Permutations and Linear Arrangements

13 questions

Four distinct books—two mathematics books and two history books—are to be arranged in a single line on a shelf. If the two mathematics books must stand next to each other, in how many different linear arrangements can the four books be placed?

Show answer & explanation

Answer: 12

Answer

12 linear arrangements

To find the total number of linear arrangements where the two mathematics books are adjacent, treat the two mathematics books as a single combined block. This leaves 3 items to arrange (the mathematics block and the 2 individual history books), which can be arranged in

3! = 6

ways. Within the block, the 2 mathematics books can be ordered in

2! = 2

ways. Multiplying the independent arrangements yields

6 \times 2 = 12

total arrangements.

Step-by-Step Solution

Group the adjacent items into a single unit and count total units to arrange

3 units (1 grouped math unit + 2 individual history books), which can be arranged in

3! = 6

ways

Treating items that must be adjacent as a single block simplifies the linear arrangement into distinct available positions.

Determine internal permutations within the grouped unit

2! = 2

ways to order the two mathematics books inside their block

The mathematics books can switch positions with each other within their designated block.

Multiply external and internal arrangements

6 \times 2 = 12

total linear arrangements

By the Fundamental Counting Principle, total arrangements equal the product of unit arrangements and internal arrangements.

Key Concept

Linear Permutations with Adjacent Restrictions (Tie-Together Method)

Question 2Question →

A panel of 6 distinct experts—3 scientists, 2 economists, and 1 moderator—are to sit in a single row of 6 chairs for a discussion. If the 3 scientists must all sit in adjacent chairs and the 2 economists cannot sit in adjacent chairs, how many different seating arrangements are possible?

108

144

Show answer & explanation

Answer: 72

Answer

72 seating arrangements

To find the number of valid seating arrangements, we first group the 3 scientists together as 1 unit, leaving us with 2 non-economist entities (the scientist block and the moderator). These 2 entities can be arranged in 2! = 2 ways. Placing the 2 distinct economists into the 3 available gaps around these entities ensures they are not adjacent, yielding P(3, 2) = 6 choices. Finally, multiplying by the 3! = 6 internal arrangements of the scientists yields a total of 2 * 6 * 6 = 72 valid arrangements.

Step-by-Step Solution

Group the 3 scientists into a single block

S

, and treat the moderator

M

as an individual unit.

There are 2 non-economist units: block

S

and moderator

M

Grouping elements that must be adjacent allows us to treat them temporarily as a single entity.

Calculate the arrangements of the non-economist units.

The 2 non-economist units can be arranged in

2! = 2

ways.

Linear arrangement of 2 distinct entities.

Insert the 2 economists into the available gaps created by the non-economist units.

For any arrangement of

S

and

M

(e.g., _

S

M

_), there are 3 available gaps. The 2 distinct economists can be placed in these gaps in

P(3,2) = 3 \times 2 = 6

ways.

To ensure no two economists sit together, each economist must occupy a separate gap.

Account for the internal arrangements of the 3 scientists within block

S

The 3 distinct scientists can be arranged among themselves in

3! = 6

ways.

Order matters among distinct individuals within a grouped block.

Apply the fundamental counting principle to compute total arrangements.

2 \times 6 \times 6 = 72

total arrangements.

Multiply the independent choices made in steps 2, 3, and 4.

Key Concept

Permutations with Adjacency and Non-Adjacency Restrictions

Question 3Question →

A chef is arranging 6 distinct dishes—3 appetizers and 3 main courses—in a single straight row on a serving table for a culinary presentation. If all 3 appetizers must be placed adjacent to one another and all 3 main courses must also be placed adjacent to one another, in how many different linear arrangements can the dishes be displayed?

144

720

Show answer & explanation

Answer: 72

Answer

To find the number of valid linear arrangements, treat the 3 appetizers as a single group and the 3 main courses as another single group. There are

2! = 2

ways to arrange these two groups on the table (appetizers first or main courses first). Within the appetizer group, the 3 distinct dishes can be arranged in

3! = 6

ways. Within the main course group, the 3 distinct dishes can also be arranged in

3! = 6

ways. By the Fundamental Counting Principle, the total number of linear arrangements is

2! \times 3! \times 3! = 2 \times 6 \times 6 = 72

Step-by-Step Solution

Group the dishes into distinct blocks based on the given adjacency restriction.

