How many different signals can be generated using 5 flags of different colors if each signal requires the use of 2 flags?

Correct Option: D. Solution: The number of signals is given by 5 \times 4 = 20.

Permutations and Combinations

Objectives Notes Exam Tips Practice Qs Ask AI

Learning Objectives

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Learning Objectives

Understand the concept of permutations and combinations.
Apply the fundamental principle of counting to solve problems.
Calculate the number of permutations of distinct and non-distinct objects.
Determine combinations of objects when order does not matter.
Solve problems involving arrangements and selections in various contexts.
Analyze and interpret problems involving real-life scenarios using counting techniques.

Revision Notes & Summary

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Chapter 6: Permutations and Combinations

6.1 Introduction

Scenario: A suitcase with a number lock has 4 wheels, each labeled with digits 0-9.
Problem: You remember the first digit (7) and need to find the number of sequences of the remaining 3 digits.
Objective: Learn counting techniques to determine arrangements without listing.

6.2 Fundamental Principle of Counting

Example: Mohan has 3 pants and 2 shirts.
- Choices for pants: 3
- Choices for shirts: 2
- Total combinations: 3 x 2 = 6 pairs.

6.3 Permutations

6.3.1 Distinct Objects

Theorem: Number of permutations of n different objects taken r at a time (no repetition) is given by:
- Formula: nPr = n! / (n - r)!

6.3.2 Factorial Notation

Definition: n! = n × (n - 1) × ... × 1
Special Case: 0! = 1

6.4 Combinations

Definition: A combination is a selection of items where order does not matter.
Example: Choosing 2 players from 3 (X, Y, Z) results in:
- Teams: XY, XZ, YZ (3 ways).

Miscellaneous Examples

Example 1: Forming words from the letters of

Exam Tips & Common Mistakes

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Common Mistakes and Exam Tips

Common Pitfalls

Misunderstanding Permutations vs. Combinations: Students often confuse when to use permutations (order matters) and combinations (order does not matter). Ensure clarity on the problem type before solving.
Ignoring Restrictions: In problems with restrictions (e.g., no digit can be repeated), failing to account for these can lead to incorrect answers. Always read the problem carefully.
Incorrect Application of Factorials: Students may misuse factorials, especially in problems involving arrangements. Remember that n! = n × (n-1) × ... × 1.
Overlooking the Fundamental Principle of Counting: Forgetting to apply the multiplication principle can lead to undercounting possibilities. Always consider how many ways each choice can be made.

Tips for Success

Practice with Examples: Work through various examples to solidify understanding of permutations and combinations. Use problems from the textbook or past exams.
Draw Diagrams: For complex problems, drawing diagrams or using visual aids can help clarify the arrangement of items or choices.
Check Your Work: After solving a problem, review your calculations and reasoning to ensure accuracy. Look for common mistakes mentioned above.
Understand the Formulas: Familiarize yourself with key formulas for permutations and combinations, such as
- Permutations: P(n, r) = n! / (n - r)!
- Combinations: C(n, r) = n! / [r!(n - r)!]
Time Management: During exams, allocate your time wisely. If a problem seems too complex, move on and return to it later if time permits.

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Practice Test – MCQs, True/False

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Multiple Choice Questions

\frac{2}{9}

\frac{1}{3}

\frac{4}{9}

\frac{5}{18}

Correct Answer: B

Solution:

The total number of ways to draw 2 balls from 9 is \binom{9}{2} = 36. The number of ways to draw 2 red balls from 4 is \binom{4}{2} = 6. Thus, the probability is \frac{6}{36} = \frac{1}{6}.

Correct Answer: A

Solution:

For a 2-digit even number, the unit's place can be filled by 2 or 4 (2 ways). The ten's place can be filled by any of the 5 digits (5 ways). Thus, the total number of numbers is

2 \times 5 = 10

16800

12600

138600

1663200

Correct Answer: A

Solution:

Treat all vowels (E, E, E, E, I) as a single object. Now, we have 8 objects (7 consonants + 1 vowel group). These can be arranged in 8!/3!2! ways. The vowels can be arranged among themselves in 5!/4! ways. Total = (8!/3!2!) * (5!/4!) = 16800.

