Combinations Practice Questions — JAMB UTME

Question 1Question →

A committee of $3$ members is to be selected from a group of $7$ people. In how many different ways can this committee be formed?

Show answer & explanation

Answer: 35

Answer

35

Selecting a committee of

3

members from

7

people requires calculating the number of combinations, given by

^7C_3 = \frac{7!}{3!4!} = 35

.

Step-by-Step Solution

1

Identify total elements and group size

n = 7

and

r = 3

Since the arrangement or order of members in the committee does not matter, this is a selection problem (combinations).

2

Apply the combinations formula

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

^7C_3 = \frac{7!}{3!4!}

This formula counts the distinct subsets of size

r

that can be chosen from

n

items.

3

Evaluate the factorial expression

^7C_3 = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35

Expanding

7!

as

7 \times 6 \times 5 \times 4!

allows cancelling

4!

, leaving

\frac{210}{6} = 35

.

Key Concept

Combinations formula

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

Estimated Time:45s

Question 2Question →

A convex polygon has $54$ diagonals. How many distinct triangles can be formed by joining any three of its vertices?

Show answer & explanation

Answer: 220

Answer

220 distinct triangles

Solving the equation for the number of diagonals

\frac{n(n-3)}{2} = 54

yields

n = 12

vertices. The number of triangles that can be formed by selecting any 3 of these 12 vertices is given by

\binom{12}{3} = \frac{12 \times 11 \times 10}{6} = 220

.

Step-by-Step Solution

1

Determine the number of vertices

n

of the polygon using the diagonals formula.

n = 12

The number of diagonals

D

in an

n

-sided convex polygon is given by

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

. Setting

\frac{n(n-3)}{2} = 54

gives

n^2 - 3n - 108 = 0

. Factoring

(n - 12)(n + 9) = 0

yields

n = 12

since the number of vertices must be positive.

2

Calculate the number of distinct triangles formed by choosing 3 vertices from 12.

220

Each set of 3 distinct vertices forms one unique triangle. The order in which the vertices are chosen does not matter, so we use combinations:

\binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220

.

Key Concept

Combinations applied to geometric figures and polygon properties

Estimated Time:2m 0s

Question 3Question →

A school committee of $4$ members is to be selected from $6$ male teachers and $4$ female teachers. If the committee must contain exactly $2$ male teachers and $2$ female teachers, how many different committees can be formed?

Show answer & explanation

Answer: 90

Answer

The total number of different committees that can be formed is

90

.

Selecting 2 female teachers from 4 yields

\binom{4}{2} = 6

ways. Selecting 2 male teachers from 6 yields

\binom{6}{2} = 15

ways. By the multiplication principle, the total number of ways to form the committee is

6 \times 15 = 90

.

Step-by-Step Solution

1

Find the number of ways to select 2 female teachers out of 4

$\binom{4}{2} = 6$

Selection order does not matter within the committee, so combination formula $nCr$ applies.

2

Find the number of ways to select 2 male teachers out of 6

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

Selection order does not matter within the committee, so combination formula $nCr$ applies.

3

Multiply the possibilities for selecting male and female members

6 \times 15 = 90

According to the fundamental counting principle, independent group selections are multiplied.

Key Concept

Combinations with restricted subset selections (Product Rule of Counting)

Question 4Question →

$10$ distinct points are marked on the circumference of a circle. How many different triangles can be formed by connecting any $3$ of these points as vertices?

45

120

360

720

Show answer & explanation

Answer: $120$

Answer

The number of distinct triangles that can be formed is

120

.

To form a triangle, any

3

points must be chosen from the

10

available points. Because no three points on a circle are collinear, every choice of

3

points forms a unique triangle. Since the order of choosing vertices does not alter the triangle, we use combinations:

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

.

Step-by-Step Solution

1

Identify the total number of items

n

and the subset size

r

.

n = 10

points on the circle, and

r = 3

points required to form a triangle.

Any set of

3

non-collinear points uniquely determines a triangle. Since all points lie on a circle, no three points are collinear.

2

Determine whether order matters.

