Combinations and Group Selections Practice Questions — GMAT

Question 1Question →

A project committee of $4$ members is to be selected from a pool of $n$ senior engineers and $6$ junior engineers, where $n \ge 4$ . The committee must contain at least one senior engineer and at least one junior engineer. If there are exactly $310$ different possible ways to select the committee, what is the value of $n$ ?

4

5

6

7

8

Show answer & explanation

Answer: 5

Answer

The value of

n

is 5.

Using complementary counting, the total number of ways to pick any 4 members from the

(n+6)

total engineers is

\binom{n+6}{4}

. The condition requires at least one senior engineer and at least one junior engineer. The invalid cases are when all 4 are junior engineers (which can occur in

\binom{6}{4} = 15

ways) or all 4 are senior engineers (which can occur in

\binom{n}{4}

ways). Setting up the equation:

\binom{n+6}{4} - \binom{6}{4} - \binom{n}{4} = 310

, which simplifies to

\binom{n+6}{4} - \binom{n}{4} = 325

. Testing

n = 5

gives

\binom{11}{4} - \binom{5}{4} = 330 - 5 = 325

, which matches the given condition.

Step-by-Step Solution

1

Set up the combination formula for total unrestricted selections

Total selections from

(n + 6)

engineers choosing 4 is

\binom{n+6}{4}

.

Order of selection does not matter when forming a committee.

2

Identify the restricted (invalid) committee configurations

Committees with 0 senior engineers:

\binom{6}{4} = 15

. Committees with 0 junior engineers:

\binom{n}{4}

.

The committee must contain at least one member from each category.

3

Formulate the equation using complementary counting

\binom{n+6}{4} - \binom{n}{4} - 15 = 310 \implies \binom{n+6}{4} - \binom{n}{4} = 325

.

Subtracting invalid configurations from total configurations yields the valid configurations.

4

Test values for

n

For

n = 5

:

\binom{11}{4} - \binom{5}{4} = 330 - 5 = 325

.

Since

325 = 325

,

n = 5

satisfies the given constraint exactly.

Key Concept

Group selections with constraints using complementary counting:

\text{Valid} = \text{Total} - \text{Restricted}

.

Question 2Question →

A task force must select a 4-member subcommittee from a pool consisting of 6 architects and $n$ urban planners. If there are exactly 111 different possible subcommittees that include at least 1 urban planner, what is the value of $n$ ?

2

3

4

5

9

Show answer & explanation

Answer: 3

Answer

The value of

n

is 3.

Using the complementary counting method, the number of subcommittees with at least 1 urban planner equals the total possible 4-member subcommittees from the pool of

(6+n)

people minus the subcommittees formed entirely of architects. The number of all-architect subcommittees is

\binom{6}{4} = 15

. Adding this to 111 gives

\binom{6+n}{4} = 126

. Since

\binom{9}{4} = 126

, the total pool size is 9 people. Subtracting the 6 architects gives

n = 3

.

Step-by-Step Solution

1

Apply the complementary counting principle

Number of valid subcommittees = (Total subcommittees) - (Subcommittees with 0 urban planners)

It is easier to subtract subcommittees containing only architects from the total number of subcommittees than to sum cases with 1, 2, 3, or 4 urban planners.

2

Calculate the number of subcommittees consisting only of architects

\binom{6}{4} = \frac{6 \times 5}{2 \times 1} = 15

Choosing 4 members from the 6 available architects gives 15 possible all-architect subcommittees.

3

Set up the combination equation for the total pool size

\binom{6+n}{4} - 15 = 111 \implies \binom{6+n}{4} = 126

Adding 15 to both sides isolates the combination expression representing selections from the total pool of

6+n

people.

4

Solve for the total pool size and isolate

n

\binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 \implies 6+n = 9 \implies n = 3

Evaluating combinations shows that selecting 4 items from 9 gives 126. Subtracting the 6 architects leaves 3 urban planners.

Key Concept

Group Selection with Restrictions and Complementary Counting

Estimated Time:2m 0s

Question 3Question →

A student must choose $3$ books to read from a reading list of $6$ distinct books. How many different combinations of $3$ books can the student select?

15

18

20

60

120

Show answer & explanation

Answer: $20$

Answer

The student can select

20

different combinations of

3

books.

To find the number of ways to select

3

books from

6

distinct books when order does not matter, use the combination formula

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

. Evaluating

\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20

. Thus,

20

is the correct number of combinations.

