Permutations Practice Questions — JAMB UTME

In how many different ways can a chairperson and a secretary be selected from a committee of $5$ members?

10

20

25

120

Show answer & explanation

Answer: $20$

Answer

The number of ways to select a chairperson and a secretary from

5

members is

20

.

Selecting

2

individuals for distinct positions (chairperson and secretary) from

5

candidates is an ordered selection problem. The number of ways is given by the permutation formula

_5P_2 = 5 \times 4 = 20

.

Step-by-Step Solution

1

Identify whether order matters

Order matters because the roles of chairperson and secretary are distinct.

When distinct roles are assigned, the arrangement is a permutation rather than a combination.

2

Apply the permutation formula

_nP_r = \frac{n!}{(n-r)!}

for

n=5

and

r=2

_5P_2 = \frac{5!}{(5-2)!} = \frac{5!}{3!} = 5 \times 4 = 20

There are

5

choices for chairperson and

4

remaining choices for secretary.

Key Concept

Permutation of

n

distinct items taken

r

at a time

Estimated Time:45s

Question 2Question →

In how many different ways can the letters of the word SUCCESS be arranged such that the three 'S's do not all come together?

360

420

60

35

Show answer & explanation

Answer: 360

Answer

The letters of the word SUCCESS can be arranged in 360 ways such that the three 'S's do not all come together.

The correct answer is calculated using complementary counting. First, the total unrestricted permutations of SUCCESS (7 letters with 3 'S's and 2 'C's) is

\frac{7!}{3!2!} = 420

. Next, treating the three 'S's as one single block leaves 5 items to arrange with 2 'C's, giving

\frac{5!}{2!} = 60

ways where the 'S's are together. Subtracting 60 from 420 yields 360.

Step-by-Step Solution

1

Calculate the total number of unrestricted arrangements of the word SUCCESS.

The word SUCCESS has 7 letters in total: 3 'S's, 2 'C's, 1 'U', and 1 'E'. Total arrangements

N_{total} = \frac{7!}{3! \times 2!} = \frac{5040}{6 \times 2} = 420

.

Repeated letters must be accounted for by dividing the factorial of the total count by the factorials of the counts of repeated letters.

2

Calculate the number of arrangements where the three 'S's are all together.

Treat the three 'S's as a single entity (SSS). We now arrange 5 entities: (SSS), U, C, C, E. Since 'C' appears twice,

N_{together} = \frac{5!}{2!} = \frac{120}{2} = 60

.

Grouping restricted identical items into a single block allows us to find the subset of arrangements where they stay together.

3

Subtract the number of 'together' arrangements from the total arrangements.

N_{not\_together} = N_{total} - N_{together} = 420 - 60 = 360

.

The complementary counting principle gives the number of ways where the restriction is satisfied.

Key Concept

Permutations with Repeated Elements and Complementary Counting

Estimated Time:1m 30s

Question 3Question →

Find the positive integer value of $n$ such that $^{n}P_4 = 42 \times {^{n}P_2}$ .

Show answer & explanation

Answer: 9

Answer

The positive integer value of

n

is 9.

Expanding

^{n}P_4

as

n(n-1)(n-2)(n-3)

and

^{n}P_2

as

n(n-1)

allows dividing out

n(n-1)

, leading to

(n-2)(n-3) = 42

. Expanding and factoring gives

n^2 - 5n - 36 = 0

, which yields

n = 9

as the only valid positive integer.

Step-by-Step Solution

1

Apply the permutation formula

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

^{n}P_4 = n(n-1)(n-2)(n-3)

and

^{n}P_2 = n(n-1)

By definition of permutations, selecting

r

items from

n

distinct items without replacement.

2

Substitute the expansions into the given relation

n(n-1)(n-2)(n-3) = 42 n(n-1)

Direct substitution into

^{n}P_4 = 42 \times {^{n}P_2}

.