Order does not matter because choosing points

A, B, C

produces the same triangle as choosing

B, C, A

.

Selection of vertices for a geometric shape is a combination problem, not a permutation problem.

3

Apply the combination formula

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

.

\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120

Evaluating the combination gives the total number of distinct triangles.

Key Concept

Combinations for Geometric Formations

Estimated Time:1m 30s

Question 5Question →

In a mathematics examination, a student is required to answer $5$ questions out of $8$ available questions. If the first $2$ questions are compulsory, in how many different ways can the student select the questions to answer?

Show answer & explanation

Answer: 20

Answer

The student can select the questions in 20 different ways.

With 2 compulsory questions, the student only has to select 3 more questions from the remaining 6 questions. The number of ways to select 3 items from 6 without regard to order is given by ^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20.

Step-by-Step Solution

1

Deduct compulsory questions from both the required total and available total

The student must choose 3 additional questions from the remaining 6 questions.

Compulsory questions are fixed and provide only 1 selection choice.

2

Apply the combination formula ^6C_3 to calculate selection ways

^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20

The order in which the student chooses the examination questions does not alter the group of questions selected.

Key Concept

Combinations with restricted/compulsory choices

Question 6Question →

A fruit vendor has $7$ distinct types of fresh fruits on display. A customer wants to buy a gift basket containing exactly $4$ different types of fruits. How many different combinations of fruits can the customer choose?

Show answer & explanation

Answer: 35

Answer

The customer can choose the fruits in 35 different ways.

The number of ways to choose

r = 4

items from

n = 7

distinct items without regard to order is given by

^7C_4 = \frac{7!}{4!3!} = \frac{210}{6} = 35

.

Step-by-Step Solution

1

Identify total elements (

n

) and selected subset size (

r

).

n = 7

and

r = 4

.

Since the selection order inside the fruit basket does not matter, combinations (

nCr

) must be used.

2

Substitute values into the combination formula

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

.

^7C_4 = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!}

.

This evaluates the total ways to choose 4 items from 7 without repetition or ordering.

3

Simplify the factorials and compute the numerical result.

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

.

Canceling out

4!

from numerator and denominator gives

(7 \times 6 \times 5) / 6 = 35

.

Key Concept

Combinations (

nCr

)

Estimated Time:45s

Question 7Question →

A panel of $5$ members is to be selected from $10$ eligible candidates. If $2$ specific candidates refuse to serve on the panel together, in how many different ways can the panel be formed?

Show answer & explanation

Answer: 196

Answer

The panel can be formed in 196 different ways.

To find the number of valid panels, use complementary counting. First, compute the total number of ways to pick any 5 candidates from 10 without restrictions:

\binom{10}{5} = 252

. Next, find the number of invalid panels that contain both restricted candidates; since 2 candidates are already placed, pick the remaining 3 members from the remaining 8 candidates:

\binom{8}{3} = 56

. Subtracting these invalid panels from the total gives

252 - 56 = 196

.

Step-by-Step Solution

1

Calculate the total possible combinations without any restrictions.

\binom{10}{5} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252

The combinations formula

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

applies since the order of selection does not matter.

2

Determine the number of invalid combinations where both specific candidates are included.

\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56

If both specific candidates are already placed on the panel, 3 more members must be chosen from the remaining 8 candidates.

3

Apply complementary counting to subtract invalid selections from total selections.

252 - 56 = 196

Subtracting the restricted combinations from the total combinations gives the number of valid panel configurations.

Key Concept

Combinations with mutual exclusion (Complementary Counting)

Question 8Question →

A project team of $5$ members is to be formed from $6$ senior engineers and $4$ junior engineers. If the team must contain strictly more senior engineers than junior engineers, in how many different ways can the team be selected?

180

252

186

3600

Show answer & explanation

Answer: $186$

Answer

186

To form a

5

-member team where senior engineers outnumber junior engineers, we consider three mutually exclusive cases: (5 senior, 0 junior), (4 senior, 1 junior), and (3 senior, 2 junior). Evaluating

^6C_5 \times ^4C_0 = 6

,

^6C_4 \times ^4C_1 = 60

, and

^6C_3 \times ^4C_2 = 120

, and adding them together yields

6 + 60 + 120 = 186

total ways.