Step-by-Step Solution

1

Identify the counting method

Since the order in which the books are selected does not matter, use the combinations formula

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

.

Selection of a group without regard to order requires combinations.

2

Substitute the given values into the formula

Here

n = 6

and

k = 3

, so

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

.

Apply

n=6

available items and

k=3

chosen items.

3

Simplify the expression

\frac{120}{6} = 20.

Perform the basic arithmetic division.

Key Concept

Combinations and Group Selections

Estimated Time:45s

Question 4Question →

A committee of $5$ members is to be formed from a pool of $6$ men and $5$ women. The committee must include at least $2$ men and at least $2$ women. However, two specific individuals in the pool, one man and one woman, refuse to serve on the committee together. How many different valid $5$ -member committees can be formed?

Show answer & explanation

Answer: 280

Answer

The correct answer is 280.

The correct answer is 280, found by taking the total possible committees satisfying the gender constraint (350) and subtracting the invalid committees containing both conflicting members (70).

Step-by-Step Solution

1

Calculate total combinations satisfying gender constraints without individual restriction

350 valid gender-balanced committees (200 from 3M/2W and 150 from 2M/3W)

Establishes the total baseline number of committee selections before removing forbidden pairings

2

Calculate forbidden combinations containing both specific conflicting individuals

70 forbidden combinations (40 with 3M/2W overall and 30 with 2M/3W overall)

Identifies committee selections that violate the condition that the two individuals cannot serve together

3

Subtract forbidden combinations from total baseline combinations

280 valid committee selections

Applying complementary counting (

350 - 70

) yields the exact number of allowable committees

Key Concept

Combinations with group constraints and complementary counting

Question 5Question →

A research team needs to select a subcommittee of 3 scientists from a department containing 7 scientists. How many different 3-member subcommittees can be selected?

Show answer & explanation

Answer: 35

Answer

35 different subcommittees can be selected.

The total number of ways to choose a committee of 3 members from a group of 7 without regard to order is given by the combination formula

C(7,3) = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35

.

Step-by-Step Solution

1

Determine whether the selection depends on order.

Order does not matter since all 3 members of the subcommittee have equal roles.

When order does not matter in group selection, combinations (

nCr

) must be used rather than permutations (

nPr

).

2

Apply the combination formula

C(n,k) = \frac{n!}{k!(n-k)!}

with

n=7

and

k=3

.

C(7,3) = \frac{7 \times 6 \times 5}{3 \times 2 \times 1}

This counts the unique groups of 3 that can be chosen from a pool of 7.

3

Calculate the numerical result.

35

Dividing

7 \times 6 \times 5 = 210

by

3 \times 2 \times 1 = 6

yields 35.

Key Concept

Combinations and Group Selections

Question 6Question →

In a plane, there are $n$ points such that no three points are collinear. If the number of distinct triangles that can be formed using these points as vertices is exactly $5$ times the number of distinct line segments that can be formed by joining pairs of these points, what is the value of $n$ ?

7

15

16

17

32

Show answer & explanation

Answer: $17$

Answer

The correct value of

n

is

17

.

The number of distinct triangles formed by

n

points (no three collinear) is

\binom{n}{3} = \frac{n(n-1)(n-2)}{6}

, and the number of line segments is

\binom{n}{2} = \frac{n(n-1)}{2}

. Setting

\binom{n}{3} = 5 \binom{n}{2}

gives

\frac{n(n-1)(n-2)}{6} = \frac{5n(n-1)}{2}

. Dividing both sides by

\frac{n(n-1)}{2}

simplifies the equation to

\frac{n-2}{3} = 5

, yielding

n - 2 = 15

, so

n = 17

.

Step-by-Step Solution

1

Express the number of triangles and line segments using combinations.

The number of triangles formed by selecting

3

non-collinear points from

n

points is

\binom{n}{3} = \frac{n(n-1)(n-2)}{6}

. The number of line segments formed by selecting

2

points from

n

points is

\binom{n}{2} = \frac{n(n-1)}{2}

.

Order does not matter when selecting vertices for triangles or endpoints for line segments.

2

Set up the given equation relating the two quantities.

\frac{n(n-1)(n-2)}{6} = 5 \times \frac{n(n-1)}{2}

The problem states that the number of triangles is

5

times the number of line segments.

3

Simplify the equation for

n \ge 3

.

Divide both sides by

\frac{n(n-1)}{2}

to get

\frac{n-2}{3} = 5

, which simplifies to

n - 2 = 15

.

