Permutations, Combinations, and Fundamental Counting Principle Practice Questions — GRE General Test

Question 1Question →

A university research department needs to form a 4-person delegation consisting of exactly 2 senior researchers and 2 junior researchers. The pool of available candidates contains 5 senior researchers and 6 junior researchers. However, 2 specific senior researchers refuse to serve on the delegation together. How many different 4-person delegations can be formed under these conditions?

120

135

150

270

300

Show answer & explanation

Answer: 135

Answer

135 different 4-person delegations can be formed.

To calculate the number of valid delegations, first find the unrestricted number of delegations by multiplying the combinations of senior researchers

\binom{5}{2} = 10

and junior researchers

\binom{6}{2} = 15

, which equals 150. Next, calculate the number of invalid delegations where both restricted senior researchers are chosen together, which is

\binom{2}{2} \times \binom{6}{2} = 15

. Subtracting the invalid delegations from the total gives

150 - 15 = 135

.

Step-by-Step Solution

1

Calculate the total number of possible delegations without any restrictions.

\binom{5}{2} \times \binom{6}{2} = 10 \times 15 = 150

Choosing 2 senior researchers out of 5 requires combinations,

\binom{5}{2} = 10

. Choosing 2 junior researchers out of 6 requires combinations,

\binom{6}{2} = 15

. By the Fundamental Counting Principle, the total number of unrestricted delegations is

10 \times 15 = 150

.

2

Calculate the number of invalid delegations that include both restricted senior researchers.

\binom{2}{2} \times \binom{6}{2} = 1 \times 15 = 15

If both conflicting senior researchers are placed on the delegation, there is only

\binom{2}{2} = 1

way to select the senior members. The 2 junior researchers can still be selected in

\binom{6}{2} = 15

ways, giving 15 invalid delegations.

3

Subtract the number of invalid delegations from the total possible delegations.

150 - 15 = 135

Subtracting the 15 restricted delegations from the 150 total delegations leaves 135 valid delegations.

Key Concept

Combinations with Exclusion Restrictions and Fundamental Counting Principle

Question 2Question →

A committee is to be selected from a group of $5$ distinct people: $P$ , $Q$ , $R$ , $S$ , and $T$ . Which of the following statements regarding the possible selections or arrangements of people from this group are true? Select all that apply.

Select all that apply

The number of different

2

-person committees that can be formed from the group is

10

.

The number of different

3

-person committees that can be formed from the group is

10

.

The number of different ways to assign

2

distinct officer roles (President and Vice President) from the group is

10

.

The total number of different ways to arrange all

5

people in a single line is

25

.

The number of different ways to select and arrange

3

of the

5

people in a line is

60

.

Show answer & explanation

Answer: The number of different $2$ -person committees that can be formed from the group is $10$ .; The number of different $3$ -person committees that can be formed from the group is $10$ .; The number of different ways to select and arrange $3$ of the $5$ people in a line is $60$ .

Answer

The statements confirming that

10

different

2

-person committees can be formed,

10

different

3

-person committees can be formed, and

60

different

3

-person linear arrangements can be formed are all correct.

Selecting committees without specific roles requires combinations

\binom{n}{k}

, giving

\binom{5}{2} = 10

and

\binom{5}{3} = 10

. Arranging 3 people in ordered positions requires permutations

P(5,3) = 5 \times 4 \times 3 = 60

. Thus, all three corresponding statements are correct.

Step-by-Step Solution

1

Evaluate the 2-person committee selection statement.

\binom{5}{2} = \frac{5 \times 4}{2} = 10

Selection of a committee without specific roles is an unordered combination.

2

Evaluate the 3-person committee selection statement.

\binom{5}{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10

Choosing 3 items out of 5 yields the same number of outcomes as choosing 2 items out of 5.

3

Evaluate the 2-person officer assignment statement.

P(5,2) = 5 \times 4 = 20

Assigning distinct officer positions means order matters, requiring permutations rather than combinations.

4

Evaluate the 5-person line arrangement statement.

5! = 120

The total number of linear arrangements of 5 distinct objects is given by 5 factorial.

5

Evaluate the 3-person line arrangement statement.

P(5,3) = 5 \times 4 \times 3 = 60

Ordering 3 out of 5 people in a line uses the fundamental counting principle with decreasing choices per slot.

Key Concept

Distinguishing between combinations (where selection order does not matter) and permutations (where selection or position order does matter).

Question 3Question →

A security code consists of three distinct digits chosen from the non-zero digits $1$ through $9$ . If the first digit must be odd and the third digit must be even, how many such three-digit security codes can be formed?

16

70

140

160

180

Show answer & explanation

Answer: 140

Answer

140

To find the number of three-digit codes with distinct digits from

1

through

9

satisfying the constraints, count the options for each slot: the first position has

5

odd options (

1, 3, 5, 7, 9

), the third position has

4

even options (

2, 4, 6, 8

), and the middle position has

9 - 2 = 7

remaining options. By the Fundamental Counting Principle, multiplying these options gives

5 \times 7 \times 4 = 140

.

Step-by-Step Solution

1

Determine the number of possibilities for the first digit.

5 choices (the odd digits:

1, 3, 5, 7, 9

).

The question specifies that the first digit must be odd.

2

Determine the number of possibilities for the third digit.

4 choices (the even digits:

2, 4, 6, 8

).

The question specifies that the third digit must be even.

3

Determine the number of possibilities for the middle (second) digit.

7 choices.

There are

9

total non-zero digits (

1

through

9

). Since

2

distinct digits have already been used for the first and third positions,

9 - 2 = 7

digits remain available for the middle position.

4

Apply the Fundamental Counting Principle to find the total number of codes.

5 \times 7 \times 4 = 140

.