Step-by-Step Solution

1

Identify the conditions and valid cases for team composition

The team must have

5

members. The number of senior engineers must be strictly greater than junior engineers. The valid combinations of (senior, junior) engineers are:

(5, 0)

,

(4, 1)

, and

(3, 2)

.

Any other breakdown (such as 2 senior and 3 junior) would result in senior engineers not outnumbering junior engineers.

2

Calculate the combinations for Case 1 (

5

senior,

0

junior)

^6C_5 \times ^4C_0 = 6 \times 1 = 6

Choosing

5

out of

6

senior engineers and

0

out of

4

junior engineers.

3

Calculate the combinations for Case 2 (

4

senior,

1

junior)

^6C_4 \times ^4C_1 = 15 \times 4 = 60

Choosing

4

out of

6

senior engineers and

1

out of

4

junior engineers.

4

Calculate the combinations for Case 3 (

3

senior,

2

junior)

^6C_3 \times ^4C_2 = 20 \times 6 = 120

Choosing

3

out of

6

senior engineers and

2

out of

4

junior engineers.

5

Sum the combinations across all mutually exclusive cases

Total ways =

6 + 60 + 120 = 186

By the addition principle of counting, the total number of ways is the sum of ways from each disjoint case.

Key Concept

Conditional combinations using the formula

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

and the addition principle for disjoint cases.

Question 9Question →

A sports delegation of $6$ athletes is to be selected from a pool of $7$ sprinters and $5$ distance runners. If the delegation must contain at least $4$ sprinters, in how many different ways can the delegation be formed?

Show answer & explanation

Answer: 462

Answer

The total number of different ways to form the delegation is 462.

To select at least 4 sprinters in a delegation of 6 athletes, we must evaluate three mutually exclusive cases: selecting 4 sprinters and 2 distance runners (

{}^7\text{C}_4 \times {}^5\text{C}_2 = 350

), selecting 5 sprinters and 1 distance runner (

{}^7\text{C}_5 \times {}^5\text{C}_1 = 105

), and selecting 6 sprinters and 0 distance runners (

{}^7\text{C}_6 \times {}^5\text{C}_0 = 7

). Adding these yields

350 + 105 + 7 = 462

total ways.

Step-by-Step Solution

1

Determine all valid combinations of sprinters and distance runners satisfying the condition of having at least 4 sprinters in a group of 6.

Three valid cases: (4 sprinters, 2 distance runners), (5 sprinters, 1 distance runner), and (6 sprinters, 0 distance runners).

The delegation requires 6 members and at least 4 sprinters.

2

Calculate the combinations for each case using

{}^n\text{C}_r = \frac{n!}{r!(n-r)!}

.

Case 1:

{}^7\text{C}_4 \times {}^5\text{C}_2 = 350

; Case 2:

{}^7\text{C}_5 \times {}^5\text{C}_1 = 105

; Case 3:

{}^7\text{C}_6 \times {}^5\text{C}_0 = 7

.

Apply the product rule of counting for selecting sprinters and distance runners independently within each case.

3

Sum the results of the mutually exclusive cases.

350 + 105 + 7 = 462.

Apply the addition principle of counting for mutually exclusive scenarios.

Key Concept

Combinations with restrictions and the addition principle of counting

Estimated Time:2m 0s

Question 10Question →

A container contains $6$ red counters, $4$ blue counters, and $3$ green counters. In how many different ways can a selection of $5$ counters be made if the selection must contain at least $2$ red counters, at least $1$ blue counter, and at most $1$ green counter?

Show answer & explanation

Answer: 750

Answer

The total number of different ways to make the selection is 750.

Breaking the problem into disjoint cases based on the number of green counters (0 or 1) and applying the combination formula for red, blue, and green counters in each valid configuration yields

240 + 510 = 750

total ways.