Since

n \ge 3

,

n(n-1) \ne 0

, so dividing by common terms is valid.

4

Solve for

n

.

n = 15 + 2 = 17

Adding

2

to both sides isolates

n

.

Key Concept

Combinations in Geometry

Question 7Question →

An executive board consisting of 8 distinct members needs to form a subcommittee of 4 members. However, board members Alex and Blair refuse to serve on the subcommittee together unless board member Morgan is also selected. How many different 4-member subcommittees can be formed under these conditions?

Show answer & explanation

Answer: 60

Answer

60

The solution uses complementary counting. First, calculate the total possible 4-member subcommittees from 8 members without restrictions, which is

\binom{8}{4} = 70

. Second, identify the restricted scenario that is not allowed: Alex and Blair are both selected, but Morgan is excluded. In this invalid scenario, 2 spots are taken by Alex and Blair, Morgan is excluded from consideration, leaving 2 spots to be filled from the remaining 5 board members, which equals

\binom{5}{2} = 10

invalid subcommittees. Subtracting the invalid subcommittees from the total gives

70 - 10 = 60

valid subcommittees.

Step-by-Step Solution

1

Calculate the total number of ways to choose a 4-member subcommittee from 8 members without restrictions.

The total unrestricted combinations is

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

.

Since the order of selecting members into a subcommittee does not matter, use the combination formula

\binom{n}{k}

.

2

Determine the condition under which a subcommittee selection is invalid.

A subcommittee is invalid if and only if both Alex and Blair are selected AND Morgan is excluded.

Alex and Blair agree to serve together only if Morgan is also present. Thus, having Alex and Blair together without Morgan violates the condition.

3

Calculate the number of invalid subcommittees.

The number of invalid subcommittees is

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

.

Alex and Blair take 2 of the 4 spots, and Morgan cannot take any spot. The remaining 2 spots must be filled from the remaining

8 - 3 = 5

members.

4

Subtract the invalid subcommittees from the total unrestricted subcommittees using complementary counting.

70 - 10 = 60

valid subcommittees.

Complementary counting yields the total number of subcommittees that satisfy the restriction.

Key Concept

Combinations with Restrictions and Complementary Counting

Estimated Time:2m 0s

Question 8Question →

A corporate committee needs to form a 5-member project team selected from a department consisting of 4 data analysts, 5 software developers, and 4 cybersecurity experts. The team must include exactly 1 cybersecurity expert, at least 1 data analyst, and at least 1 software developer. How many different 5-member teams can be formed under these conditions?

124

240

480

504

1,120

Show answer & explanation

Answer: 480

Answer

480

To form a valid 5-member team, we first choose 1 cybersecurity expert from 4 available options in

\binom{4}{1} = 4

ways. The remaining 4 members must be selected from 4 data analysts and 5 software developers such that at least 1 of each role is included. The three mutually exclusive ways to fill these 4 spots are: 1 data analyst and 3 developers (40 ways), 2 data analysts and 2 developers (60 ways), or 3 data analysts and 1 developer (20 ways). Summing these gives 120 valid combinations for the remaining members. Multiplying by the 4 choices for the cybersecurity expert yields

4 \times 120 = 480

total distinct teams.

Step-by-Step Solution

1

Determine the number of ways to select the required cybersecurity expert

4 ways

Selecting exactly 1 cybersecurity expert from a total of 4 available experts is given by the combination

\binom{4}{1} = 4

.

2

Identify the possible valid distributions for the remaining 4 team members

Three valid cases: (1 DA, 3 SD), (2 DA, 2 SD), and (3 DA, 1 SD)

The team requires 5 members in total. With 1 cybersecurity expert selected, 4 positions remain. Since the team must contain at least 1 data analyst (DA) and at least 1 software developer (SD), we list all valid splits of 4 members.

3

Calculate the combinations for each valid distribution of data analysts and software developers

Case 1: 40 ways; Case 2: 60 ways; Case 3: 20 ways

Case 1 (1 DA, 3 SD):

\binom{4}{1} \times \binom{5}{3} = 4 \times 10 = 40

.
Case 2 (2 DA, 2 SD):

\binom{4}{2} \times \binom{5}{2} = 6 \times 10 = 60

.
Case 3 (3 DA, 1 SD):

\binom{4}{3} \times \binom{5}{1} = 4 \times 5 = 20

.