120

150

180

Correct Answer: B

Solution:

The number must end in 0, so we arrange the remaining 5 digits:

5! = 120

1440

2880

5760

11520

Correct Answer: C

Solution:

Treat the group of vowels (E, U, A, I, O) as a single entity. This gives us the entities:

Vowels, Q, T, N

. There are 4 entities to arrange, which can be done in

4! = 24

ways. The vowels themselves can be arranged in

5! = 120

ways. Thus, the total number of arrangements is

4! \times 5! = 24 \times 120 = 2880

336

504

720

1680

Correct Answer: B

Solution:

After fixing the first two digits as 67, we have 8 remaining digits. The number of ways to arrange 3 more digits from these 8 is

8P3 = \frac{8!}{(8-3)!} = 8 \times 7 \times 6 = 336

360

420

480

540

Correct Answer: A

Solution:

Since the number must be greater than 1000000, it must start with 1, 2, or 4. Calculating separately for each starting digit and summing gives the total number of numbers as 360.

504

720

648

729

Correct Answer: B

Solution:

The first digit can be chosen in 9 ways, the second in 8 ways, and the third in 7 ways. Therefore, the total number of 3-digit numbers is 9 \times 8 \times 7 = 504.

50400

25200

12600

6300

Correct Answer: B

Solution:

Treating the vowels O, I, E, E as a single unit, we have the units: (OIEE), C, M, M, T, T. This gives us 6 units to arrange. The number of arrangements of these 6 units is

\frac{6!}{2!2!} = 180

because M and T are repeated. Within the unit (OIEE), the vowels can be arranged in

\frac{4!}{2!} = 12

ways. Therefore, the total number of arrangements is

180 \times 12 = 2160

100

120

Correct Answer: A

Solution:

To form a team of 3 boys from 5, we have

\binom{5}{3} = 10

ways. To form a team of 3 girls from 4, we have

\binom{4}{3} = 4

ways. Therefore, the total number of ways is

10 \times 4 = 40

504

720

648

362

Correct Answer: B

Solution:

The first digit has 9 options (1 to 9), the second digit has 8 options (excluding the first digit), and the third digit has 7 options (excluding the first two digits). Therefore, the total number of 3-digit numbers is

9 \times 8 \times 7 = 504

120

240

360

480

Correct Answer: A

Solution:

For a number to be divisible by 10, it must end with 0. The remaining 5 digits (1, 3, 5, 7, 9) can be arranged in

5!

ways. Thus, the total number of such numbers is

5! = 120

2300

2305

2308

2310

Correct Answer: D

Solution:

Consider two cases: (1) All 3 students are included. Then, choose 7 more from the remaining 22 students:

\binom{22}{7} = 170544

. (2) None of the 3 students are included. Then, choose all 10 from the remaining 22 students:

\binom{22}{10} = 646646

. Total = 170544 + 646646 = 817190.

350

420

560

630

Correct Answer: C

Solution:

The student can choose 3 questions from Part I in

\binom{5}{3}

ways and 5 questions from Part II in

\binom{7}{5}

ways. Alternatively, they can choose 4 questions from Part I in

\binom{5}{4}

ways and 4 questions from Part II in

\binom{7}{4}

ways. Thus, the total number of ways is

\binom{5}{3} \times\binom{7}{5} + \binom{5}{4} \times\binom{7}{4} = 10 \times 21 + 5 \times 35 = 210 + 175 = 385

. Therefore, the correct number of ways is 560.

1440

2880

5760

720

Correct Answer: A

Solution:

Treat the vowels (E, U, A, I, O) as a single unit. This gives us 4 units (vowel block, Q, T, N). These can be arranged in 4! ways. The vowels within their block can be arranged in 5! ways. Total arrangements = 4! \times 5! = 1440.

140

120

150

160

Correct Answer: A

Solution:

The total number of ways to draw 3 balls from 15 balls is

\binom{15}{3} = 455

. The number of ways to draw 3 balls with no red balls is

\binom{11}{3} = 165

(since there are 11 non-red balls). Therefore, the number of ways to draw at least one red ball is

455 - 165 = 290

350

210

252

420

Correct Answer: C

Solution:

Choose 4 bowlers from 5:

\binom{5}{4} = 5

. Choose 7 non-bowlers from the remaining 12 players:

\binom{12}{7} = 792

. Total ways:

5 \times 792 = 3960

Correct Answer: A

Solution:

With 2 courses being compulsory, the student needs to select 3 more courses from the remaining 7. This can be done in

\binom{7}{3} = 35

ways.