Step-by-Step Solution

1

Determine the allowable counts for Green (

G

), Red (

R

), and Blue (

B

) counters

Green counters can be 0 or 1. If

G=0

,

R+B=5

with

R \geq 2, B \geq 1

. If

G=1

,

R+B=4

with

R \geq 2, B \geq 1

.

The constraints state

G \leq 1

,

R \geq 2

, and

B \geq 1

for a total of 5 counters.

2

Calculate combinations for Case 1 (

G = 0

)

Ways for

(2R, 3B, 0G) = 60

;

(3R, 2B, 0G) = 120

;

(4R, 1B, 0G) = 60

. Total for Case 1 = 240.

Apply the combinations formula

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

to each sub-case.

3

Calculate combinations for Case 2 (

G = 1

)

Ways for

(2R, 2B, 1G) = 270

;

(3R, 1B, 1G) = 240

. Total for Case 2 = 510.

Apply combinations to choose 1 Green counter along with the valid Red and Blue combinations.

4

Sum all mutually exclusive cases

Total selection ways =

240 + 510 = 750

.

According to the addition principle of counting, the totals of mutually exclusive cases are added together.

Key Concept

Combinations with Multiple Conditional Constraints

Question 11Question →

A research committee of $5$ members is to be formed from $6$ male scientists and $5$ female scientists. The committee must contain strictly more male scientists than female scientists. If one particular male scientist and one particular female scientist refuse to serve on the same committee together, in how many different ways can the committee be selected?

231

281

1,281

256

Show answer & explanation

Answer: 231

Answer

The total number of valid ways to select the committee is 231.

The correct response of 231 is found by calculating the total committees having strictly more men than women (281) and subtracting the subset of those committees where both restricted individuals are chosen together (50).

Step-by-Step Solution

1

Calculate the unconstrained ways to form a 5-member committee with strictly more males than females.

Three possible gender distributions exist:
- 5 Males, 0 Females:

\binom{6}{5} \times \binom{5}{0} = 6 \times 1 = 6

- 4 Males, 1 Female:

\binom{6}{4} \times \binom{5}{1} = 15 \times 5 = 75

- 3 Males, 2 Females:

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

Total unconstrained ways =

6 + 75 + 200 = 281

.

Establishing the total pool of valid gender-ratio committees before applying the exclusion condition.

2

Calculate the number of forbidden committee selections where both the restricted male scientist and restricted female scientist are included together.

Since 1 male and 1 female are already chosen, 3 remaining members must be selected from the remaining 5 males and 4 females:
- For 4 Males, 1 Female total (need 3 more males, 0 more females):

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

- For 3 Males, 2 Females total (need 2 more males, 1 more female):

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

Total forbidden ways =

10 + 40 = 50

.

To enforce the restriction that both specific members cannot serve together, calculate all committee configurations containing both.

3

Subtract the forbidden selections from the unconstrained selections.

Valid ways =

281 - 50 = 231

.

Applying the subtraction principle of combinations to yield the final allowed arrangements.

Key Concept

Combinations with gender ratio constraints and mutual exclusion restrictions

Estimated Time:2m 30s

Question 12Question →

A plane contains $12$ distinct points such that exactly $5$ of them lie on the same straight line, while no other subset of three points is collinear. How many distinct straight lines can be formed by joining pairs of these points?

Show answer & explanation

Answer: 57

Answer

The total number of distinct straight lines that can be formed is 57.

The total number of distinct straight lines is obtained by taking all possible pairs of points from 12, which is

^{12}C_2 = 66

, subtracting the pairs formed among the 5 collinear points (

^5C_2 = 10

), and adding 1 back to account for the single straight line on which those 5 points lie. This gives

66 - 10 + 1 = 57

.

Step-by-Step Solution

1

Calculate the total possible lines by selecting any 2 points out of 12.

^{12}C_2 = \frac{12 \times 11}{2} = 66

Any pair of non-collinear points uniquely determines a straight line.

2

Calculate the combinations of 2 points selected from the 5 collinear points.

^5C_2 = \frac{5 \times 4}{2} = 10

These 10 point pairs all lie on the exact same single straight line instead of forming 10 distinct lines.