4

Sum the valid remaining member distributions and apply the Fundamental Counting Principle

480 total teams

Total remaining combinations =

40 + 60 + 20 = 120

. Total team combinations =

\binom{4}{1} \times 120 = 4 \times 120 = 480

.

Key Concept

Group selections with multiple category constraints and the Fundamental Counting Principle

Estimated Time:2m 30s

Question 9Question →

A restaurant culinary team is designing a tasting menu consisting of $4$ distinct dishes selected from a pool of $5$ appetizers and $6$ main courses. The tasting menu must contain at least $1$ appetizer and at least $2$ main courses. Furthermore, a specific appetizer, Dish A, can only be included on the menu if a specific main course, Dish B, is also included. How many different valid tasting menus can be formed?

180

200

240

250

310

Show answer & explanation

Answer: 200

Answer

200 different valid tasting menus can be formed.

The total number of valid tasting menus is

200

. This is calculated by breaking down the total requirement into two mutually exclusive dish distributions: 1 appetizer with 3 main courses (giving

90

valid menus) and 2 appetizers with 2 main courses (giving

110

valid menus), accounting for the restriction that selecting Dish A requires Dish B.

Step-by-Step Solution

1

Determine the valid distribution of dishes between appetizers and main courses.

Since

4

dishes must be selected with at least

1

appetizer and at least

2

main courses, the only possible distributions are: Case 1 (

1

appetizer,

3

main courses) and Case 2 (

2

appetizers,

2

main courses).

Selecting

3

appetizers and

1

main course violates the constraint of having at least

2

main courses.

2

Calculate the valid combinations for Case 1 (1 appetizer and 3 main courses).

Subcase 1a (Dish A not selected): Choose

1

appetizer from the remaining

4

appetizers and

3

main courses from all

6

main courses:

\binom{4}{1} \times \binom{6}{3} = 4 \times 20 = 80

.
Subcase 1b (Dish A selected): Dish B must also be selected. Choose no additional appetizers from the remaining

4

, and choose

2

more main courses from the remaining

5

main courses:

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

.
Total for Case 1:

80 + 10 = 90

.

Separating into cases based on whether Dish A is included properly enforces the conditional constraint.

3

Calculate the valid combinations for Case 2 (2 appetizers and 2 main courses).

Subcase 2a (Dish A not selected): Choose

2

appetizers from the remaining

4

appetizers and

2

main courses from all

6

main courses:

\binom{4}{2} \times \binom{6}{2} = 6 \times 15 = 90

.
Subcase 2b (Dish A selected): Dish B must also be selected. Choose

1

more appetizer from the remaining

4

appetizers and

1

more main course from the remaining

5

main courses:

\binom{4}{1} \times \binom{5}{1} = 4 \times 5 = 20

.
Total for Case 2:

90 + 20 = 110

.

Evaluates all valid menus having

2

appetizers and

2

main courses.

4

Sum the valid combinations from both cases.

Total valid menus =

90 + 110 = 200

.

The two main distribution cases are mutually exclusive.

Key Concept

Combinations with multiple category constraints and conditional dependencies

Question 10Question →

A university research department must select a 4-member committee from a group of 6 biologists and 4 chemists. If the committee must contain at least 1 biologist and at least 1 chemist, how many different 4-member committees can be formed?

Show answer & explanation

Answer: 194

Answer

194

Using complementary counting, the total number of ways to pick 4 members out of 10 is 210. The scenarios that break the rule are committees with all biologists (15 ways) or all chemists (1 way). Subtracting these 16 restricted selections from 210 results in 194 valid committees.

Step-by-Step Solution

1

Find total unrestricted combinations

210 total committees

Choosing any 4 people out of the pool of 10 without restrictions yields 10C4 = 210.

2

Calculate restricted/invalid cases

16 invalid committees

All-biologist groups account for 6C4 = 15 combinations, and all-chemist groups account for 4C4 = 1 combination.

3

Apply complementary counting

194 valid committees

Subtracting the 16 invalid combinations from the total 210 combinations gives 194.

Key Concept

Combinations with Restrictions (Complementary Counting)

Question 11Question →

A technology startup needs to form a 4-member project team selected from a pool of 5 front-end developers and 5 back-end developers. However, the selection is subject to one condition: if a specific front-end developer, Elena, is selected for the team, then a specific back-end developer, Marcus, must also be selected for the team. How many different 4-member teams can be formed under this condition?

Show answer & explanation

Answer: 154

Answer

154 different 4-member teams can be formed under the given condition.