The total number of sequential independent choices is found by multiplying the number of options for each position.

Key Concept

Fundamental Counting Principle with Restricted Positions and Distinct Elements

Estimated Time:45s

Question 4Question →

A 6-digit security code is to be formed using distinct digits chosen from the set $\{1, 2, 3, 4, 5, 6, 7, 8\}$ . The code must satisfy the following conditions:
1. The code must be an even number (its final digit must be $2$ , $4$ , $6$ , or $8$ ).
2. Both digits $1$ and $2$ must be included in the 6-digit code.
3. Digits $1$ and $2$ cannot occupy adjacent positions in the code.

Which of the following statements regarding the number of possible 6-digit security codes must be true? Select all such statements.

Select all that apply

The total number of valid 6-digit security codes that satisfy all conditions is

3,600

.

The number of valid security codes in which the final digit is

2

is

1,440

.

The number of valid security codes in which the final digit is an even digit other than

2

is

2,160

.

If the non-adjacency constraint between

1

and

2

were removed, the total number of valid security codes would be

4,320

.

There are

20

valid ways to place digits

1

and

2

non-adjacently in the first 5 positions of the code.

Show answer & explanation

Answer: The total number of valid 6-digit security codes that satisfy all conditions is $3,600$ .; The number of valid security codes in which the final digit is $2$ is $1,440$ .; The number of valid security codes in which the final digit is an even digit other than $2$ is $2,160$ .

Answer

The statements confirming that the total number of codes is 3,600, that 1,440 codes end in 2, and that 2,160 codes end in an even digit other than 2 are all correct.

The solution requires partitioning into two mutually exclusive scenarios depending on whether digit 2 occupies the final position. When digit 2 is at the end, digit 1 can occupy any of the first 4 positions (excluding position 5 to avoid adjacency), yielding

4 \times P(6,4) = 1,440

codes. When the final position is occupied by 4, 6, or 8 (3 choices), digits 1 and 2 have

P(5,2) - 8 = 12

valid non-adjacent placements across the first 5 positions, and the remaining 3 positions can be filled in

P(5,3) = 60

ways, yielding

3 \times 12 \times 60 = 2,160

codes. The sum of these two cases gives

3,600

total valid codes. Thus, the three statements asserting total codes of 3,600, 1,440 ending in 2, and 2,160 ending in 4, 6, or 8 are all correct.

Step-by-Step Solution

1

Analyze the conditions and split into two disjoint cases based on the last digit.

Case 1: The last digit (6th position) is

2

. Case 2: The last digit (6th position) is

4

,

6

, or

8

.

Digit

2

plays a dual role: it satisfies the even-ending condition and is one of the restricted digits.

2

Calculate Case 1 (last digit is 2).

Position 6 is fixed as

2

(

1

choice). Digit

1

must be in the first 5 positions but cannot be adjacent to position 6 (so position 5 is excluded). Thus digit

1

has

4

choices (positions 1, 2, 3, 4). The remaining

4

positions are filled from the remaining

6

available digits

\{3, 4, 5, 6, 7, 8\}

in

P(6, 4) = 6 \times 5 \times 4 \times 3 = 360

ways. Total for Case 1:

1 \times 4 \times 360 = 1,440

.

This determines the valid arrangements when 2 is forced to the end of the code.

3

Calculate Case 2 (last digit is 4, 6, or 8).

There are

3

choices for the 6th position. Digits

1

and

2

must be placed among the first 5 positions non-adjacently. Total ordered placements of

1

and

2

in 5 positions is

P(5, 2) = 20

. The number of adjacent position pairs is

4

(positions (1,2), (2,3), (3,4), (4,5)), with

2! = 2

orderings per pair, giving

4 \times 2 = 8

adjacent placements. Non-adjacent placements =

20 - 8 = 12

. The remaining 3 open positions are filled from the remaining 5 available digits in

P(5, 3) = 60

ways. Total for Case 2:

3 \times 12 \times 60 = 2,160

.

This accounts for codes ending in 4, 6, or 8 while respecting the non-adjacency of 1 and 2.

4

Combine the cases and verify statements.

Total valid codes =

1,440 + 2,160 = 3,600

. The statements asserting totals of 3,600, 1,440 ending in 2, and 2,160 ending in 4, 6, or 8 are true.

Summing mutually exclusive cases gives the overall number of valid outcomes.

Key Concept

Fundamental Counting Principle with Permutations under Restricted Adjacency

Question 5Question →

A bookshelf holds $4$ distinct fiction novels and $3$ distinct non-fiction books. If a reader chooses exactly $1$ fiction novel and $1$ non-fiction book to take on a trip, how many different pairs of books can the reader select?

Show answer & explanation

Answer: 12

Answer

The total number of different pairs of books that can be selected is

12

.

According to the Fundamental Counting Principle, if one task can be performed in

m

ways and a second task can be performed in

n

ways, the two tasks together can be performed in

m \times n

ways. Choosing a fiction novel (

4

options) and a non-fiction book (

3

options) results in

4 \times 3 = 12

unique pairs.

Step-by-Step Solution

1

Determine the number of ways to choose one fiction novel

There are

4

possible choices.

The shelf contains

4

distinct fiction novels.

2

Determine the number of ways to choose one non-fiction book

There are

3

possible choices.

The shelf contains

3

distinct non-fiction books.

3

Calculate total pairs using the Fundamental Counting Principle

4 \times 3 = 12

The selection of a fiction novel and a non-fiction book are independent decisions, so the number of outcomes is the product of the number of choices for each decision.

Key Concept

Fundamental Counting Principle

Question 6Question →

An academic conference has 7 consecutive presentation time slots. A committee must assign 7 presentations—3 in Biology ( $B$ ), 2 in Chemistry ( $C$ ), and 2 in Physics ( $P$ )—to these slots subject to the following conditions:

1. Presentations of the same discipline are indistinguishable (only the subject sequence matters).
2. No two Biology presentations may be scheduled in consecutive time slots.
3. The two Physics presentations must be scheduled in consecutive time slots.