3

Adjust the count to find the net number of distinct straight lines.

66 - 10 + 1 = 57

Subtracting the 10 overcounted combinations and adding back 1 for the single line passing through all 5 collinear points gives the correct total.

Key Concept

Combinations with Collinear Constraints

Question 13Question →

A convex polygon has $44$ diagonals. What is the total number of distinct triangles that can be formed by joining any three vertices of this polygon?

Show answer & explanation

Answer: 165

Answer

165

The number of diagonals of an

n

-sided convex polygon is given by

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

. Setting this equal to

44

yields

n(n-3) = 88

, which simplifies to

n^2 - 3n - 88 = 0

. Factoring gives

(n-11)(n+8) = 0

, so

n = 11

. The total number of distinct triangles formed by choosing any 3 vertices from an 11-sided polygon is

\binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165

.

Step-by-Step Solution

1

Set up the equation for the number of diagonals in terms of the number of vertices

n

\frac{n(n-3)}{2} = 44

Choosing any 2 vertices from

n

vertices gives

\binom{n}{2}

total connecting line segments. Subtracting the

n

boundary sides leaves the diagonals.

2

Solve the quadratic equation for

n

n^2 - 3n - 88 = 0 \implies (n-11)(n+8) = 0 \implies n = 11

A polygon must have a positive integer number of vertices, so

n = 11

.

3

Compute the number of distinct triangles using combinations

\binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165

In a convex polygon, no three vertices are collinear, so every unique combination of 3 vertices forms a distinct triangle.

Key Concept

Application of combinations to geometry (polygon diagonals and triangle selection)

Estimated Time:2m 0s

Question 14Question →

A student library committee is selecting $5$ distinct books from a shelf containing $7$ novel titles and $5$ biography titles. If $2$ specific novel titles are mutually exclusive (they cannot both be selected together in the same combination), in how many ways can the selection of $5$ books be made such that at least $3$ novel titles are included?

426

436

516

546

Show answer & explanation

Answer: 436

Answer

436

To find the number of valid ways, we first find the total number of ways to select at least

3

novels from

7

novels and

5

biographies, which equals

546

. Next, we determine how many of these combinations contain both of the restricted novels while still having at least

3

novels overall. There are

110

such invalid selections. Subtracting

110

from

546

gives

436

valid ways.

Step-by-Step Solution

1

Calculate the total combinations with at least 3 novels without any restriction.

Case 1 (

3

Novels,

2

Biographies):

\binom{7}{3} \times \binom{5}{2} = 35 \times 10 = 350

.
Case 2 (

4

Novels,

1

Biography):

\binom{7}{4} \times \binom{5}{1} = 35 \times 5 = 175

.
Case 3 (

5

Novels,

0

Biographies):

\binom{7}{5} \times \binom{5}{0} = 21 \times 1 = 21

.
Total unrestricted combinations =

350 + 175 + 21 = 546

.

Establishing the total pool of choices that satisfy the constraint of selecting at least 3 novels.

2

Calculate the invalid combinations where both restricted novels are selected together AND at least 3 novels are included.

If both restricted novels are selected, we have already chosen

2

novels. To reach a total of

5

books with at least

3

novels, we must pick

3

additional books from the remaining

10

books (

5

remaining novels and

5

biographies), EXCLUDING the scenario where all

3

additional books are biographies (which would leave us with only

2

novels in total).
Combinations of

3

books from

10

remaining books =

\binom{10}{3} = 120

.
Combinations of

3

biographies from

5

biographies =

\binom{5}{3} = 10

.
Invalid combinations =

120 - 10 = 110

.

To apply the mutual exclusion constraint correctly, we must subtract only those invalid selections that also satisfy the condition of having at least 3 novels.

3

Subtract invalid combinations from total unrestricted combinations.

Valid combinations =

546 - 110 = 436

.

Subtracting the forbidden overlapping outcomes yields the net valid choices.

Key Concept

Combinations with multiple constraints and mutual exclusion

Question 15Question →

A music festival coordinator needs to select $4$ bands to perform from a pool of $8$ available bands. If $2$ specific bands insist on either both being selected or neither being selected, in how many different ways can the $4$ bands be chosen?