Create Block A (containing all 3 appetizers) and Block M (containing all 3 main courses).

Because items of the same category must remain adjacent, each category functions as a single compound element.

Calculate the number of ways to arrange the 2 blocks relative to each other.

Number of block arrangements =

2! = 2

The two blocks can be placed in order (Block A then Block M) or (Block M then Block A).

Calculate the internal permutations within each block.

Internal arrangements for appetizers =

3! = 6

, and internal arrangements for main courses =

3! = 6

Each category consists of 3 distinct dishes that can be permuted among themselves.

Apply the Fundamental Counting Principle to find total linear arrangements.

Total arrangements =

2! \times 3! \times 3! = 2 \times 6 \times 6 = 72

The decision steps are independent, so their possible outcomes are multiplied.

Key Concept

Permutations with Block Restrictions (Grouping Method)

Practice More

Try solving a problem where only one specific subset of items must remain together while the rest can be arranged freely.

Alternative Method

Alternatively, place the first appetizer in slot 1 (3 choices). The remaining appetizers must take slots 2 and 3 (2 choices and 1 choice). Then slots 4, 5, 6 must be filled by main courses (3 choices, 2 choices, 1 choice). This gives

3 \times 2 \times 1 \times 3 \times 2 \times 1 = 36

ways when appetizers are placed first. Symmetrically, placing main courses in slots 1 to 3 gives another 36 ways, totaling

36 + 36 = 72

ways.

Estimated Time:1m 30s

Question 4Question →

A museum curator is arranging $5$ distinct marble statues and $2$ distinct bronze statues in a single row along a gallery wall. If the $2$ bronze statues must not be placed next to each other, how many different linear arrangements of all $7$ statues are possible?

Show answer & explanation

Answer: 3600

Answer

3,600

Using complementary counting, the total unrestricted arrangements of

7

distinct statues is

7! = 5,040

. The number of arrangements where the

2

bronze statues are placed together is determined by treating them as a single block:

6! \times 2! = 1,440

. Subtracting these forbidden arrangements from the total yields

5,040 - 1,440 = 3,600

valid linear arrangements.

Step-by-Step Solution

Find total arrangements without restriction.

7! = 5,040

There are 7 distinct statues in total to arrange in a line.

Find arrangements where the 2 bronze statues are adjacent.

6! × 2! = 1,440

Grouping the 2 bronze statues into 1 block yields 6 items to order (6!), and the 2 bronze statues can swap positions inside the block (2!).

Apply complementary counting to find non-adjacent arrangements.

5,040 - 1,440 = 3,600

Subtracting the adjacent arrangements from total arrangements gives all valid arrangements.

Key Concept

Linear arrangements with non-adjacency restrictions using complementary counting.

Estimated Time:1m 30s

Question 5Question →

Five speakers—Alice, Bob, Carol, David, and Eva—are scheduled to give consecutive presentations at a conference. If Alice must present first, in how many different linear orders can all five speakers present?

120

Show answer & explanation

Answer: 24

Answer

The total number of different presentation orders is 24.

Since Alice's position is fixed as the first speaker, there is only

1

way to fill the first slot. The remaining

4

speakers can be arranged in the remaining

4

slots in

4! = 4 \times 3 \times 2 \times 1 = 24

ways. Thus, the total number of linear arrangements is

1 \times 24 = 24

Step-by-Step Solution

Identify the fixed element and available positions.

Alice must fill the 1st position, leaving 1 choice for slot 1.

The question specifies that Alice must present first.

Calculate the number of ways to arrange the remaining speakers.

The remaining 4 speakers (Bob, Carol, David, Eva) can be arranged in

4! = 4 \times 3 \times 2 \times 1 = 24

ways.

Linear permutations of

n

distinct objects use

n!

Apply the Fundamental Counting Principle.

Total arrangements =

1 \times 24 = 24

Combine the independent choices for each position.

Key Concept

Permutations with Fixed Positions

Question 6Question →

Eight executive team members—3 vice presidents, 3 directors, and 2 managers—are to stand in a single line for a company photograph. If no two vice presidents can stand next to each other, and the 2 managers must stand next to each other, how many different linear arrangements of the eight team members are possible?