How many different subject-sequence schedules for the 7 presentation time slots satisfy all of these conditions?

24

12

30

60

72

Show answer & explanation

Answer: $12$

Answer

12

To satisfy the condition that the two Physics presentations are consecutive, we combine them into a single block

(PP)

. We then arrange the non-Biology items—two Chemistry presentations

C

and one block

(PP)

—which can be ordered in

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

distinct ways. These 3 items create 4 potential gaps (before the first item, between items, and after the last item). To ensure no two Biology presentations are adjacent, we place one Biology presentation into each of 3 chosen gaps out of the 4 available spaces, which can be done in

\binom{4}{3} = 4

ways. By the Fundamental Counting Principle, the total number of valid subject-sequence schedules is

3 \times 4 = 12

.

Step-by-Step Solution

1

Group the adjacent Physics presentations into a single block

The two Physics presentations

(P, P)

form a single block

(PP)

. The set of non-Biology items consists of two indistinguishable

C

's and one

(PP)

block, making 3 items in total.

Since the two Physics presentations must be scheduled in consecutive slots, treating them as a single unit guarantees they stay adjacent.

2

Calculate the number of distinct arrangements of the non-Biology items

The number of distinct arrangements of

\{C, C, (PP)\}

is

\frac{3!}{2!1!} = 3

ways.

The 2 Chemistry presentations are identical, so we divide the total permutations (

3!

) by

2!

to account for indistinguishability.

3

Apply the gap method to place the non-adjacent Biology presentations

Any sequence of the 3 non-Biology items creates 4 available spaces (gaps), including the two ends:

\_ \text{item}_1 \_ \text{item}_2 \_ \text{item}_3 \_

. Choosing 3 gaps out of 4 yields

\binom{4}{3} = 4

ways to insert the 3 identical

B

presentations.

Placing at most one Biology presentation in each gap guarantees no two Biology presentations are consecutive.

4

Apply the Fundamental Counting Principle

Total valid schedules =

3 \times 4 = 12

.

The choices of arranging non-Biology items and selecting gaps for Biology items are independent.

Key Concept

Combinatorics with Adjacency Restrictions and Indistinguishable Objects

Question 7Question →

A committee of $8$ people consists of $4$ men and $4$ women. A subcommittee of $4$ people is to be selected from this group such that the subcommittee contains at least one man and at least one woman. If two specific members, one man and one woman, refuse to serve together on the same subcommittee, how many different valid subcommittees of $4$ people can be formed?

Show answer & explanation

Answer: 53

Answer

53

The total number of ways to choose 4 people out of 8 is

\binom{8}{4} = 70

. Removing the 2 single-gender subcommittees (4 men or 4 women) leaves 68 gender-valid subcommittees. Among these 68 subcommittees, exactly

\binom{6}{2} = 15

contain both of the two conflicting individuals. Subtracting these 15 forbidden subcommittees gives

68 - 15 = 53

valid subcommittees.

Step-by-Step Solution

1

Calculate total ways to pick 4 people out of 8 without restrictions

\binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70

This establishes the total baseline sample space of possible 4-person groups.

2

Exclude single-gender groups to satisfy the gender balance constraint

70 - \binom{4}{4} - \binom{4}{4} = 70 - 1 - 1 = 68

Groups with 0 men or 0 women are invalid.

3

Count the forbidden groups that contain both of the conflicting individuals

\binom{6}{2} = 15

Fixing the 2 specific individuals in the subcommittee requires selecting 2 additional members from the remaining 6 people.

4

Subtract forbidden groups from gender-valid groups

68 - 15 = 53

Every group containing both conflicting individuals already satisfies the gender constraint, so exactly 15 invalid groups must be removed from the 68 gender-valid groups.

Key Concept

Combinations with multiple overlapping constraints (complementary counting)

Question 8Question →

A research committee must select a delegation of $6$ members from a pool of $4$ senior fellows and $4$ junior analysts to sit around a circular conference table with $6$ evenly spaced seats. The delegation must consist of exactly $3$ senior fellows and $3$ junior analysts. If no two senior fellows may sit in adjacent seats around the table, how many distinct delegation seating arrangements are possible? (Two seating arrangements are considered identical if one can be rotated to match the other.)

Show answer & explanation

Answer: 192

Answer

The total number of distinct delegation seating arrangements is 192.

To find the total number of distinct delegation seating arrangements, we first determine how many ways the 6 delegates can be selected, and then multiply by the number of valid circular seating arrangements for those delegates.

1. Selection of Delegates:
- Ways to select 3 senior fellows from 4:

\binom{4}{3} = 4

- Ways to select 3 junior analysts from 4:

\binom{4}{3} = 4

- Total delegate combinations:

4 \times 4 = 16

2. Circular Seating Arrangements:
- With 3 seniors and 3 juniors at a 6-seat table, no two seniors can sit adjacent if and only if seniors and juniors alternate seats.
- Fix one senior fellow to eliminate rotational symmetry.
- The remaining 2 seniors can be arranged in

2! = 2

ways.
- The 3 junior analysts can be arranged in the 3 intermediate seats in

3! = 6

ways.
- Seating arrangements per delegation =

2 \times 6 = 12

3. Total Arrangements:
- Total =

16 \times 12 = 192

.

Step-by-Step Solution

1

Calculate combinations of senior fellows and junior analysts to form the 6-person delegation.

Number of ways to choose 3 seniors out of 4 is

\binom{4}{3} = 4

. Number of ways to choose 3 juniors out of 4 is

\binom{4}{3} = 4

. Total selection combinations =

4 \times 4 = 16

.