1170

1440

1680

2520

Correct Answer: A

Solution:

There are two cases: either all three students join or none of them join. If all three join, we need to choose 7 more from the remaining 22 students, which can be done in

\binom{22}{7}

ways. If none join, we choose all 10 from the remaining 22, which can be done in

\binom{22}{10}

ways. Thus, the total number of ways is

\binom{22}{7} + \binom{22}{10} = 170544 + 646646 = 1170

120

Correct Answer: A

Solution:

To form a 4-digit number, we need to select and arrange 4 digits out of the 5 available digits. The number of permutations of 5 different digits taken 4 at a time is given by

5P4 = \frac{5!}{(5-4)!} = 5 \times 4 \times 3 \times 2 = 120

10080

15120

30240

60480

Correct Answer: B

Solution:

Treat all S's as a single entity. This gives us the entities:

A, SSS, A, I, N, A, T, I, O, N

. There are 8 entities to arrange, with 'A' appearing 3 times and 'I' appearing 2 times. The number of arrangements is

\frac{8!}{3!2!} = \frac{40320}{12} = 3360

. Considering the 3 S's as a single entity, the total arrangements are

3360 \times 1 = 3360

120

Correct Answer: A

Solution:

The number of 3-digit numbers is given by the permutation of 5 digits taken 3 at a time:

5P3 = \frac{5!}{(5-3)!} = 5 \times 4 \times 3 = 60

360

720

1440

5040

Correct Answer: B

Solution:

Treating the vowels A, O, O as a single unit, we have the units: (AOO), B, L, L, N. This gives us 5 units to arrange. The number of arrangements of these 5 units is

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

because L is repeated. Within the unit (AOO), the vowels can be arranged in

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

ways. Therefore, the total number of arrangements is

60 \times 3 = 180

120

Correct Answer: A

Solution:

To form a 3-digit even number, the last digit must be even (2, 4, or 6). Choose the last digit in 3 ways, then choose the first digit in 5 ways, and the second digit in 4 ways:

3 \times 5 \times 4 = 60

14400

720

120

5040

Correct Answer: A

Solution:

First, arrange the 5 men:

5! = 120

ways. There are 6 gaps around the men where the women can be seated. Choose 3 out of these 6 gaps for the women:

\binom{6}{3} = 20

ways. Arrange the women in these gaps:

3! = 6

ways. Total arrangements:

120 \times 20 \times 6 = 14400

1440

1680

1920

2160

Correct Answer: B

Solution:

The word 'EQUATION' has 5 vowels (E, U, A, I, O) and 3 consonants (Q, T, N). To form a 5-letter word starting and ending with a vowel, choose 2 vowels for the start and end in 5P2 ways and arrange the remaining 3 letters using the remaining 6 letters (3 vowels + 3 consonants) in 6P3 ways. Thus, the total number of words is 5P2 \times 6P3 = 20 \times 120 = 2400.

220

560

840

1260

Correct Answer: B

Solution:

To form a committee with exactly 3 girls, we choose 3 girls from the 4 available, and 4 boys from the 9 available. The number of ways to choose the girls is

\binom{4}{3} = 4

, and the number of ways to choose the boys is

\binom{9}{4} = 126

. Therefore, the total number of ways is

4 \times 126 = 504

720

504

336

216

Correct Answer: C

Solution:

Since the first digit is fixed as 7, we have 9 remaining digits (0-9 excluding 7) to choose from for the remaining 3 positions. The number of permutations of 9 digits taken 3 at a time is given by 9P3 = 9 \times 8 \times 7 = 504. Therefore, the correct number of sequences is 336.

2100

2520

5040

1260

Correct Answer: B

Solution:

There are 5 vowels and 21 consonants. Choose 2 vowels from 5 and 2 consonants from 21, then arrange them: 5C2 * 21C2 * 4! = 10 * 210 * 24 = 2520.

210

140

105

Correct Answer: D

Solution:

The student must select 3 questions from Part I and 5 questions from Part II, or 4 questions from Part I and 4 questions from Part II. The number of ways is

\binom{5}{3} \times \binom{7}{5} + \binom{5}{4} \times \binom{7}{4} = 10 \times 21 + 5 \times 35 = 210 + 175 = 105

210

140

105

Correct Answer: A

Solution:

The student must select at least 3 questions from each part. Possible selections are: (3 from Part I, 5 from Part II), (4 from Part I, 4 from Part II), (5 from Part I, 3 from Part II). The number of ways for each case is:

\binom{5}{3} \times \binom{7}{5} = 10 \times 21 = 210

;

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

;

\binom{5}{5} \times \binom{7}{3} = 1 \times 35 = 35

. Total = 210 + 175 + 35 = 420.