30

60

70

390

Show answer & explanation

Answer: $30$

Answer

The total number of ways to choose the bands under the given condition is

30

.

To satisfy the condition that the two specific bands are either both chosen or neither chosen, we evaluate two distinct cases. Case 1 (both chosen) requires choosing

2

additional bands from the remaining

6

, yielding

{}^6C_2 = 15

ways. Case 2 (neither chosen) requires choosing all

4

bands from the remaining

6

, yielding

{}^6C_4 = 15

ways. Adding both cases gives

15 + 15 = 30

ways.

Step-by-Step Solution

1

Analyze Case 1: Both specific bands are selected.

If both specific bands are included, we only need to select

2

more bands from the remaining

6

bands. The number of ways is

{}^6C_2 = \frac{6 \times 5}{2 \times 1} = 15

.

Since order does not matter in forming a group of performers, we use combinations.

2

Analyze Case 2: Neither of the specific bands is selected.

If neither of the

2

specific bands is chosen, all

4

bands must be selected from the remaining

6

bands. The number of ways is

{}^6C_4 = {}^6C_2 = 15

.

Excluding the

2

specific bands leaves

6

candidate bands to choose

4

from.

3

Sum the mutually exclusive cases.

\text{Total ways} = 15 + 15 = 30.

The two scenarios are disjoint, so the addition principle applies.

Key Concept

Combinations with conditional restrictions

Estimated Time:1m 30s

Question 16Question →

A committee of $5$ members is to be selected from $6$ doctors and $4$ nurses. In how many ways can this committee be formed if it must include at least $3$ doctors?

186

120

720

3600

Show answer & explanation

Answer: 186

Answer

186 ways

The required committee must contain at least 3 doctors out of 5 total members. The three possible scenarios are: 3 doctors and 2 nurses (

^6C_3 \times ^4C_2 = 120

), 4 doctors and 1 nurse (

^6C_4 \times ^4C_1 = 60

), and 5 doctors and 0 nurses (

^6C_5 \times ^4C_0 = 6

). Summing these gives

120 + 60 + 6 = 186

.

Step-by-Step Solution

1

Identify the possible valid committee compositions given the condition 'at least 3 doctors'.

Three mutually exclusive cases exist for a 5-member committee: (3 doctors, 2 nurses), (4 doctors, 1 nurse), or (5 doctors, 0 nurses).

The total size of the committee is 5, so selecting more doctors reduces the required number of nurses.

2

Calculate the combinations for Case 1: 3 doctors and 2 nurses.

^6C_3 \times ^4C_2 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} \times \frac{4 \times 3}{2 \times 1} = 20 \times 6 = 120

Order of selection within each group does not matter.

3

Calculate the combinations for Case 2: 4 doctors and 1 nurse.

^6C_4 \times ^4C_1 = \frac{6 \times 5}{2 \times 1} \times 4 = 15 \times 4 = 60

Selecting 4 doctors out of 6 is equivalent to choosing which 2 doctors to exclude.

4

Calculate the combinations for Case 3: 5 doctors and 0 nurses.

^6C_5 \times ^4C_0 = 6 \times 1 = 6

Choosing 5 doctors out of 6 yields 6 possibilities, and choosing 0 nurses yields 1.

5

Sum the number of ways from all valid cases.

120 + 60 + 6 = 186

By the addition principle of counting, the total ways to form the committee is the sum of ways across mutually exclusive cases.

Key Concept

Combinations with constraints (addition and multiplication principles)

Question 17Question →

In a mathematics examination paper consisting of $8$ questions, a candidate is required to answer $5$ questions in total. If the candidate must answer the first $2$ questions, in how many ways can the candidate select the remaining questions?

Show answer & explanation

Answer: 20

Answer

The candidate can select the questions in 20 ways.

Because the first 2 questions are mandatory, the choice is reduced to picking 3 additional questions from the remaining 6 questions. Since selection order is irrelevant, the number of distinct ways is $\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.