Show answer & explanation

Answer: 2880

Answer

2,880

To satisfy both constraints simultaneously, first treat the 2 managers as a single block, which has

2! = 2

internal arrangements. Next, arrange the non-vice-president units—consisting of the 3 distinct directors and the 1 manager block—giving 4 units total, which can be ordered in

4! = 24

ways. These 4 units create 5 distinct slot locations (at the ends and between adjacent units). To ensure no two vice presidents are adjacent, place the 3 distinct vice presidents into 3 of these 5 slots in

P(5, 3) = 60

ways. Multiplying these independent decisions yields

2 \times 24 \times 60 = 2,880

Step-by-Step Solution

Group the elements that must remain adjacent and calculate internal permutations

The 2 managers form 1 block with

2! = 2

internal arrangements.

Since the 2 managers must stand next to each other, treating them as a single block ensures they are never separated.

Arrange all non-restricted base units in a line

The 3 directors and 1 manager block yield

4! = 24

linear arrangements.

Establishing the sequence of non-vice-president units creates the fixed slots into which the vice presidents will later be inserted.

Calculate the available slot arrangements for the separated elements

4 base units create 5 available slots. Permuting 3 vice presidents into 5 slots yields

P(5, 3) = 60

ways.

Placing at most one vice president per slot guarantees that no two vice presidents are placed adjacently.

Apply the Fundamental Counting Principle to determine total arrangements

Total arrangements =

2 \times 24 \times 60 = 2,880

The choices for internal block arrangement, base unit ordering, and slot placement are independent sequential events.

Key Concept

Linear Permutations with Simultaneous Grouping and Non-Adjacency Constraints

Question 7Question →

Seven distinct paintings—3 landscapes and 4 portraits, one of which is a portrait of the founder—are to be displayed side-by-side in a single row along a gallery wall. If all 3 landscapes must be displayed together as a contiguous block, and the portrait of the founder cannot be placed at either end of the 7-painting row, in how many different linear arrangements can the 7 paintings be displayed?

288

432

576

720

Show answer & explanation

Answer: 432

Answer

432

The correct answer is 432. Treating the 3 landscape paintings as a single block leaves 5 items (the landscape block, the founder's portrait, and 3 other portraits) to arrange. There are 5! = 120 total arrangements of these 5 items. The founder's portrait occupies an end slot of the row if it is placed in either the 1st position (4! = 24 ways) or the 5th position (4! = 24 ways) among the 5 items. Subtracting these 48 invalid arrangements gives 120 - 48 = 72 valid block placements. Finally, accounting for the 3! = 6 internal arrangements of the landscapes inside their block yields 72 × 6 = 432 total linear arrangements.

Step-by-Step Solution

Group the 3 landscape paintings into a single block unit.

We now have 5 items to arrange linearly: the 1 landscape block, the founder's portrait, and the 3 other portraits.

Grouping elements that must remain together simplifies the arrangement problem into smaller independent choices.

Calculate the total number of linear arrangements of these 5 items without restrictions on the founder's portrait.

5! = 120 arrangements.

5 distinct objects can be ordered in 5 factorial ways.

Determine the number of arrangements where the founder's portrait is placed at either end of the row.

2 × 4! = 48 arrangements.

The founder's portrait is at an end of the 7-painting row if and only if it is in position 1 (first item) or position 5 (last item) among the 5 items. Fixing it at position 1 leaves 4! = 24 ways for the remaining items, and fixing it at position 5 gives another 4! = 24 ways.

Subtract the invalid end-position arrangements from the total 5-item arrangements.

120 - 48 = 72 valid arrangements of the 5 items.

Using complementary counting isolates the cases where the founder's portrait is not at either end.

Multiply by the number of internal arrangements of the 3 landscape paintings within their block.

72 × 3! = 72 × 6 = 432.

The 3 distinct landscapes within the single block can be ordered internally in 3! = 6 ways for each overall arrangement.

Key Concept

Permutations with Block Constraints and Complementary Restriction Counting

Question 8Question →

Six distinct letters— $A, B, C, D, E,$ and $F$ —are to be arranged in a single line. How many different linear arrangements are possible such that letter $A$ appears somewhere to the left of letter $B$ , and letters $C$ and $D$ are not adjacent to each other?