220

180

120

Correct Answer: A

Solution:

Choose 3 girls from 4 and 4 boys from 9. The number of ways is

\binom{4}{3} \times \binom{9}{4} = 4 \times 126 = 504

336

720

1680

5040

Correct Answer: A

Solution:

The first two digits are fixed as 67. The remaining three digits must be chosen from the 8 digits (0, 1, 2, 3, 4, 5, 8, 9) without repetition. The number of ways to choose and arrange 3 digits from 8 is

\binom{8}{3} \times 3! = 336

1024

31104

649740

Correct Answer: C

Solution:

There are 4 kings in a deck. Choose 1 king in

\binom{4}{1} = 4

ways. The remaining 4 cards must be chosen from the other 48 cards (non-kings), which can be done in

\binom{48}{4} = 194580

ways. Therefore, the total number of combinations is

4 \times 194580 = 778320

16800

33600

67200

134400

Correct Answer: A

Solution:

Treat all vowels (I, E, E, E, E) as a single unit. This gives us 8 units to arrange (7 consonants + 1 vowel unit). These 8 units can be arranged in \frac{8!}{3!2!} ways. The 5 vowels can be arranged in \frac{5!}{4!} = 5 ways. Thus, the total number of arrangements is \frac{8!}{3!2!} \times 5 = 16800.

2880

1440

720

3600

Correct Answer: A

Solution:

There are 4 even places for women, which can be arranged in

4!

ways, and 5 places for men, which can be arranged in

5!

ways. Total arrangements:

4! \times 5! = 24 \times 120 = 2880

Correct Answer: C

Solution:

Mohan can choose a pant in 3 ways and a shirt in 2 ways. Therefore, the total number of pairs is 3 \times 2 = 6.

14400

7200

3600

1800

Correct Answer: A

Solution:

First, arrange the 5 girls in

5!

ways. This creates 6 gaps (before, between, and after the girls) for the boys. Choose 3 out of these 6 gaps in

\binom{6}{3}

ways to place the boys. The boys can be arranged in

3!

ways. Total arrangements =

5! \times \binom{6}{3} \times 3! = 120 \times 20 \times 6 = 14400

150

180

200

210

Correct Answer: D

Solution:

To form a committee with at least 2 women, consider the cases: (i) 2 women and 2 men, (ii) 3 women and 1 man, (iii) 4 women. Calculate: (i) \binom{5}{2} \times \binom{6}{2} = 10 \times 15 = 150, (ii) \binom{5}{3} \times \binom{6}{1} = 10 \times 6 = 60, (iii) \binom{5}{4} = 5. Total = 150 + 60 + 5 = 215.

210

231

441

105

Correct Answer: B

Solution:

The number of chords is the number of ways to choose 2 points from 21, which is

\binom{21}{2} = \frac{21 \times 20}{2} = 210

16800

1663200

30240

25200

Correct Answer: A

Solution:

Treat all vowels (I, E, E, E, E) as a single entity. This gives us 8 entities (7 consonants + 1 vowel group). The number of arrangements is

\frac{8!}{3! \times 2!} \times \frac{5!}{4!} = 16800

5148

4996

4824

5108

Correct Answer: A

Solution:

First, select 1 king from the 4 available kings, which can be done in

\binom{4}{1} = 4

ways. Then, select 4 cards from the remaining 48 cards, which can be done in

\binom{48}{4}

ways. Thus, the total number of combinations is

4 \times \binom{48}{4} = 4 \times 194580 = 5148

2880

1440

720

360

Correct Answer: B

Solution:

There are 4 even places in which the 4 women can be seated, which can be done in

4!

ways. The 5 men can be seated in the remaining 5 places, which can be done in

5!

ways. Therefore, the total number of arrangements is

4! \times 5! = 24 \times 120 = 2880

2598960

4992

5108

Correct Answer: C

Solution:

Choose 1 king from 4 and 4 other cards from 48. The number of ways is

\binom{4}{1} \times \binom{48}{4} = 4 \times 194580 = 778320

Correct Answer: B

Solution:

Since 2 courses are compulsory, the student needs to choose 3 more courses from the remaining 7. The number of ways is given by

\binom{7}{3} = 35

Correct Answer: C

Solution:

The number of ways to select 3 boys from 5 is

\binom{5}{3} = 10

and the number of ways to select 3 girls from 4 is

\binom{4}{3} = 4

. Therefore, the total number of ways is

10 \times 4 = 40

Correct Answer: B

Solution:

The number of ways to form the committee is given by the combination of 5 people taken 3 at a time:

\binom{5}{3} = 10

820

792

756

728

Correct Answer: B

Solution:

We calculate the number of ways to select 5 balls with at least 2 red balls: Case 1: 2 red balls, 3 non-red balls:

\binom{8}{2} \times \binom{10}{3} = 28 \times 120 = 336

. Case 2: 3 red balls, 2 non-red balls:

\binom{8}{3} \times \binom{10}{2} = 56 \times 45 = 2520

. Case 3: 4 red balls, 1 non-red ball:

\binom{8}{4} \times \binom{10}{1} = 70 \times 10 = 700

. Case 4: 5 red balls:

\binom{8}{5} = 56

. Total = 336 + 2520 + 700 + 56 = 792.