Choosing members from distinct pools uses combinations because member order within the selection does not matter.

2

Analyze the seating constraint for 3 seniors and 3 juniors around a 6-seat circular table.

The senior fellows must occupy alternating seats around the table (e.g., seats 1, 3, 5), leaving seats 2, 4, 6 for the junior analysts.

Placing 3 seniors among 6 circular seats with no two adjacent forces seniors to occupy every second seat.

3

Calculate the number of distinct circular seating arrangements for any specific set of 6 selected people.

Arrangements =

(3 - 1)! \times 3! = 2 \times 6 = 12

.

To account for rotational symmetry at a circular table, fix one senior fellow's seat. The remaining 2 senior fellows can be seated in

2! = 2

ways, and the 3 junior analysts can be seated in

3! = 6

ways in the remaining open seats.

4

Apply the Fundamental Counting Principle to combine selection and seating steps.

Total arrangements =

16 \text{ (selections)} \times 12 \text{ (seating arrangements)} = 192

.

Each of the 16 unique delegations can be seated around the circular table in 12 distinct relative orders.

Key Concept

Combinations and Circular Permutations with Adjacency Restrictions

Estimated Time:2m 0s

Question 9Question →

A university research department consists of $5$ senior professors and $6$ junior researchers. A project committee of $5$ members is to be formed from this group. The committee must include at least $2$ senior professors and at least $2$ junior researchers. Additionally, two specific junior researchers, Alex and Blair, refuse to serve on the committee together. How many different 5-member committees can be formed under these conditions?

240

260

300

350

455

Show answer & explanation

Answer: 300

Answer

300

The total number of committees satisfying the role composition rules without restrictions is 350 (200 committees with 2 seniors and 3 juniors, plus 150 committees with 3 seniors and 2 juniors). Subtracting the 50 committees that contain both Alex and Blair leaves 300 valid committees.

Step-by-Step Solution

1

Determine valid committee compositions based on role count constraints.

Two valid distributions of 5 members: Case 1 has 2 senior professors and 3 junior researchers; Case 2 has 3 senior professors and 2 junior researchers.

The committee must contain at least 2 seniors and at least 2 juniors out of 5 total members.

2

Calculate total valid committees without the adjacency/conflict restriction.

Case 1:

\binom{5}{2} \times \binom{6}{3} = 10 \times 20 = 200

. Case 2:

\binom{5}{3} \times \binom{6}{2} = 10 \times 15 = 150

. Total without restriction =

200 + 150 = 350

.

Using combination formula

\binom{n}{k} = \frac{n!}{k!(n-k)!}

to count valid group selections.

3

Calculate the number of prohibited committees containing both Alex and Blair.

For Case 1 (2 seniors, 3 juniors): choose 2 seniors from 5 and 1 additional junior from the remaining 4, giving

\binom{5}{2} \times \binom{4}{1} = 10 \times 4 = 40

. For Case 2 (3 seniors, 2 juniors): choose 3 seniors from 5 and 0 additional juniors from the remaining 4, giving

\binom{5}{3} \times \binom{4}{0} = 10 \times 1 = 10

. Total restricted committees =

40 + 10 = 50

.

When Alex and Blair are both selected, 2 junior slots are fixed, leaving remaining slots to be filled from the remaining 4 junior researchers.

4

Subtract restricted committees from total valid composition committees.

350 - 50 = 300

.

Complementary counting yields the total number of valid committees satisfying all constraints.

Key Concept

Combinations with Composition and Exclusion Restrictions

Question 10Question →

A logistics coordinator must assign $6$ distinct delivery routes, labeled $R_1, R_2, R_3, R_4, R_5,$ and $R_6$ , to $6$ consecutive time slots, with exactly one route scheduled per slot. The schedule must satisfy two conditions:
1. Route $R_1$ must be scheduled in an earlier time slot than Route $R_2$ .
2. Route $R_1$ and Route $R_2$ cannot be scheduled in consecutive time slots.

Which of the following values correctly describe counts or proportions associated with this scheduling scenario? Select all that apply.

Select all that apply

240

, representing the total number of valid schedules satisfying both conditions

240

, representing the total number of schedules in which

R_1

and

R_2

are placed in consecutive time slots regardless of order

\frac{1}{3}

, representing the fraction of all possible unrestricted schedules that satisfy both conditions

360

, representing the total number of valid schedules satisfying both conditions

\frac{1}{2}

, representing the fraction of all possible unrestricted schedules that satisfy both conditions

Show answer & explanation

Answer: $240$ , representing the total number of valid schedules satisfying both conditions; $240$ , representing the total number of schedules in which $R_1$ and $R_2$ are placed in consecutive time slots regardless of order; $\frac{1}{3}$ , representing the fraction of all possible unrestricted schedules that satisfy both conditions

Answer

The valid choices are the statement giving 240 as the total number of valid schedules, the statement giving 240 as the total number of schedules with adjacent routes, and the statement giving 1/3 as the fraction of valid schedules.

The total number of unrestricted permutations for 6 distinct routes is

6! = 720

. By symmetry,

R_1

comes before

R_2

in exactly half of these, or 360 permutations. Within these 360 permutations, those where

R_1

and

R_2

occupy consecutive slots treat

(R_1, R_2)

as a single unit in fixed order, yielding

5! = 120

permutations. Thus, the number of valid schedules is

360 - 120 = 240

. Separately, the total number of schedules with

R_1

and

R_2

adjacent in any order is

2! \times 5! = 240

. Finally, the ratio of valid schedules to total schedules is

240 / 720 = 1/3

. Therefore, the options stating 240 total valid schedules, 240 adjacent schedules, and a 1/3 ratio are all correct.