Step-by-Step Solution

1

Determine the remaining number of questions needed

3 questions

Since 2 out of the required 5 questions are compulsory, the candidate must choose 3 more.

2

Determine the available pool of remaining questions

6 questions

Subtracting the 2 compulsory questions from the 8 total questions leaves 6 questions available.

3

Calculate the combinations using the nCr formula

20 ways

The order in which questions are selected does not matter, so we use combinations: $\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.

Key Concept

Combinations with restricted or fixed choices

Question 18Question →

A international conference delegation of $6$ members is to be selected from $5$ diplomats and $4$ translators. In how many distinct ways can the delegation be formed if it must include at least $4$ diplomats?

Show answer & explanation

Answer: 34

Answer

The total number of distinct ways to form the delegation is 34.

To form a delegation of 6 with at least 4 diplomats out of 5 diplomats and 4 translators, we evaluate two mutually exclusive cases: selecting 4 diplomats and 2 translators (5C4 * 4C2 = 30 ways) and selecting 5 diplomats and 1 translator (5C5 * 4C1 = 4 ways). Summing these gives 34 distinct ways.

Step-by-Step Solution

1

Determine the valid combinations of diplomats and translators.

Two valid cases exist: (4 diplomats, 2 translators) and (5 diplomats, 1 translator).

The total delegation size is 6 and it must contain at least 4 diplomats out of the 5 available.

2

Compute the selection ways for Case 1 (4 diplomats and 2 translators).

5 * 6 = 30 ways

Selecting 4 diplomats out of 5 is 5C4 = 5 ways, and selecting 2 translators out of 4 is 4C2 = 6 ways.

3

Compute the selection ways for Case 2 (5 diplomats and 1 translator).

1 * 4 = 4 ways

Selecting 5 diplomats out of 5 is 5C5 = 1 way, and selecting 1 translator out of 4 is 4C1 = 4 ways.

4

Sum the combinations from both mutually exclusive cases.

30 + 4 = 34 ways

By the addition principle of counting, mutually exclusive scenarios are added together.

Key Concept

Combinations with constraints and mutually exclusive cases

Question 19Question →

For a science exhibition, a student coordinator needs to choose $4$ project displays from a pool of $6$ chemistry projects and $4$ physics projects. If the selection must include at least $2$ chemistry projects, in how many distinct ways can the projects be selected?

170

185

210

1,200

Show answer & explanation

Answer: $185$

Answer

The total number of distinct ways to select the projects is

185

.

The correct total number of ways is

185

. The condition requiring at least

2

chemistry projects splits the total selection into three mutually exclusive scenarios: selecting

2

chemistry and

2

physics projects (

90

ways),

3

chemistry and

1

physics project (

80

ways), or

4

chemistry projects (

15

ways). Adding these together yields

90 + 80 + 15 = 185

.

Step-by-Step Solution

1

Identify the distinct valid cases for selecting

4

projects with at least

2

chemistry projects.

Case 1:

2

chemistry and

2

physics; Case 2:

3

chemistry and

1

physics; Case 3:

4

chemistry and

0

physics.

The phrase 'at least 2 chemistry projects' means the number of chemistry projects can be

2

,

3

, or

4

.

2

Calculate the combinations for Case 1 (

2

chemistry,

2

physics).

$\binom{6}{2} \times \binom{4}{2} = 15 \times 6 = 90$ ways.

We select

2

projects out of

6

chemistry projects and

2

out of

4

physics projects using the combination formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$.

3

Calculate the combinations for Case 2 (

3

chemistry,

1

physics).

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

We select

3

projects out of

6

chemistry projects and

1

out of

4

physics projects.

4

Calculate the combinations for Case 3 (

4

chemistry,

0

physics).

$\binom{6}{4} \times \binom{4}{0} = 15 \times 1 = 15$ ways.

We select

4

projects out of

6

chemistry projects and

0

out of

4

physics projects.

5

Sum the number of ways from all mutually exclusive cases.

$90 + 80 + 15 = 185$ ways.

By the addition principle of counting, the total number of ways is the sum of the ways from each independent case.

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