Use the gap method to count valid arrangements.

A 5-digit number is formed using the digits 1–5 (each once). How many such numbers have 1 and 2 in the first two positions (in either order)?

Odd one out: Three of these statements about restricted permutations are correct. Which one is NOT?

Revisit your thinking

Earlier you estimated what fraction of 720 arrangements have Alice and Bob together.

The answer: glue A and B $\Rightarrow$ 5 units, $5! = 120$ arrangements, times $2! = 2$ internal orders $= 240$. That is $\dfrac{240}{720} = \dfrac{1}{3}$ of all arrangements. The intuition many people have is "about half" — the true fraction is smaller because there are only $\dfrac{2}{6} = \dfrac{1}{3}$ of the slots where A and B can be placed together.

auto-saved

Multiple choice

+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 42 marks

Q1. In how many ways can 4 boys and 3 girls be arranged in a row if the girls must sit together? (2 marks)

auto-saved

ApplyBand 42 marks

Q2. How many arrangements of the letters of the word LEADER have the two E's together? (2 marks)

auto-saved

AnalyseBand 52 marks

Q3. In how many ways can 6 people be arranged in a row if two particular people must not sit together? (2 marks)

auto-saved

Comprehensive answers (click to reveal)

Activity answers:

1. Glue 3 girls as 1 unit. Units: {GGG}, B, B, B, B = 5 units. $5! \times 3! = 120 \times 6 = 720$.

2. LEADER: 6 letters. Glue EE: 5 units {EE}, L, A, D, R. Arrangements = $5! \times 1 = 120$ (EE unit has only 1 internal order since the two E's are identical). So answer = $120$. (If treating the two E's as distinguishable: $5! \times 2! = 240$.)

3. Total $6! = 720$. Together: glue 2 people $\Rightarrow$ $5! \times 2! = 240$. Not together $= 720 - 240 = 480$.

4. Gap method: arrange 5 others in $5! = 120$ ways, creating 6 gaps. Place X and Y: $^6P_2 = 30$. Total $= 120 \times 30 = 3600$.

5. Place 1 and 2 in positions 1–2: $2! = 2$ orders. Arrange 3, 4, 5 in positions 3–5: $3! = 6$. Total $= 2 \times 6 = 12$.

Q1 (2 marks): Glue 3 girls as 1 unit: 5 units, $5! = 120$ [1]. Internal arrangements of girls: $3! = 6$ [1]. Total $= 120 \times 6 = \mathbf{720}$.

Q2 (2 marks): Glue EE: 5 units (L, EE, A, D, R). $5! = 120$ unit arrangements [1]. The two E's are identical so the EE unit has only 1 internal order [1]. Total $= 120$. (Full marks also awarded for $5! \times 2! = 240$ if treating E's as distinguishable — accept with justification.)

Q3 (2 marks): Total $= 6! = 720$ [0.5]. Together $= 5! \times 2! = 240$ [1]. Not together $= 720 - 240 = \mathbf{480}$ [0.5].

Boss battle · The Seating Enforcer

earn bronze · silver · gold

Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering restricted permutation questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

← Lesson 3 · Permutations (nPr) Lesson 5 · Circular Permutations →

Module overview · Extension 1 · Checkpoint 1