3

Simplify by dividing out common non-zero terms

(n-2)(n-3) = 42

Since

n \ge 4

,

n(n-1) \neq 0

and can be safely divided from both sides.

4

Form and solve the quadratic equation

n^2 - 5n + 6 = 42 \implies n^2 - 5n - 36 = 0 \implies (n-9)(n+4) = 0

Expanding terms and factoring the resulting quadratic expression.

5

Determine the valid root

n = 9

Permutation total items

n

must satisfy

n \ge r \ge 0

, rejecting the negative root

n = -4

.

Key Concept

Algebraic equations involving permutations

Estimated Time:1m 15s

Question 4Question →

Six members of a board of directors are to be seated around a circular conference table. If two specific members refuse to sit next to each other, how many different seating arrangements are possible?

72

48

120

240

Show answer & explanation

Answer: 72

Answer

72

To find the number of circular seating arrangements where two specific members do not sit next to each other, use the complement rule. First, the total unrestricted circular arrangements of 6 members is

(6 - 1)! = 5! = 120

. Next, calculate the arrangements where the two members sit together by treating them as 1 unit (giving 5 units in total). The circular arrangements of these 5 units is

(5 - 1)! = 4! = 24

, and the 2 members can swap seats in

2! = 2

ways, yielding

24 \times 2 = 48

arrangements together. Subtracting this from the total yields

120 - 48 = 72

.

Step-by-Step Solution

1

Calculate total circular arrangements without restrictions

Total arrangements =

(6 - 1)! = 5! = 120

The number of ways to arrange

n

distinct items in a circle is

(n - 1)!

.

2

Calculate arrangements where the two specific members sit together

Restricted arrangements =

(5 - 1)! \times 2! = 4! \times 2 = 24 \times 2 = 48

Treat the two members as a single block, giving 5 units to arrange around a circle in

(5-1)!

ways, and multiply by

2!

for internal ordering of the pair.

3

Subtract the together arrangements from total arrangements

Ways apart =

120 - 48 = 72

Complementary counting gives the number of arrangements where the two members do not sit next to each other.

Key Concept

Circular Permutations with Restrictions

Estimated Time:1m 30s

Question 5Question →

In how many different ways can $5$ boys and $3$ girls be seated in a straight row such that all $3$ girls must sit together?

Show answer & explanation

Answer: 4320

Answer

The total number of ways to seat

5

boys and

3

girls in a row such that all

3

girls sit together is

4320

.

To arrange

5

boys and

3

girls so that the girls are always together, treat the

3

girls as

1

single unit. Combined with the

5

boys, there are

6

units to arrange in a straight line, which can be done in

6! = 720

ways. Within their group, the

3

girls can be arranged in

3! = 6

ways. By the multiplication principle, the total number of seating arrangements is

720 \times 6 = 4320

.

Step-by-Step Solution

1

Group the restricted items into a single block

The

3

girls form

1

unit. Combined with the

5

boys, there are

5 + 1 = 6

units to arrange.

Since all

3

girls must sit together, treating them as a single block ensures they are not separated.

2

Calculate the linear arrangements of the combined units

The

6

units can be arranged in

6! = 720

ways.

The number of distinct ways to arrange

n

items in a line is

n!

.

3

Calculate internal arrangements of the girls' block

The

3

girls can be arranged among themselves in

3! = 6

ways.

The

3

girls inside the block are distinct individuals and can swap positions.

4

Apply the fundamental counting principle

Total arrangements =

6! \times 3! = 720 \times 6 = 4320

.

The total number of arrangements is the product of external block arrangements and internal block arrangements.

Key Concept

Permutations with grouping constraints (string method)

Question 6Question →

Calculate the total number of distinct four-digit numbers that can be formed using the digits $1, 2, 3, 4, 5, 6,$ and $7$ without repetition, such that the resulting number is divisible by either $4$ or $5$ .

Show answer & explanation

Answer: 320

Answer

The total number of such four-digit numbers is

320

.