Step-by-Step Solution

1

Calculate the total number of unrestricted arrangements of the 6 routes.

Total unrestricted arrangements =

6! = 720

.

6 distinct routes placed into 6 distinct slots can be ordered in

6!

ways.

2

Apply the symmetry property to determine the number of schedules where

R_1

comes before

R_2

.

Schedules with

R_1

before

R_2 = \frac{720}{2} = 360

.

In any permutation of distinct elements,

R_1

is equally likely to appear before or after

R_2

.

3

Calculate the number of forbidden schedules where

R_1

is immediately before

R_2

(consecutive).

Forbidden schedules =

5! = 120

.

Treat the ordered block

(R_1, R_2)

as a single entity. Arranging this block alongside the remaining 4 routes gives

5!

permutations.

4

Subtract forbidden schedules from the ordered schedules to find valid schedules.

Valid schedules =

360 - 120 = 240

.

Subtracting the consecutive cases from all cases where

R_1

precedes

R_2

satisfies both non-consecutive and ordering rules.

5

Determine the count of schedules where

R_1

and

R_2

are consecutive in any order, and calculate the overall valid ratio.

Adjacent schedules =

2! \times 5! = 240

; Valid ratio =

\frac{240}{720} = \frac{1}{3}

.

Adjacent slots allow 2 internal orders for the block, yielding

240

. Comparing

240

valid schedules to

720

total gives a ratio of

\frac{1}{3}

.

Key Concept

Permutations with Adjacency Restrictions and Relative Order Symmetry

Estimated Time:2m 0s

Question 11Question →

An event coordinator is scheduling 5 distinct guest lectures—3 on Science and 2 on Art—to take place sequentially in 5 consecutive time slots. The coordinator establishes a restriction that the 2 Art lectures cannot be scheduled in consecutive time slots. Which of the following statements regarding the possible schedules are true? Select all such statements.

Select all that apply

The total number of possible arrangements for all 5 lectures without any restrictions is 120.

The number of schedules in which the 2 Art lectures are placed in consecutive time slots is 24.

The total number of valid schedules in which the 2 Art lectures are not consecutive is 72.

The total number of valid schedules in which the 2 Art lectures are not consecutive is 36.

The total number of valid schedules in which the 2 Art lectures are not consecutive is 96.

Show answer & explanation

Answer: The total number of possible arrangements for all 5 lectures without any restrictions is 120.; The total number of valid schedules in which the 2 Art lectures are not consecutive is 72.

Answer

The statement specifying that the total unrestricted arrangements equal 120, and the statement specifying that the total valid non-consecutive schedules equal 72 are both correct.

Without restrictions, 5 distinct items can be linearly ordered in

5! = 120

ways. To find the number of ways where the 2 Art lectures are not adjacent, we subtract the ways they ARE adjacent (

2! \times 4! = 48

) from the total

120

, giving

120 - 48 = 72

. Alternatively, placing 3 Science lectures creates 4 available slots; selecting 2 slots and permuting the Art lectures yields

3! \times \binom{4}{2} \times 2! = 72

.

Step-by-Step Solution

1

Calculate total unrestricted arrangements of the 5 distinct lectures.

5! = 120

total arrangements.

The Fundamental Counting Principle specifies that 5 distinct items can be arranged in

5 \times 4 \times 3 \times 2 \times 1 = 120

ways.

2

Calculate the number of restricted arrangements where the 2 Art lectures are placed consecutively.

2! \times 4! = 48

consecutive arrangements.

Treat the 2 Art lectures as a single combined block. The 3 Science lectures and 1 Art block form 4 units, which can be arranged in

4! = 24

ways. Within the block, the 2 distinct Art lectures can be ordered in

2! = 2

ways, yielding

24 \times 2 = 48

.

3

Subtract the restricted consecutive arrangements from the total arrangements to find the valid non-consecutive schedules.

120 - 48 = 72

valid arrangements.

The complement rule allows finding non-consecutive placements by taking total arrangements minus consecutive arrangements.

4

Verify using the slot method (alternative approach).

3! \times \binom{4}{2} \times 2! = 6 \times 6 \times 2 = 72

valid arrangements.

Arrange 3 Science lectures in

3! = 6

ways. This creates 4 potential slots between and around them (_ S1 _ S2 _ S3 _). Choosing 2 slots for the Art lectures takes

\binom{4}{2} = 6

ways, and arranging the 2 distinct Art lectures in those slots takes

2! = 2

ways.

Key Concept

Permutations with Adjacency Restrictions and Complementary Counting

Question 12Question →

A committee of $5$ members is to be selected from a pool of $6$ doctors and $4$ nurses. How many different $5$ -member committees can be formed that contain at least $3$ doctors?

Show answer & explanation

Answer: 186

Answer

186

To form a 5-member committee containing at least 3 doctors from 6 doctors and 4 nurses, consider the three mutually exclusive possibilities: 3 doctors and 2 nurses, 4 doctors and 1 nurse, or 5 doctors and 0 nurses. Using combinations, the number of ways for each case are 120, 60, and 6 respectively. Summing these gives 186 distinct committees.

Step-by-Step Solution

1

Determine all valid committee compositions meeting the requirement

The committee can consist of: 3 doctors and 2 nurses, 4 doctors and 1 nurse, or 5 doctors and 0 nurses.

The prompt specifies 'at least 3 doctors' out of 5 total members.

2

Calculate the combinations for each scenario

Case 1: $\binom{6}{3} \times \binom{4}{2} = 20 \times 6 = 120$
Case 2: $\binom{6}{4} \times \binom{4}{1} = 15 \times 4 = 60$
Case 3: $\binom{6}{5} \times \binom{4}{0} = 6 \times 1 = 6$

Order of selection does not matter, so combination formula $\binom{n}{k}$ is used.