Show answer & explanation

Answer: 240

Answer

240

To find the number of valid linear arrangements, apply symmetry and the complement rule. First, in half of all

6! = 720

arrangements (360 arrangements), letter A appears to the left of letter B. Next, find the number of arrangements where A is to the left of B AND letters C and D are adjacent. Treating C and D as one combined block gives

5! \times 2! = 240

total arrangements where C and D are adjacent. By symmetry, letter A appears to the left of letter B in half of these cases (

\frac{240}{2} = 120

). Subtracting these 120 restricted arrangements from the 360 total arrangements where A precedes B gives

360 - 120 = 240

Step-by-Step Solution

Determine the number of linear arrangements in which letter A appears somewhere to the left of letter B.

360 arrangements

For 6 distinct letters, there are

6! = 720

total linear arrangements. By symmetry, letter A appears before letter B in exactly half of all arrangements:

\frac{720}{2} = 360

Determine the number of arrangements where letter A is to the left of letter B AND letters C and D are adjacent.

120 arrangements

Treating C and D as a single block yields 5 items to arrange, which can be done in

5! = 120

ways. The block itself has

2! = 2

internal orderings, giving

120 \times 2 = 240

arrangements where C and D are adjacent. By symmetry, letter A appears before letter B in half of these arrangements:

\frac{240}{2} = 120

Subtract the arrangements where C and D are adjacent from the total arrangements where A is to the left of B.

240 arrangements

The number of arrangements where letter A is to the left of letter B and letters C and D are not adjacent is

360 - 120 = 240

Key Concept

Permutations with Relative Position and Non-Adjacency Restrictions

Question 9Question →

A security system generates unique 7-digit access codes using each of the digits $1, 2, 3, 4, 5, 6,$ and $7$ exactly once. How many such codes can be formed in which all odd digits appear in strictly ascending order from left to right and the digit $2$ appears somewhere to the left of the digit $4$ ?

Show answer & explanation

Answer: 105

Answer

The total number of valid 7-digit security access codes is 105.

Out of the total

7! = 5,040

unrestricted linear arrangements of the 7 distinct digits, the 4 odd digits can be ordered in

4! = 24

ways, but only 1 of these orderings satisfies the strictly ascending condition. Furthermore, the digits 2 and 4 can be ordered in

2! = 2

ways, with digit 2 appearing to the left of digit 4 in exactly 1 of those orderings. Therefore, the number of valid arrangements is given by

\frac{7!}{4! \times 2!} = \frac{5,040}{24 \times 2} = \frac{5,040}{48} = 105

Step-by-Step Solution

Calculate slot selections for the odd digits

35 ways

From 7 positions, selecting 4 positions for the odd digits {1, 3, 5, 7} can be done in

\binom{7}{4} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35

ways. Since the odd digits must be ordered in strictly ascending order (1, then 3, then 5, then 7), there is only 1 valid relative order for any selection of 4 slots.

Determine valid arrangements for the even digits

3 ways

The remaining 3 slots are occupied by the even digits {2, 4, 6}. The total number of linear arrangements of 3 distinct digits is

3! = 6

. By symmetry, in exactly half of these arrangements, digit 2 appears to the left of digit 4, yielding

\frac{6}{2} = 3

valid ways.

Multiply independent choices to find the total arrangements

105 codes

Applying the Fundamental Counting Principle,

35 \text{ (slot choices for odd digits)} \times 1 \text{ (ordering of odd digits)} \times 3 \text{ (arrangements of even digits)} = 105

Key Concept

Permutations with Relative Order Restrictions

Estimated Time:2m 0s

Question 10Question →

A jewelry store window displays $6$ distinct luxury watches— $3$ gold watches and $3$ silver watches—in a single straight line from left to right. If no two gold watches can be placed adjacent to each other, how many different linear arrangements of the $6$ watches are possible?