5040

504

10000

210

Correct Answer: A

Solution:

The number of ways to choose 4 letters from 10 is calculated as a permutation since order matters and no repetition is allowed. The formula for permutations is given by

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

. Thus,

10P4 = \frac{10!}{(10-4)!} = \frac{10 \times 9 \times 8 \times 7}{1} = 5040

Correct Answer: D

Solution:

The number of signals is given by 5 \times 4 = 20.

120

240

Correct Answer: A

Solution:

The word 'AGAIN' has 5 letters with 'A' repeated twice. The number of permutations is

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

120

240

480

Correct Answer: C

Solution:

Treat the two specific books as a single unit. This gives us 4 units to arrange (3 individual books + 1 unit of 2 books). These 4 units can be arranged in 4! ways. The two books within the unit can be arranged in 2! ways. Thus, the total number of arrangements is 4! \times 2! = 24 \times 2 = 48.

350

210

420

280

Correct Answer: B

Solution:

To form a committee of 10 members with exactly 4 women, we choose 4 women from 5 and 6 men from 7. The number of ways to choose 4 women from 5 is

\binom{5}{4} = 5

. The number of ways to choose 6 men from 7 is

\binom{7}{6} = 7

. Therefore, the total number of ways is

5 \times 7 = 35

5040

10000

3024

210

Correct Answer: C

Solution:

The number of 4-letter codes is given by the permutation of 10 letters taken 4 at a time:

10P4 = \frac{10!}{(10-4)!} = \frac{10 \times 9 \times 8 \times 7}{1} = 3024

14400

7200

3600

1800

Correct Answer: A

Solution:

First, seat the 5 girls in 5! ways. The 3 boys can be seated in the 6 gaps between the girls in 6P3 ways. Total ways = 5! \times 6P3 = 14400.

120

240

360

Correct Answer: A

Solution:

For a 3-digit even number, the last digit must be even. We have 3 choices for the last digit (2, 4, 6). After choosing the last digit, we have 5 remaining digits to fill the first two places. The number of ways to choose the first digit is 5, and the second digit is 4. Thus, the total number of even numbers is

3 \times 5 \times 4 = 60

True or False

Correct Answer: True

Solution:

Treat all the vowels (I, E, E, E, E) as a single object. This gives 8 objects to arrange (including the single vowel object), which can be arranged in

\frac{8!}{3!2!}

ways. The vowels can be arranged among themselves in

\frac{5!}{4!}

ways. Thus, the total number of arrangements is

\frac{8!}{3!2!} \times \frac{5!}{4!} = 16800

Correct Answer: True

Solution:

The number of 3-digit numbers that can be formed from 9 distinct digits is

9 \times 8 \times 7 = 504

, as the first digit can be chosen in 9 ways, the second in 8 ways, and the third in 7 ways.

Correct Answer: True

Solution:

In permutations, the order of arrangement of objects is important. This is contrasted with combinations, where the order is not important.

Correct Answer: False

Solution:

The number of 3-digit even numbers that can be formed is calculated by choosing an even digit (2, 4, 6) for the unit place (3 choices), and any of the 6 digits for the hundreds and tens places (6 choices each), giving a total of

3 \times 6 \times 6 = 108

Correct Answer: True

Solution:

The number of ways to select 3 boys from 5 is

^5C_3 = 10

and the number of ways to select 3 girls from 4 is

^4C_3 = 4

. Thus, the total number of ways to form the team is

10 \times 4 = 40

Correct Answer: True

Solution:

The English alphabet consists of 26 letters, of which 5 are vowels (A, E, I, O, U). Therefore, the remaining 21 letters are consonants.

Correct Answer: True

Solution:

First, arrange the 5 girls in 5! ways. Then, place the 3 boys in the 6 available gaps between the girls in

\binom{6}{3} \times 3!

ways. The total number of arrangements is 5! \times \binom{6}{3} \times 3! = 14400.

Correct Answer: True

Solution:

This problem involves selecting cards with a specific condition, which is a typical application of combinations.

Correct Answer: True

Solution:

The word 'AGAIN' has 5 letters where 'A' is repeated twice. The number of distinct permutations is

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

Correct Answer: True

Solution:

There are 12 face cards in a deck. The number of ways to choose 4 face cards from these 12 is

\binom{12}{4} = 495

Correct Answer: True

Solution:

Treating all vowels as a single object and arranging the remaining letters gives the number of arrangements as 16800.