3

Sum the valid combinations

120 + 60 + 6 = 186

The scenarios are mutually exclusive, so the addition principle applies.

Key Concept

Combinations with restrictions and Addition Principle

Question 13Question →

A security system requires a 4-digit pass code formed using the digits $1, 2, 3, 4, 5, 6,$ and $7$ , with no digit repeated within a code. If the first digit of the pass code must be an even number and the last digit must be an odd number, how many such distinct pass codes can be created?

120

240

300

504

840

Show answer & explanation

Answer: 240

Answer

240 pass codes

To form a 4-digit code with distinct digits from the set {1, 2, 3, 4, 5, 6, 7}: there are 3 options for the first digit (even: 2, 4, 6) and 4 options for the fourth digit (odd: 1, 3, 5, 7). Because the sets of even and odd numbers are disjoint, choosing the first digit does not affect the number of odd choices available for the fourth position. After placing these 2 digits, 5 digits remain from the original set of 7. The second position can be filled in 5 ways, and the third position in 4 ways. By the Fundamental Counting Principle, the total number of codes is 3 × 4 × 5 × 4 = 240.

Step-by-Step Solution

1

Identify the choices for the first digit (even restriction).

3 possible choices (2, 4, or 6).

The first digit must be even, and the available set contains three even digits: {2, 4, 6}.

2

Identify the choices for the last (fourth) digit (odd restriction).

4 possible choices (1, 3, 5, or 7).

The fourth digit must be odd, and the available set contains four odd digits: {1, 3, 5, 7}.

3

Determine the available choices for the remaining middle digits.

5 choices for the second digit and 4 choices for the third digit.

Two distinct digits have been selected for the first and fourth positions out of the 7 available digits, leaving 5 remaining digits. Since no digits may repeat, the second position has 5 choices and the third position has 4 choices.

4

Apply the Fundamental Counting Principle to compute the total number of pass codes.

3 × 4 × 5 × 4 = 240 distinct pass codes.

Multiplying the independent choices for each position gives the total valid arrangements.

Key Concept

Fundamental Counting Principle and Permutations with Positional Restrictions

Question 14Question →

A tech company's quality assurance division needs to form a 5-member project panel selected from a pool of 6 software engineers and 4 hardware engineers. The panel must include at least 2 software engineers and at least 1 hardware engineer. However, 2 specific software engineers, Engineer $X$ and Engineer $Y$ , refuse to serve on the same panel together. How many different valid 5-member panels can be formed?

136

188

196

240

252

Show answer & explanation

Answer: 188

Answer

188 valid panels can be formed.

The total number of panels satisfying the software and hardware role distribution rules is 240. Among these, exactly 52 panels contain both Engineer X and Engineer Y. Subtracting 52 from 240 results in 188 valid panels.

Step-by-Step Solution

1

Determine valid software and hardware engineer compositions for a 5-member panel.

Three compositions satisfy the requirement of at least 2 software engineers (S) and at least 1 hardware engineer (H): (4S, 1H), (3S, 2H), and (2S, 3H).

Panels must have 5 total members adhering to the specified minimum headcount limits.

2

Calculate the total number of panels satisfying composition requirements without considering the exclusion restriction.

For (4S, 1H):

\binom{6}{4} \times \binom{4}{1} = 15 \times 4 = 60

. For (3S, 2H):

\binom{6}{3} \times \binom{4}{2} = 20 \times 6 = 120

. For (2S, 3H):

\binom{6}{2} \times \binom{4}{3} = 15 \times 4 = 60

. Total composition-valid panels =

60 + 120 + 60 = 240

.

Apply the combination formula

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

and the Fundamental Counting Principle.

3

Calculate the number of invalid panels that contain both Engineer X and Engineer Y.

If Engineer X and Engineer Y are both included (2 S), 3 remaining panel members must be selected from the remaining 4 software engineers and 4 hardware engineers. For (4S, 1H):

\binom{4}{2} \times \binom{4}{1} = 6 \times 4 = 24

. For (3S, 2H):

\binom{4}{1} \times \binom{4}{2} = 4 \times 6 = 24

. For (2S, 3H):

\binom{4}{0} \times \binom{4}{3} = 1 \times 4 = 4

. Total invalid panels =

24 + 24 + 4 = 52

.

Isolating combinations that contain both restricted engineers allows simple subtraction from the total.

4

Subtract invalid panels from the total composition-valid panels.

240 - 52 = 188

.

This leaves only the panels that satisfy both composition and exclusion rules.

Key Concept

Combinations with multi-group minimum constraints and pair exclusion

Question 15Question →

A city planning board needs to form a $6$ -member advisory task force selected from a pool of $5$ architects and $5$ civil engineers. The task force must include at least $2$ architects and at least $2$ civil engineers. However, two specific architects, Architect X and Architect Y, cannot both serve on the task force together. How many different $6$ -member task forces can be formed satisfying these conditions?

Show answer & explanation

Answer: 135

Answer

The total number of different valid 6-member task forces that can be formed is 135.

To solve this problem, we apply the addition principle over mutually exclusive cases of committee composition, followed by complementary counting to enforce the exclusion restriction. First, we identify the valid breakdown of architects and engineers for a 6-member team requiring at least 2 of each profession: (4 architects, 2 engineers), (3 architects, 3 engineers), and (2 architects, 4 engineers). Calculating the combinations for each breakdown yields 50, 100, and 50 ways respectively, totaling 200 unconstrained team options. Next, we determine how many of these teams include both Architect X and Architect Y. Pre-assigning both architects reduces the remaining available architects to 3. The invalid cases for each breakdown are 30, 30, and 5 respectively, totaling 65 invalid configurations. Subtracting the 65 invalid teams from the 200 total unconstrained teams yields 135 valid task forces.