Show answer & explanation

Answer: 144

Answer

144

To place items such that no two gold watches are adjacent, we use the gap method. First, arrange the 3 distinct silver watches, which can be done in

3! = 6

ways. Placing these 3 silver watches creates 4 potential spaces (one at each end and two between the silver watches). Next, select 3 of these 4 spaces to place the gold watches, which can be done in

\binom{4}{3} = 4

ways. Finally, arrange the 3 distinct gold watches within the selected spaces in

3! = 6

ways. Multiplying these possibilities gives

6 \times 4 \times 6 = 144

total distinct arrangements.

Step-by-Step Solution

Calculate the arrangements of the 3 distinct silver watches.

3! = 6 ways

The 3 silver watches are distinct and can be arranged among themselves in 3! ways.

Determine the available positions (gaps) for the gold watches.

4 available gaps

Placing 3 silver watches in a line creates 4 potential spaces (before the first, between adjacent pairs, and after the last) to ensure non-adjacency.

Select positions and arrange the 3 distinct gold watches.

C(4, 3) × 3! = 4 × 6 = 24 ways

Choosing 3 out of 4 gaps gives C(4, 3) = 4 combinations, and ordering the 3 distinct gold watches in those selected gaps gives 3! = 6 arrangements.

Multiply the possibilities from all independent steps using the Fundamental Counting Principle.

6 × 24 = 144

The total number of valid linear arrangements is the product of the number of ways to complete each step.

Key Concept

Linear Permutations with Non-Adjacent Restrictions (Gap Method)

Estimated Time:2m 0s

Question 11Question →

A corporate board of 7 members—consisting of 4 senior executives (including the CEO and the COO) and 3 junior associates—is to be seated in a single row of 7 chairs for a press conference. If the 3 junior associates must sit in adjacent chairs, and the CEO and the COO cannot sit next to each other, in how many different linear arrangements can the 7 members be seated?

288

432

576

720

1,440

Show answer & explanation

Answer: 432

Answer

432

To satisfy the condition that the 3 junior associates sit together, we treat them as 1 block with

3! = 6

internal orderings. Combining this block with the 4 senior executives yields 5 entities, which can be arranged in

5! = 120

ways, giving

120 \times 6 = 720

total arrangements with the juniors seated together. To enforce that the CEO and COO cannot sit together, we subtract the arrangements where they do sit together: treating the CEO and COO as a block gives 4 entities to arrange (

4! = 24

), with

2! = 2

ways to arrange CEO and COO, and

3! = 6

ways for the junior block, totaling

24 \times 2 \times 6 = 288

restricted cases. Subtracting 288 from 720 gives 432 valid arrangements.

Step-by-Step Solution

Group the 3 junior associates into a single block.

The 3 junior associates can be arranged internally within their block in

3! = 6

ways. Treating this block as 1 single element along with the 4 senior executives gives a total of 5 items to arrange.

The condition specifies that all 3 junior associates must sit in adjacent chairs.

Calculate total arrangements where junior associates sit together without CEO/COO restrictions.

The 5 items (1 block + 4 senior executives) can be arranged in

5! = 120

ways. Including internal block arrangements gives

120 \times 6 = 720

ways.

Applying the Fundamental Counting Principle to the 5 units and the 3 internal positions.

Calculate unwanted arrangements where the CEO and COO sit next to each other (with junior associates together).

Group the CEO and COO into a second block with

2! = 2

internal arrangements. Now there are 4 items to arrange (Junior block, CEO-COO block, and 2 other senior executives). Total unwanted arrangements:

4! \times 3! \times 2! = 24 \times 6 \times 2 = 288

ways.

To find valid non-adjacent arrangements, subtract adjacent CEO-COO arrangements from total junior-grouped arrangements.

Subtract the unwanted arrangements from the total grouped arrangements.

720 - 288 = 432

valid seating arrangements.

Complementary counting provides a direct solution.

Key Concept

Linear Permutations with Grouping and Complementary Non-Adjacency Constraints

Question 12Question →

A secure access code is formed by arranging all $6$ of the distinct characters $A, E, B, C, D,$ and $F$ in a single line. If the two vowels ( $A$ and $E$ ) cannot be adjacent to each other, and the code cannot begin with the consonant $B$ , how many different valid access codes can be formed?

240

252

360

408

480

Show answer & explanation

Answer: 408

Answer

408 valid access codes can be formed.