Correct Answer: False

Solution:

The number of ways to select 2 black balls is

\binom{5}{2} = 10

and to select 3 red balls is

\binom{6}{3} = 20

. Thus, the total number of ways is

10 \times 20 = 200

Correct Answer: True

Solution:

The number of permutations of 5 different objects taken 3 at a time is calculated as 5 \times 4 \times 3 = 60.

Correct Answer: True

Solution:

The number of combinations of 12 different objects taken 2 at a time is given by

^{12}C_2 = \frac{12!}{2!(12-2)!} = \frac{12 \times 11}{2} = 66

Correct Answer: True

Solution:

The numbers greater than 1000000 can start with 1, 2, or 4. Calculating separately for each starting digit and summing gives 360 possible numbers.

Correct Answer: True

Solution:

First, seat the 5 men in 5! ways. There are 4 even positions available for the women, which can be filled in

4!

ways. Therefore, the total number of arrangements is

5! \times 4! = 120 \times 24 = 2880

Correct Answer: False

Solution:

The word 'ARRANGEMENT' has 11 letters with repetitions: A(2), R(2), N(2), G(1), E(2), M(1), T(1). The number of distinct arrangements is

\frac{11!}{2! \times 2! \times 2! \times 1! \times 2! \times 1! \times 1!} = 83160

Correct Answer: True

Solution:

To form a 6-digit number divisible by 10, the last digit must be 0. The remaining 5 digits can be arranged in 5! ways. Therefore, the number of such numbers is

5! = 120

Correct Answer: False

Solution:

Since repetition is not allowed, the number of possible sequences is given by permutations, not powers. The correct calculation is 10P4, not 10^4.

Correct Answer: True

Solution:

The 5 men can be arranged in 5! ways. The 4 women can be arranged in the 4 even positions in 4! ways. Therefore, the total number of arrangements is

5! \times 4! = 14400

Correct Answer: True

Solution:

Since the first digit is known to be 7, there are 9 remaining digits (0 to 9 excluding 7) to choose from for the remaining 3 positions. The number of sequences is given by the permutation formula:

9P3 = \frac{9!}{(9-3)!} = 9 \times 8 \times 7 = 504

Correct Answer: True

Solution:

The formula for permutations of n different objects taken r at a time is given by nPr. For 5 different objects taken 3 at a time, it is 5P3.

Correct Answer: True

Solution:

The number of combinations of 3 players taken 2 at a time is 3C2 = 3.

Correct Answer: True

Solution:

The formula for permutations of

n

objects taken

r

at a time without repetition is indeed

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

Correct Answer: True

Solution:

Handshakes between two people are combinations because the order does not matter. The number of handshakes is given by 12C2.

Correct Answer: True

Solution:

The first two digits are fixed as 67. The remaining 3 digits can be chosen from the remaining 8 digits (0-9 excluding 6 and 7) without repetition, calculated as

8 \times 7 \times 6 = 336

Correct Answer: False

Solution:

In permutations, the order of arrangement is crucial, as permutations are arrangements in a definite order.

Correct Answer: True

Solution:

With 4 flags of different colors, the number of ways to arrange 2 flags is

4 \times 3 = 12

, as the first flag can be any of the 4 and the second can be any of the remaining 3.

Correct Answer: False

Solution:

To form a 6-digit number divisible by 10, the last digit must be 0. The remaining 5 digits can be arranged in

5! = 120

ways.

Correct Answer: True

Solution:

This is the formula for permutations where

n

objects are taken

r

at a time without repetition.

Correct Answer: False

Solution:

The problem requires calculating combinations with specific conditions, which involves more complex calculations than a simple combination formula.

Correct Answer: True

Solution:

To form a committee with exactly 1 man and 2 women, we choose 1 man from 2 men in

\binom{2}{1}

ways and 2 women from 3 women in

\binom{3}{2}

ways. Thus, the number of ways is

\binom{2}{1} \times \binom{3}{2} = 2 \times 3 = 6

Correct Answer: True

Solution:

To form a committee of 7 with exactly 3 girls, choose 3 girls from 4 and 4 boys from 9. The number of ways is

\binom{4}{3} \times \binom{9}{4} = 4 \times 126 = 504

Correct Answer: False

Solution:

The number of ways to arrange the letters of 'ASSASSINATION' with all S's together is not directly provided in the excerpts. However, a similar problem for another word was solved, resulting in 16800 arrangements when all vowels were together.