The total number of valid four-digit numbers is found by adding the count of numbers divisible by

5

(

120

) to the count of numbers divisible by

4

(

200

). Since a number ending in

5

is odd, it cannot be divisible by

4

, making the two conditions mutually exclusive. Thus, the total count is

120 + 200 = 320

.

Step-by-Step Solution

1

Calculate the number of four-digit numbers divisible by

5

.

For a number to be divisible by

5

, its units digit must be

5

(since

0

is not available). There is

1

choice for the units digit. The remaining

3

positions are filled from the remaining

6

available digits in

^{6}P_{3} = 6 \times 5 \times 4 = 120

ways.

Divisibility by

5

requires the last digit to be

5

.

2

Calculate the number of four-digit numbers divisible by

4

.

A number is divisible by

4

if its last two digits form a multiple of

4

. Using distinct digits from

\{1, 2, 3, 4, 5, 6, 7\}

, the valid two-digit endings are

12, 16, 24, 32, 36, 52, 56, 64, 72,

and

76

(

10

valid pairs). For each pair, the first two positions are filled from the remaining

5

digits in

^{5}P_{2} = 5 \times 4 = 20

ways. Thus, total ways =

10 \times 20 = 200

.

Divisibility by

4

depends entirely on the last two digits.

3

Check for overlap (numbers divisible by both

4

and

5

).

A number divisible by

5

must end in

5

, which is an odd digit. All multiples of

4

must end in an even digit. Hence, no number is divisible by both

4

and

5

in this set. The overlap is

0

.

The two events are mutually exclusive.

4

Apply the addition principle of counting.

Total =

120 + 200 - 0 = 320

.

Add the counts of the two mutually exclusive sets.

Key Concept

Restricted Permutations and Mutually Exclusive Events

Estimated Time:1m 30s

Question 7Question →

Find the number of distinct arrangements of the letters of the word $\text{PARALLEL}$ such that no two letters $\text{'L'}$ are adjacent.

Show answer & explanation

Answer: 1200

Answer

1200

To ensure no two letters 'L' are adjacent, we use the gap method. First, arrange the 5 non-L letters (P, A, R, A, E). Because 'A' repeats twice, there are 5! / 2! = 60 distinct arrangements. These 5 letters form 6 available gaps (including the ends). Selecting 3 of these 6 gaps to insert the 3 identical 'L's can be done in C(6, 3) = 20 ways. Multiplying these gives 60 × 20 = 1200 valid arrangements.

Step-by-Step Solution

1

Count the frequency of each letter in the word PARALLEL.

The word has 8 letters: 1 P, 2 A's, 1 R, 3 L's, and 1 E.

Recognizing repeated elements is essential for permutations with duplicates.

2

Arrange the non-restricted letters (P, A, R, A, E).

Number of arrangements = 5! / 2! = 60.

The letter 'A' is repeated twice, so we divide 5! by 2!.

3

Calculate the number of available gaps for placing the 3 'L's so that no two are adjacent.

5 arranged letters create 6 gap positions. Choosing 3 gaps gives C(6, 3) = (6 × 5 × 4) / (3 × 2 × 1) = 20 ways.

Placing at most one 'L' per gap guarantees that no two 'L's will be adjacent.

4

Multiply the number of arrangements of non-L letters by the gap choices.

Total arrangements = 60 × 20 = 1200.

By the fundamental counting principle, total arrangements equal the product of independent choices.

Key Concept

Permutations with repeated elements and non-adjacency constraints using the Gap Method

Estimated Time:1m 30s

Question 8Question →

How many distinct five-digit even numbers greater than $50,000$ can be formed using the digits $2, 3, 5, 6, 7,$ and $8$ if no digit may be repeated in any number?

240

288

480

40

Show answer & explanation

Answer: 240

Answer

The total number of distinct five-digit even numbers greater than 50,000 is 240.