The total number of ways to select a 4-member team from 10 developers without restrictions is

\binom{10}{4} = 210

. The conditional restriction requires that whenever Elena is on the team, Marcus must also be on the team. The only selection that violates this rule is one where Elena is included but Marcus is excluded. For this invalid scenario, Elena takes 1 spot and Marcus is removed from consideration, leaving

\binom{8}{3} = 56

ways to pick the remaining 3 members from the remaining 8 developers. Subtracting these 56 invalid teams from the 210 total teams gives 154 valid teams.

Step-by-Step Solution

1

Calculate the total possible 4-member teams from the 10 developers without any restrictions.

Total combinations =

\binom{10}{4} = 210

.

Choosing any 4 members out of a total pool of 10 developers (5 front-end and 5 back-end).

2

Identify and calculate the number of invalid team combinations.

Invalid combinations =

\binom{8}{3} = 56

.

The conditional rule is violated only if Elena is selected while Marcus is excluded. Selecting Elena fills 1 spot and excluding Marcus leaves 8 developers from which to choose the remaining 3 team members.

3

Subtract the invalid combinations from the total unconstrained combinations.

210 - 56 = 154

.

Using complementary counting yields the number of valid team selections directly and efficiently.

Key Concept

Combinations with Conditional Restrictions (Complementary Counting)

Estimated Time:1m 30s

Question 12Question →

A committee of $4$ delegates is to be selected from a group of $5$ diplomats from Country X and $4$ diplomats from Country Y. If $1$ specific diplomat from Country X and $1$ specific diplomat from Country Y cannot both serve on the committee together, how many different $4$ -member committees can be formed?

70

84

91

105

126

Show answer & explanation

Answer: 105

Answer

105

To find the number of valid 4-member committees, calculate the total combinations of choosing 4 delegates from 9 diplomats,

\binom{9}{4} = 126

. Then subtract the restricted outcomes where both specific diplomats are selected. If both specific diplomats are included, the remaining 2 positions on the committee must be chosen from the remaining 7 diplomats, which can be done in

\binom{7}{2} = 21

ways. Subtracting these invalid outcomes gives

126 - 21 = 105

.

Step-by-Step Solution

1

Calculate the total number of ways to choose 4 delegates from the total pool of 9 diplomats without any restrictions.

\binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126

The total pool consists of 5 diplomats from Country X + 4 diplomats from Country Y = 9 diplomats.

2

Calculate the number of invalid committee selections where both specific diplomats are selected together.

\binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21

If the 2 specific diplomats are both placed on the committee, 2 remaining seats must be filled from the remaining 7 diplomats.

3

Subtract the invalid combinations from the total combinations to find the number of valid committees.

126 - 21 = 105

Complementary counting gives the number of committees where both specific diplomats do not serve together.

Key Concept

Combinations with restrictions using complementary counting

Estimated Time:2m 0s

Question 13Question →

A cybersecurity firm must select a 4-tool diagnostic suite from a pool of 6 network scanners and 4 code analyzers. If the suite must contain at least 1 network scanner and at least 1 code analyzer, how many different 4-tool suites can be selected?

Show answer & explanation

Answer: 194

Answer

The total number of different 4-tool suites that can be selected is 194.

The total number of ways to select any 4 tools from the pool of 10 is

\binom{10}{4} = 210

. The only selections that violate the requirement of having at least one tool of each type are choosing 4 scanners and 0 analyzers (

\binom{6}{4} = 15

ways) or 0 scanners and 4 analyzers (

\binom{4}{4} = 1

way). Subtracting these 16 invalid outcomes from the total gives

210 - 16 = 194

valid 4-tool suites.

Step-by-Step Solution

1

Calculate the total unconstrained selections

210

There are

6 + 4 = 10

total tools, and 4 must be chosen.

\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210

.

2

Determine the number of invalid combinations violating the constraint

16 invalid combinations

The constraint requires at least 1 scanner and 1 analyzer. Invalid suites consist entirely of scanners (

\binom{6}{4} = 15

) or entirely of analyzers (

\binom{4}{4} = 1

).

3

Apply complementary counting to find the valid combinations

194 valid combinations

Subtract the invalid combinations from total unconstrained combinations:

210 - (15 + 1) = 194

.

Key Concept

Combinations with Group Selections and Complementary Counting

Combinations and Group Selections

Answer

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Common Mistakes

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Common Mistakes

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Common Mistakes