Permutations and Combinations — Class 11 Maths Notes

Permutations and Combinations · Class 11 Maths · 9 topics.

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Topics covered in Permutations and Combinations

  1. 1.Introduction to Permutations and Combinations

    Permutations and combinations are fundamental concepts in mathematics, particularly in the field of combinatorics. These concepts are used to count the number of ways in which a set of items can be arranged or selected. They have wide applications in various fields including mathematics, computer science, statistics, and everyday life situations.

    Permutations

    Permutations refer to the different ways in which a set of items can be arranged. The order of the items matters in permutations.

    Formula for Permutations: If we have 𝑛n items and we want to arrange 𝑟r of them, the number of permutations is given by:

    𝑃(𝑛,𝑟)=𝑛!(𝑛−𝑟)!P(n,r)=(n−r)!n!​

    Here, 𝑛!n! (n factorial) is the product of all positive integers up to 𝑛n.

    Example: Imagine you have 3 books and you want to arrange 2 of them on a shelf. How many ways can you do this?

    𝑃(3,2)=3!(3−2)!=3!1!=3×2×11=6P(3,2)=(3−2)!3!​=1!3!​=13×2×1​=6

    So, there are 6 different ways to arrange 2 out of the 3 books.

    Combinations

    Combinations refer to the different ways in which a set of items can be selected, where the order does not matter.

    Formula for Combinations: If we have 𝑛n items and we want to select 𝑟r of them, the number of combinations is given by:

    𝐶(𝑛,𝑟)=𝑛!𝑟!(𝑛−𝑟)!C(n,r)=r!(n−r)!n!​

    Example: Imagine you have 3 books and you want to select 2 of them. How many ways can you do this?

    𝐶(3,2)=3!2!(3−2)!=3!2!⋅1!=3×2×12×1×1=3C(3,2)=2!(3−2)!3!​=2!⋅1!3!​=2×1×13×2×1​=3

    So, there are 3 different ways to select 2 out of the 3 books.

    Real-life Applications

    1. Seating Arrangements: Permutations can be used to determine the number of ways people can be seated in a row.
    2. Lottery Tickets: Combinations can be used to calculate the odds of winning by selecting the right numbers.
    3. Password Generation: Permutations help in understanding the number of possible passwords that can be created with a set of characters.

    Step-by-Step Example

    Let’s consider a practical problem:

    Problem: You have 4 different fruits: apple, banana, cherry, and date. How many ways can you select 2 fruits from these 4?

    Solution:

    1. Identify the total number of items (𝑛n) and the number of items to be selected (𝑟r).

      • 𝑛=4n=4
      • 𝑟=2r=2

    2. Use the combination formula: 𝐶(4,2)=4!2!(4−2)!=4×3×2×12×1×2×1=244=6C(4,2)=2!(4−2)!4!​=2×1×2×14×3×2×1​=424​=6

    So, there are 6 ways to select 2 fruits from 4.

  2. 2.Fundamental Principle of Counting

    Overview

    The Fundamental Principle of Counting is a basic rule in combinatorics that helps us determine the number of ways in which a sequence of events can occur. It is the foundation for understanding more complex concepts like permutations and combinations. This principle is also known as the Multiplication Principle.

    Definition

    The Fundamental Principle of Counting states that if there are 𝑚m ways of performing one task and 𝑛n ways of performing another task, then there are 𝑚×𝑛m×n ways of performing both tasks together.

    Simple Example

    Suppose you have 2 shirts (blue and red) and 3 pairs of pants (black, white, and gray). How many different outfits can you create by pairing one shirt with one pair of pants?

    • Number of ways to choose a shirt: 2 (blue or red)
    • Number of ways to choose a pair of pants: 3 (black, white, or gray)

    Using the Fundamental Principle of Counting:

    Total outfits=2×3=6Total outfits=2×3=6

    So, there are 6 different outfits you can create.

    Detailed Example

    Let's consider a more detailed example to understand this principle better.

    Example: A restaurant offers 3 choices of appetizers, 4 choices of main courses, and 2 choices of desserts. How many different meals can you create if you choose one item from each category?

    1. Appetizers:
      • Choices: 3 (e.g., salad, soup, breadsticks)
    2. Main Courses:
      • Choices: 4 (e.g., pasta, burger, pizza, sandwich)
    3. Desserts:
      • Choices: 2 (e.g., ice cream, cake)

    Using the Fundamental Principle of Counting, the total number of different meals is calculated as:

    Total meals=3×4×2=24Total meals=3×4×2=24

    Therefore, you can create 24 different meals by selecting one item from each category.

    Real-life Applications

    1. Travel Plans: If you have 2 options for flight routes, 3 options for hotels, and 2 options for car rentals, you can calculate the total number of different travel plans.
    2. Password Generation: If a password must include 3 characters where each character can be a letter (26 options) or a digit (10 options), you can determine the number of possible passwords.
    3. Menu Combinations: Restaurants can use this principle to calculate the number of possible meal combinations they can offer.