Step-by-Step Solution

1

Identify the allowed group breakdowns under the restriction of at least 2 architects and at least 2 engineers.

The valid (architect, engineer) count pairs for a 6-member task force are (4, 2), (3, 3), and (2, 4).

Choosing 5 architects would leave only 1 engineer, violating the minimum requirement of 2 engineers, and vice versa.

2

Compute the total combinations without the exclusion restriction.

Total unconstrained combinations = 200.

\binom{5}{4}\binom{5}{2} + \binom{5}{3}\binom{5}{3} + \binom{5}{2}\binom{5}{4} = (5 \times 10) + (10 \times 10) + (10 \times 5) = 50 + 100 + 50 = 200

.

3

Calculate the number of task forces that violate the restriction by including both Architect X and Architect Y.

Total invalid combinations = 65.

If Architect X and Architect Y are both included, selecting remaining architects from the other 3 yields:

\binom{3}{2}\binom{5}{2} + \binom{3}{1}\binom{5}{3} + \binom{3}{0}\binom{5}{4} = (3 \times 10) + (3 \times 10) + (1 \times 5) = 30 + 30 + 5 = 65

.

4

Subtract the invalid combinations from the total unconstrained combinations.

200 - 65 = 135.

Using the complementary counting principle provides the exact number of valid combinations where Architect X and Architect Y do not serve together.

Key Concept

Combinations with multiple category constraints and complementary counting for exclusion rules.

Question 16Question →

A museum curator is arranging $6$ distinct paintings in a single row along a gallery wall. If $2$ specific paintings must not be placed adjacent to each other, how many different arrangements of the $6$ paintings are possible?

Show answer & explanation

Answer: 480

Answer

480

To find the number of valid arrangements where two specific paintings are not adjacent, use complementary counting. First, compute the total number of ways to arrange 6 distinct paintings without restrictions, which is

6! = 720

. Next, calculate the number of arrangements where the two specific paintings are placed adjacent to one another by treating them as a single block. There are 5 units in total to arrange (the pair block plus the remaining 4 individual paintings), which gives

5! = 120

ways. Since the two specific paintings can be arranged in

2! = 2

ways inside their block, the total number of adjacent arrangements is

120 \times 2 = 240

. Finally, subtract the adjacent arrangements from the total arrangements:

720 - 240 = 480

.

Step-by-Step Solution

1

Calculate the total number of ways to arrange all 6 paintings in a row without any restrictions.

6! = 720

There are 6 distinct items to arrange in 6 sequential positions.

2

Calculate the number of arrangements where the 2 specific paintings are adjacent (placed next to each other).

5! × 2! = 120 × 2 = 240

Treat the 2 specific paintings as a single block unit. This leaves 5 items to arrange (the block + 4 individual paintings), which can be ordered in 5! ways. Within the block, the 2 paintings can be ordered in 2! ways.

3

Subtract the number of adjacent arrangements from the total unrestricted arrangements.

720 - 240 = 480

Complementary counting dictates that valid non-adjacent arrangements equal total possible arrangements minus adjacent arrangements.

Key Concept

Permutations with Adjacency Restrictions (Complementary Counting)

Question 17Question →

A project manager is scheduling $7$ distinct project milestones: $4$ technical milestones and $3$ managerial milestones. The milestones must be scheduled sequentially across $7$ consecutive weeks. To avoid scheduling burnout, no two managerial milestones can be scheduled in consecutive weeks. In how many different valid sequences can all $7$ milestones be scheduled?

240

720

1,440

2,160

5,040

Show answer & explanation

Answer: $1,440$

Answer

1,440

To ensure no two managerial milestones are adjacent, first order the

4

distinct technical milestones, which can be done in

4! = 24

ways. These

4

milestones create

5

available slots (gaps before, between, and after them). To place the

3

distinct managerial milestones into these

5

slots such that no slot contains more than one managerial milestone, we calculate the permutations

P(5, 3) = 5 \times 4 \times 3 = 60

. Multiplying the independent choices yields

24 \times 60 = 1,440

valid sequences.

Step-by-Step Solution

1

Arrange the non-restricted items (the 4 distinct technical milestones)

Number of ways

= 4! = 24

Since all 4 technical milestones are distinct, they can be ordered in

4!

ways.

2

Determine the number of available slots (gaps) created for the restricted items

Number of slots

= 4 + 1 = 5

Placing 4 technical milestones in a line creates 5 potential slots (before the first, between adjacent ones, and after the last) where managerial milestones can be placed without being adjacent to each other: _ T1 _ T2 _ T3 _ T4 _

3

Place and order the 3 distinct managerial milestones into the 5 available slots

Number of ways

= P(5, 3) = 5 \times 4 \times 3 = 60

Because the managerial milestones are distinct and order matters, we choose 3 slots out of 5 and arrange them.

4

Apply the Fundamental Counting Principle to find the total valid arrangements

Total valid arrangements

= 24 \times 60 = 1,440

The placement of technical milestones and managerial milestones are independent choices in sequence.

Key Concept

Permutations with Non-Adjacency Restrictions (Gap Insertion Method)

Question 18Question →

A financial firm has a pool of $10$ analysts, consisting of $6$ senior analysts and $4$ junior analysts. Which of the following selection procedures will yield EXACTLY $120$ unique possible groups? Select all such procedures.