Correct Answer: True

Solution:

There are

\binom{26}{2}

ways to choose 2 red cards and

\binom{26}{2}

ways to choose 2 black cards. Total =

\binom{26}{2} \times \binom{26}{2} = 29900

Correct Answer: False

Solution:

The number of sequences of 3 remaining digits is given by the permutations of 9 digits taken 3 at a time:

9P3 = \frac{9!}{(9-3)!} = 9 \times 8 \times 7 = 504

Correct Answer: True

Solution:

The number of permutations of 6 different letters taken 3 at a time is calculated as

6 \times 5 \times 4 = 120

Correct Answer: True

Solution:

The numbers can be formed by arranging the digits 1, 2, 0, 2, 4, 2, 4, ensuring the first digit is not 0. By calculating, the number of such arrangements is 360.

Correct Answer: True

Solution:

The number of different signals can be calculated as

4 \times 3 = 12

, since the upper place can be filled in 4 ways and the lower place in 3 ways.

Correct Answer: True

Solution:

The number of ways to form a committee of 3 people from 5 (2 men and 3 women) is given by

\binom{5}{3} = 10

Correct Answer: True

Solution:

According to the fundamental principle of counting, the total number of occurrences is the product of the number of ways each event can occur.

Correct Answer: True

Solution:

The number of permutations of 5 distinct objects taken 3 at a time is given by the formula

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

. For 5 objects taken 3 at a time,

5P_3 = \frac{5!}{(5-3)!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = 5 \times 4 \times 3 = 60

Correct Answer: True

Solution:

The number of ways to select 3 persons from a group of 5 (2 men + 3 women) is given by

\binom{5}{3} = 10

Correct Answer: True

Solution:

To form a 4-digit number using the digits 1, 2, 3, 4, 5 without repetition, we can choose the first digit in 5 ways, the second in 4 ways, the third in 3 ways, and the fourth in 2 ways. Thus, the total number of 4-digit numbers is

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

Correct Answer: False

Solution:

If the first digit is known and is 7, then there are 3 remaining digits to be arranged from 9 possible digits (0-9 excluding 7). The number of permutations of 9 digits taken 3 at a time is

9P_3 = \frac{9!}{(9-3)!} = 9 \times 8 \times 7 = 504

Correct Answer: True

Solution:

This formula calculates the number of ways to choose

r

objects from

n

without regard to order.

Correct Answer: True

Solution:

The total number of ways to choose 3 persons from 5 (2 men and 3 women) is

\binom{5}{3} = 10

Correct Answer: True

Solution:

The problem of forming a committee with specific constraints can be solved using combinations, as the order of selection does not matter.

Correct Answer: True

Solution:

There are 4 flags and we need to arrange 2 at a time. The number of permutations is given by 4P2 = 4 \times 3 = 12.

Correct Answer: True

Solution:

The number of ways to choose 3 boys from 5 is

\binom{5}{3} = 10

and the number of ways to choose 3 girls from 4 is

\binom{4}{3} = 4

. Therefore, the total number of ways is

10 \times 4 = 40

Correct Answer: True

Solution:

The number of permutations of 7 distinct objects taken 3 at a time is given by the formula

^7P_3 = \frac{7!}{(7-3)!} = \frac{7 \times 6 \times 5}{1} = 210

Correct Answer: True

Solution:

There are 4 suits in a deck, and each suit has 13 cards. The number of ways to choose 4 cards from one suit is

\binom{13}{4}

. Since there are 4 suits, the total number of ways is

4 \times \binom{13}{4}

Correct Answer: True

Solution:

The number of ways to choose 4 cards from a deck of 52 cards is given by

\binom{52}{4} = 270725

Correct Answer: True

Solution:

The word 'AGAIN' has 5 letters with 'A' repeating twice. The number of permutations is calculated as

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

Correct Answer: False

Solution:

The number of ways to form a committee of 3 persons from a group of 2 men and 3 women is given by the combination formula

\binom{5}{3} = 10

. However, this specific calculation was not directly provided in the excerpts.

Correct Answer: True

Solution:

The number of combinations of

n

different objects taken

r

at a time is given by

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

. For 4 objects taken 2 at a time,

4C_2 = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6

Correct Answer: False

Solution:

The number of 3-digit numbers is given by 9P3, which equals 9 × 8 × 7 = 504, but this calculation is incorrect as the correct number is 9 × 8 × 7 = 504.

Correct Answer: True

Solution:

The number of ways to choose 2 players from a group of 3 is given by

\binom{3}{2} = 3

Correct Answer: True

Solution:

The number of arrangements of the letters of the word 'INDEPENDENCE' is calculated as

\frac{12!}{3!4!2!} = 1663200

Correct Answer: False

Solution:

In permutations, the order of arrangement of objects does matter, unlike in combinations where order does not matter.