The correct total is 240. Because the first digit requirement (greater than 50,000 requires 5, 6, 7, or 8) and the last digit requirement (even requires 2, 6, or 8) share common elements (6 and 8), the problem must be split into two mutually exclusive cases: when the first digit is odd (2 choices for

d_1

, 3 for

d_5

, and

^4P_3 = 24

for the middle digits, giving

2 \times 3 \times 24 = 144

) and when the first digit is even (2 choices for

d_1

, 2 for

d_5

, and

^4P_3 = 24

for the middle digits, giving

2 \times 2 \times 24 = 96

). Adding these yields

144 + 96 = 240

.

Step-by-Step Solution

1

Identify constraints on the positions of a five-digit number

d_1 d_2 d_3 d_4 d_5

.

For the number to be greater than 50,000,

d_1 \in \{5, 6, 7, 8\}

. For the number to be even,

d_5 \in \{2, 6, 8\}

.

The first digit determines the magnitude and the last digit determines parity.

2

Analyze Case 1 where the first digit

d_1

is odd (

5

or

7

).

There are 2 choices for

d_1

. All 3 even digits remain available for

d_5

(3 choices). The remaining 3 middle positions (

d_2, d_3, d_4

) can be arranged from the 4 remaining unused digits in

^4P_3 = 24

ways. Total for Case 1 =

2 \times 3 \times 24 = 144

.

Choosing an odd first digit does not reduce the pool of available even digits for the units position.

3

Analyze Case 2 where the first digit

d_1

is even (

6

or

8

).

There are 2 choices for

d_1

. Since one even digit is used in

d_1

, 2 even digits remain for

d_5

(2 choices). The 3 middle positions are filled from the 4 remaining digits in

^4P_3 = 24

ways. Total for Case 2 =

2 \times 2 \times 24 = 96

.

Selecting an even digit for the ten-thousands place directly reduces the available choices for the units place.

4

Sum the possibilities from both mutually exclusive cases.

Total valid numbers =

144 + 96 = 240

.

Applying the addition principle of counting over disjoint cases.

Key Concept

Conditional permutations with non-disjoint position restrictions

Question 9Question →

In a science competition, $3$ distinct prizes (first, second, and third place) are to be awarded to $3$ different students chosen from a group of $5$ finalists. In how many different ways can these $3$ prizes be awarded?

Show answer & explanation

Answer: 60

Answer

60 ways

Because the prizes are distinct, the order in which the finalists are selected matters. Calculating the number of arrangements of

3

students from

5

candidates is given by

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

.

Step-by-Step Solution

1

Determine if order matters

Since the prizes are distinct (1st, 2nd, and 3rd place), the order of assignment matters, making this a permutation problem.

Assigning distinct positions to individuals requires calculating permutations rather than combinations.

2

Apply the permutation formula

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

^{5}P_{3} = \frac{5!}{(5-3)!} = \frac{5!}{2!}

There are

5

total candidates (

n = 5

) and

3

positions to fill (

r = 3

).

3

Calculate the numeric value

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

Canceling

2!

from the numerator and denominator leaves

5 \times 4 \times 3 = 60

.

Key Concept

Linear permutation of r items selected from n distinct items

Estimated Time:45s

Question 10Question →

Four different Mathematics textbooks and three different Physics textbooks are to be arranged in a line on a shelf. In how many distinct ways can the books be arranged if all four Mathematics textbooks must be kept together?

144

576

840

5,040

Show answer & explanation

Answer: 576

Answer

576 distinct ways

Treating the four Mathematics textbooks as a single unit gives 4 items to arrange on the shelf (the Mathematics unit and the three individual Physics textbooks). These 4 items can be arranged in

4! = 24

ways. Furthermore, the four distinct Mathematics textbooks within the unit can be arranged among themselves in

4! = 24

ways. By the multiplication principle of counting, the total number of distinct arrangements is

24 \times 24 = 576

.