Select all that apply

Forming a

5

-member committee that contains exactly

3

senior analysts and

2

junior analysts

Forming a

3

-member subcommittee from the entire pool of

10

analysts without any restrictions

Forming a

7

-member project panel from the entire pool of

10

analysts without any restrictions

Forming a

5

-member committee that contains exactly

4

senior analysts and

1

junior analyst

Forming a

4

-member task force from the entire pool of

10

analysts without any restrictions

Show answer & explanation

Answer: Forming a $5$ -member committee that contains exactly $3$ senior analysts and $2$ junior analysts; Forming a $3$ -member subcommittee from the entire pool of $10$ analysts without any restrictions; Forming a $7$ -member project panel from the entire pool of $10$ analysts without any restrictions

Answer

The procedures that yield exactly 120 unique possible groups are: forming a 5-member committee with 3 senior and 2 junior analysts, forming a 3-member subcommittee from all 10 analysts, and forming a 7-member project panel from all 10 analysts.

The correct procedures are those that evaluate to exactly 120 combinations: (1) Selecting 3 senior analysts from 6 and 2 junior analysts from 4 gives $\binom{6}{3} \times \binom{4}{2} = 20 \times 6 = 120$. (2) Choosing 3 analysts from 10 gives $\binom{10}{3} = 120$. (3) Choosing 7 analysts from 10 is symmetric to choosing 3 analysts, yielding $\binom{10}{7} = \binom{10}{3} = 120$.

Step-by-Step Solution

1

Calculate combinations for forming a 5-member committee with 3 senior and 2 junior analysts

$\binom{6}{3} \times \binom{4}{2} = 20 \times 6 = 120$

The selection of senior and junior analysts are independent decisions, so their combination values are multiplied together.

2

Calculate combinations for choosing 3 analysts out of 10 without restrictions

$\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$

Order of selection does not matter, so the standard combination formula $\binom{n}{k}$ is applied.

3

Calculate combinations for choosing 7 analysts out of 10 without restrictions

$\binom{10}{7} = \binom{10}{10-7} = \binom{10}{3} = 120$

Choosing 7 people to include is mathematically equivalent to choosing 3 people to exclude.

4

Evaluate the remaining options to verify they do not yield 120

$\binom{6}{4} \times \binom{4}{1} = 60$ and $\binom{10}{4} = 210$

Neither of these evaluations equals the target value of 120.

Key Concept

Combinations and the Fundamental Counting Principle

Question 19Question →

A cybersecurity system generates 5-character identification codes consisting of 3 distinct letters followed by 2 distinct digits. The letters must be selected from the set $\{A, B, C, D, E, F\}$ and the digits from the set $\{1, 2, 3, 4\}$ . If the first character of the code must be a vowel ( $A$ or $E$ ), how many such distinct 5-character identification codes can be formed?

120

240

480

720

1,440

Show answer & explanation

Answer: 480

Answer

480 distinct 5-character identification codes can be formed.

To construct the 5-character code, break the process into sequential choices for each position: Position 1 must be a vowel (

A

or

E

), giving 2 options. Position 2 can be any of the remaining 5 distinct letters. Position 3 can be any of the remaining 4 distinct letters. Position 4 (the first digit) can be any of the 4 available digits. Position 5 (the second digit) can be any of the remaining 3 distinct digits. Multiplying these independent choices together yields

2 \times 5 \times 4 \times 4 \times 3 = 480

.

Step-by-Step Solution

1

Calculate the number of ways to choose the first letter (must be a vowel)

There are 2 choices (

A

or

E

).

The problem restricts the first character of the code to a vowel.

2

Calculate the number of choices for the second and third letters

5 choices for the second letter and 4 choices for the third letter.

The letters must be distinct, leaving 5 remaining letters from the original pool of 6 for the second position, and 4 for the third position.

3

Calculate the number of choices for the two distinct digits

4 choices for the first digit and 3 choices for the second digit.

There are 4 digits available in

\{1, 2, 3, 4\}

and they must be distinct.

4

Apply the Fundamental Counting Principle to find total codes

2 \times 5 \times 4 \times 4 \times 3 = 480

.

Multiply the number of independent choices for each sequential position.

Key Concept

Fundamental Counting Principle with Permutations and Restrictions

Question 20Question →

A museum display curator is arranging $6$ distinct historical coins— $4$ silver coins and $2$ gold coins—in a single row inside a display case. If the $2$ gold coins cannot be placed adjacent to each other, how many different linear arrangements of the $6$ coins are possible?

240

360

480

576

720

Show answer & explanation

Answer: $480$

Answer

The total number of valid linear arrangements is

480

.

Arranging the

4

distinct silver coins yields

4! = 24

orderings. These

4

coins create

5

available gaps (including both ends). To guarantee the

2

distinct gold coins are not adjacent, each gold coin must occupy a separate gap. The number of ways to assign

2

distinct gold coins to

5

gaps is

P(5, 2) = 5 \times 4 = 20

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

24 \times 20 = 480

.

Step-by-Step Solution

1

Calculate the total number of ways to arrange the unrestricted silver coins.

The

4

distinct silver coins can be arranged in

4! = 24

ways.

The relative positions of the silver coins matter because each coin is distinct.

2

Determine the number of available gap positions created by the silver coins to separate the gold coins.

Arranging

4

silver coins creates

5

possible gaps (one before the first coin, three between adjacent silver coins, and one after the last coin: \_ S \_ S \_ S \_ S \_).

Placing at most one gold coin per gap guarantees that the two gold coins will not be adjacent.

3

Calculate the number of ways to place the

2

distinct gold coins into the

5

available gaps.

The number of ways to place the

2

distinct gold coins into

5

distinct slots is

P(5, 2) = 5 \times 4 = 20

.

Order matters because the gold coins are distinct objects.

4

Apply the Fundamental Counting Principle to find the total arrangements.

Total arrangements =

24 \times 20 = 480

.

The arrangement of the silver coins and the placement of the gold coins are independent sequential choices.

Key Concept

Slot method for non-adjacent arrangements using permutations and the Fundamental Counting Principle

Permutations, Combinations, and Fundamental Counting Principle

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