Correct Answer: True

Solution:

Treating the vowels EEEEI as a single unit, we have 8 units to arrange (7 consonants and 1 vowel unit). These 8 units can be arranged in

8!

ways. The vowels within the unit can be arranged in

4!

ways. Thus, the total number of arrangements is

8! \times 4! = 16800

Correct Answer: True

Solution:

The number of ways to choose 3 persons from a group of 5 (2 men and 3 women) is given by

\binom{5}{3} = 10

Correct Answer: True

Solution:

The number of permutations of 5 different digits taken 4 at a time is 5P4 = 5 \times 4 \times 3 \times 2 = 120.

Correct Answer: True

Solution:

The fundamental principle of counting is correctly stated as the product of the number of ways each event can occur, which is

m \times n

Correct Answer: False

Solution:

In combinations, the order of selection is not important. This is the main difference between combinations and permutations, where order does matter.

Permutations and Combinations

Learning Objectives

Learning Objectives

Revision Notes & Summary

Chapter 6: Permutations and Combinations

6.1 Introduction

6.2 Fundamental Principle of Counting

6.3 Permutations

6.3.1 Distinct Objects

6.3.2 Factorial Notation

6.4 Combinations

Miscellaneous Examples

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Tips for Success

AI Learning Assistant

Practice Test – MCQs, True/False

Multiple Choice Questions

Q1.A bag contains 4 red balls and 5 green balls. Two balls are drawn at random. What is the probability that both balls are red?

Correct Answer: B

Q2.How many 2-digit even numbers can be formed from the digits 1, 2, 3, 4, 5 if the digits can be repeated?

Correct Answer: A

Q3.How many ways can the letters of the word 'INDEPENDENCE' be arranged so that all the vowels occur together?

Correct Answer: A

Q4.How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7, and 9 which are divisible by 10 and no digit is repeated?

Correct Answer: B

Q5.How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?

Correct Answer: C

Q6.How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?

Correct Answer: B

Q7.How many numbers greater than 1000000 can be formed by using the digits 1, 2, 0, 2, 4, 2, 4?

Correct Answer: A

Q8.How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?

Correct Answer: B

Q9.In how many ways can the letters of the word 'COMMITTEE' be arranged so that all the vowels always come together?

Correct Answer: B

Q10.In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?

Correct Answer: A

Q11.How many 3-digit numbers can be formed using the digits 1 to 9 if no digit is repeated?

Correct Answer: B

Q12.How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7, and 9 which are divisible by 10 and no digit is repeated?

Correct Answer: A

Q13.A committee of 10 students is to be formed from a class of 25 students. There are 3 students who insist that either all or none of them be included. In how many ways can the committee be formed?

Correct Answer: D

Correct Answer: C

Q15.How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?

Correct Answer: A

Q16.A bag contains 4 red balls, 5 green balls, and 6 blue balls. In how many ways can 3 balls be drawn from the bag such that at least one of the balls is red?

Correct Answer: A

Q17.How many ways can you select a cricket team of 11 from 17 players if only 5 players can bowl and the team must include exactly 4 bowlers?

Correct Answer: C

Q18.In how many ways can a student choose a program of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?

Correct Answer: A

Q19.From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen?

Correct Answer: A

Q20.How many different 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated?

Correct Answer: A

Q21.In how many ways can the letters of the word ASSASSINATION be arranged so that all the S's are together?

Correct Answer: B

Q22.How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, and 5 if repetition of digits is not allowed?

Correct Answer: A

Q23.How many different words can be formed using all the letters of the word 'BALLOON', such that the vowels are always together?

Correct Answer: B

Q24.How many 3-digit even numbers can be formed using the digits 1, 2, 3, 4, 5, and 6 if no digit is repeated?

Correct Answer: A

Q25.How many ways can 5 men and 3 women be seated in a row so that no two women are together?

Correct Answer: A

Q26.How many different 5-letter words can be formed using the letters of the word 'EQUATION' such that the word starts and ends with a vowel?

Correct Answer: B

Q27.A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done if the committee must include exactly 3 girls?

Correct Answer: B

Q28.Consider a lock with 4 wheels, each labeled with digits from 0 to 9. If the first digit is fixed as 7, how many different sequences of the remaining 3 digits can be formed without repetition?

Correct Answer: C

Q29.How many words with two different vowels and two different consonants can be formed from the English alphabet?

Correct Answer: B

Q30.How many ways can a student select 8 questions from a question paper consisting of 12 questions divided into two parts, Part I with 5 questions and Part II with 7 questions, if at least 3 questions must be selected from each part?

Correct Answer: D

Q31.In how many ways can a student select 8 questions from a question paper consisting of 12 questions divided into two parts, Part I with 5 questions and Part II with 7 questions, if at least 3 questions must be selected from each part?

Correct Answer: A