Step-by-Step Solution

1

Group the Mathematics textbooks into a single block

1 Mathematics block + 3 individual Physics textbooks = 4 items to arrange.

Because all four Mathematics textbooks must remain together, they act as a single composite unit.

2

Calculate the arrangements of the 4 main items

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

ways.

There are 4 distinct items (the Mathematics block and 3 separate Physics books) to place in linear order.

3

Calculate internal permutations of the Mathematics textbooks inside their block

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

ways.

The 4 Mathematics textbooks can be arranged in different orders among themselves.

4

Apply the fundamental counting principle

Total arrangements =

24 \times 24 = 576

.

Multiply the number of block arrangements by the internal arrangements of the Mathematics textbooks.

Key Concept

Permutations with Restricted Grouping (Block Method)

Question 11Question →

In how many distinct ways can a president, a vice-president, and a secretary be chosen from a group of $7$ candidates, assuming no candidate can hold more than one position?

Show answer & explanation

Answer: 210

Answer

The total number of distinct ways to choose the three officers from 7 candidates is 210.

Selecting 3 distinct officers from a group of 7 candidates requires ordering 3 individuals out of 7, which equals

7 \times 6 \times 5 = 210

ways.

Step-by-Step Solution

1

Identify whether the problem involves permutations or combinations

Order is important because the positions (President, Vice-President, Secretary) are distinct.

Selecting person X as President and person Y as Secretary is different from selecting person Y as President and person X as Secretary.

2

Calculate the permutation

^7P_3

^7P_3 = 7 \times 6 \times 5 = 210

There are 7 choices for President, 6 remaining choices for Vice-President, and 5 remaining choices for Secretary.

Key Concept

Permutations of n items taken r at a time

Estimated Time:45s

Question 12Question →

In how many distinct ways can the letters of the word $\text{KADUNA}$ be arranged?

720

15

360

180

Show answer & explanation

Answer: $360$

Answer

The total number of distinct arrangements is

360

.

The word

\text{KADUNA}

consists of

6

total letters with the letter '\text{A}' repeated

2

times. Using the formula for permutations with identical items, the number of distinct arrangements is

\frac{6!}{2!} = \frac{720}{2} = 360

.

Step-by-Step Solution

1

Count the total number of letters in the word

\text{KADUNA}

and identify repetitions.

Total letters

n = 6

. The letter '\text{A}' appears

2

times, while '\text{K}', '\text{D}', '\text{U}', and '\text{N}' each appear

1

time.

Arrangements of

n

items with repeated elements require dividing

n!

by the factorial of the count of each repeated element.

2

Apply the permutation formula for repeated elements:

P = \frac{n!}{p!}

where

p

is the frequency of the repeated letter.

P = \frac{6!}{2!} = \frac{720}{2} = 360

Dividing by

2!

eliminates duplicate arrangements caused by swapping identical letters.

Key Concept

Permutations of items with repeated elements

Estimated Time:45s

Question 13Question →

A committee of $6$ delegates is to be seated around a circular table for a conference. If two specific delegates must always sit next to each other, in how many distinct ways can the delegates be arranged?

48

24

240

15

Show answer & explanation

Answer: 48

Answer

48

The correct answer is 48 because treating the 2 restricted delegates as a single unit leaves 5 items to arrange around a circle, which yields

(5-1)! = 24

arrangements. Since the 2 delegates can arrange themselves in

2! = 2

ways within their block, the total number of arrangements is

24 \times 2 = 48

.

Step-by-Step Solution

1

Group the restricted delegates into a single block

2 specified delegates are treated as 1 unit, leaving 4 remaining delegates, making a total of 5 items to arrange.

Since the two delegates must sit next to each other, treating them as a single entity ensures they remain together in all seating positions.

2

Calculate the circular arrangements of the 5 items

(5 - 1)! = 4! = 24

ways.

The number of ways to arrange

n

distinct items around a circular table is given by

(n - 1)!