To find the number of valid arrangements satisfying both conditions, we use complementary counting. First, the total number of arrangements of 6 distinct characters is 6! = 720. Next, we determine the number of invalid arrangements. Let V be the set of arrangements where A and E are adjacent: treating (AE) as a single unit gives 5! × 2! = 240 ways. Let B_first be the set of arrangements starting with B: fixing B in the first spot leaves 5! = 120 ways. The intersection of these two invalid sets (where A and E are adjacent AND the code starts with B) places B in position 1 and arranges the remaining 4 items including the (AE) block in 4! × 2! = 48 ways. By inclusion-exclusion, the total invalid arrangements equal 240 + 120 - 48 = 312. Subtracting invalid arrangements from the total gives 720 - 312 = 408 valid access codes.

Step-by-Step Solution

Calculate the total total number of unrestricted linear arrangements of the 6 distinct characters.

6! = 720 total arrangements.

6 distinct characters can be ordered in 6! different ways.

Calculate the number of arrangements where the two vowels (A and E) ARE adjacent (Set V).

|V| = 240 arrangements.

Treating (AE) as 1 block leaves 5 items to arrange in 5! = 120 ways. Since (AE) has 2! = 2 internal arrangements, |V| = 2 * 120 = 240.

Calculate the number of arrangements that BEGIN with the consonant B (Set B).

|B| = 120 arrangements.

Fixing B in the first position leaves 5 remaining positions to be filled by the remaining 5 characters in 5! = 120 ways.

Calculate the overlap: arrangements where vowels ARE adjacent AND the code BEGINS with B (Set V ∩ B).

|V ∩ B| = 48 arrangements.

With B fixed in position 1, the remaining 5 positions are filled by 4 blocks: (AE), C, D, F. These 4 blocks can be arranged in 4! = 24 ways, and (AE) has 2! = 2 internal orderings, giving 24 * 2 = 48.

Apply the Principle of Inclusion-Exclusion to find the total invalid arrangements, and subtract from the total.

720 - (240 + 120 - 48) = 720 - 312 = 408 valid arrangements.

The number of invalid arrangements is |V ∪ B| = |V| + |B| - |V ∩ B| = 312. Subtracting invalid codes from total codes yields 408.

Key Concept

Linear permutations with multiple restrictions using complementary counting and inclusion-exclusion.

Estimated Time:2m 0s

Question 13Question →

An event organizer is scheduling a conference featuring $7$ distinct speakers: $3$ technology experts, $3$ finance experts, and $1$ keynote presenter. The $7$ speakers will give back-to-back presentations on a single main stage. If all $3$ technology experts must speak consecutively in any order, and no two finance experts may speak consecutively, how many different presentation schedules are possible?

Show answer & explanation

Answer: 72

Answer

The total number of possible presentation schedules satisfying all conditions is 72.

To find the number of valid schedules, we combine the block method and the gap method. First, bundle the 3 technology experts into a single block. This block can be ordered internally in

3! = 6

ways. Next, arrange the non-finance items (the technology block and the keynote presenter), which can be ordered in

2! = 2

ways. These 2 items create 3 available gaps (one at each end and one in the middle) for the 3 finance experts. To ensure no two finance experts are consecutive, place exactly one finance expert into each of the 3 gaps, which can be done in

3! = 6

ways. Multiplying these independent choices yields

6 \times 2 \times 6 = 72

total valid schedules.

Step-by-Step Solution

Group the 3 technology experts into a single unit and calculate internal permutations

3! = 6 internal arrangements for the technology block

The technology experts must appear consecutively, so treating them as a single block ensures they remain together.

Arrange the non-finance elements (the technology block and the keynote presenter)

2! = 2 linear arrangements

There are 2 distinct items that act as separators for the finance experts.

Apply the gap method to place the 3 finance experts into non-adjacent slots

3 available gaps for 3 finance experts gives 3! = 6 ways

Placing at most one finance expert in each gap created by the non-finance items guarantees that no two finance experts are seated or scheduled consecutively.

Calculate total permutations using the Fundamental Counting Principle

6 × 2 × 6 = 72 total schedules

The decisions for internal block order, base item order, and gap assignments are independent.

Key Concept

Permutations with Combined Block and Separation (Gap Method) Restrictions

Permutations and Linear Arrangements

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Common Mistakes

Practice More

Alternative Method

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

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