.

3

Account for internal arrangements of the paired delegates

2! = 2

ways.

The two delegates within the single block can swap positions between themselves.

4

Multiply the circular arrangements by the internal arrangements

24 \times 2 = 48

total distinct arrangements.

By the fundamental counting principle, total arrangements equal the product of the independent steps.

Key Concept

Circular Permutations with Restrictions

Estimated Time:1m 15s

Question 14Question →

Given the permutation equation ${^{n}P_4} = 42 \cdot {^{n}P_2}$ , what is the positive integer value of $n$ ?

Show answer & explanation

Answer: 9

Answer

The positive integer value of

n

is

9

.

Expanding

{^{n}P_4}

as

n(n-1)(n-2)(n-3)

and

{^{n}P_2}

as

n(n-1)

transforms the equation to

n(n-1)(n-2)(n-3) = 42n(n-1)

. Dividing both sides by

n(n-1)

gives

(n-2)(n-3) = 42

, which simplifies to

n^2 - 5n - 36 = 0

. Factoring this quadratic gives

(n-9)(n+4) = 0

. Since

n

must be a positive integer greater than or equal to

4

, the correct answer is

9

.

Step-by-Step Solution

1

Write out the expanded expressions for

{^{n}P_4}

and

{^{n}P_2}

.

{^{n}P_4} = n(n-1)(n-2)(n-3)

and

{^{n}P_2} = n(n-1)

.

By definition,

{^{n}P_r} = \frac{n!}{(n-r)!} = n(n-1)\dots(n-r+1)

.

2

Set up the algebraic equation based on the given problem statement.

n(n-1)(n-2)(n-3) = 42n(n-1)

.

Substitute the expanded permutation formulas into the given identity.

3

Simplify the equation by dividing both sides by the non-zero common product

n(n-1)

.

(n-2)(n-3) = 42

.

Since

n \ge 4

,

n(n-1) \neq 0

, so we can cancel these terms from both sides.

4

Expand and rearrange the resulting expression into a standard quadratic equation.

n^2 - 5n - 36 = 0

.

Expanding

(n-2)(n-3)

gives

n^2 - 5n + 6

; subtracting

42

yields

n^2 - 5n - 36 = 0

.

5

Solve the quadratic equation for

n

and discard non-physical roots.

n = 9

.

Factoring yields

(n-9)(n+4) = 0

, so

n = 9

or

n = -4

. A permutation requires

n

to be a positive integer

\ge 4

, so

n = 9

.

Key Concept

Solving algebraic equations involving permutations

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

Question 15Question →

Find the value of $r$ such that ${^{8}P_r} = 6720$ .

Show answer & explanation

Answer: 5

Answer

The value of

r

is 5.

Expanding

{^{8}P_r}

into consecutive decreasing factors starting from 8 yields

8 \times 7 \times 6 \times 5 \times 4 = 6720

. Counting the number of factors multiplied (8, 7, 6, 5, 4) gives exactly 5 factors, so

r = 5

.

Step-by-Step Solution

1

Write the formula for permutation

{^{8}P_r}

as a product of descending integers.

{^{8}P_r} = 8 \times 7 \times 6 \times \dots \times (8 - r + 1)

By definition,

{^{n}P_r}

represents the product of

r

consecutive factors starting from

n

and decreasing by 1.

2

Perform sequential multiplication starting from 8 until reaching 6720.

Product of 5 factors:

8 \times 7 \times 6 \times 5 \times 4 = 6720

Multiplying factors gives:

8 \times 7 = 56

;

56 \times 6 = 336

;

336 \times 5 = 1680

;

1680 \times 4 = 6720

.

3

Count the number of terms in the product to find

r

.

r = 5

Since 5 consecutive integers were multiplied together to obtain 6720, the subset size

r

is 5.

Key Concept

Permutations of

n

distinct items taken

r

at a time

Estimated Time:1m 0s

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