    Step-by-Step Problem

    Problem: You are planning a party and need to decide on a theme, music, and decorations. You have 3 themes to choose from (beach, jungle, carnival), 2 types of music (pop, classical), and 4 types of decorations (balloons, streamers, lights, banners). How many different combinations can you create?

    Solution:

    1. Themes:
      • Choices: 3 (beach, jungle, carnival)
    2. Music:
      • Choices: 2 (pop, classical)
    3. Decorations:
      • Choices: 4 (balloons, streamers, lights, banners)

    Using the Fundamental Principle of Counting:

    Total combinations=3×2×4=24Total combinations=3×2×4=24

    So, there are 24 different ways to plan the party with these choices.

  3. 3.Exercise Questions

    Question 1:

    How many 3-digit numbers can be formed from the digits 1, 2, 3, 4, and 5 assuming that (i) repetition of the digits is allowed? (ii) repetition of the digits is not allowed?

    Solution:

    (i) Repetition allowed:

    For a 3-digit number, each digit can be any of the 5 given digits.

    • First digit: 5 choices (1, 2, 3, 4, 5)
    • Second digit: 5 choices
    • Third digit: 5 choices

    So, the total number of 3-digit numbers is: 5×5×5=1255×5×5=125

    (ii) Repetition not allowed:

    For a 3-digit number with no repetition:

    • First digit: 5 choices
    • Second digit: 4 choices (since one digit is already used)
    • Third digit: 3 choices

    So, the total number of 3-digit numbers is: 5×4×3=605×4×3=60

    Question 2:

    How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?

    Solution:

    For a number to be even, the last digit must be one of the even digits (2, 4, 6).

    • Last digit: 3 choices (2, 4, 6)
    • First digit: 6 choices (since digits can be repeated)
    • Second digit: 6 choices

    So, the total number of 3-digit even numbers is: 6×6×3=1086×6×3=108

    Question 3:

    How many 4-letter codes can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?

    Solution:

    We need to choose 4 letters out of 10 without repetition:

    • First letter: 10 choices
    • Second letter: 9 choices
    • Third letter: 8 choices
    • Fourth letter: 7 choices

    So, the total number of 4-letter codes is: 10×9×8×7=504010×9×8×7=5040

    Question 4:

    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?

    Solution:

    The first two digits are fixed as 67. We need to find the remaining 3 digits from the digits 0 to 9, excluding 6 and 7.

    • Remaining digits: 0, 1, 2, 3, 4, 5, 8, 9 (8 digits left)
    • Third digit: 8 choices
    • Fourth digit: 7 choices
    • Fifth digit: 6 choices

    So, the total number of telephone numbers is: 1×1×8×7×6=3361×1×8×7×6=336

    Question 5:

    A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

    Solution:

    Each toss has 2 possible outcomes (Heads or Tails).

    • First toss: 2 choices
    • Second toss: 2 choices
    • Third toss: 2 choices

    So, the total number of possible outcomes is: 2×2×2=82×2×2=8

    Question 6:

    Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?

    Solution:

    We need to select and arrange 2 flags out of 5.

    • First flag: 5 choices
    • Second flag: 4 choices

    So, the total number of different signals is: 5×4=205×4=20

  4. 4.Permutations

    Definition

    Permutations refer to the different arrangements of a given set of objects where the order matters. For example, the permutations of the set {1,2,3}{1,2,3} include {1,2,3}{1,2,3}, {1,3,2}{1,3,2}, {2,1,3}{2,1,3}, and so on. The key characteristic of permutations is that changing the order of objects creates a different permutation.

    Theorem and Formula

    If you have a set of 𝑛n distinct objects and you want to arrange 𝑟r of them, the number of permutations is given by the formula:

    𝑃(𝑛,𝑟)=𝑛!(𝑛−𝑟)!P(n,r)=(n−r)!n!​

    Here, 𝑛!n! (n factorial) is the product of all positive integers up to 𝑛n. For example, 5!=5×4×3×2×1=1205!=5×4×3×2×1=120.

    Permutation Theorem

    Theorem: The number of permutations of 𝑛n distinct objects taken 𝑟r at a time is given by:

    𝑃(𝑛,𝑟)=𝑛!(𝑛−𝑟)!P(n,r)=(n−r)!n!​

    This formula accounts for selecting 𝑟r objects from 𝑛n and arranging them in order.

    Example

    Let's go through an example step-by-step to understand how to apply the permutation formula.

    Example: How many different ways can you arrange 3 out of 5 books?

    1. Identify the values:

      • 𝑛=5n=5 (total books)
      • 𝑟=3r=3 (books to arrange)

    2. Apply the formula: 𝑃(5,3)=5!(5−3)!=5!2!=5×4×3×2×12×1=1202=60P(5,3)=(5−3)!5!​=2!5!​=2×15×4×3×2×1​=2120​=60

    So, there are 60 different ways to arrange 3 out of 5 books.

    Detailed Explanation

    Let's delve deeper into why this formula works.

    1. First position: You have 𝑛n choices for the first position because you can pick any of the 𝑛n objects.
    2. Second position: After choosing the first object, you have 𝑛−1n−1 choices left for the second position.
    3. Third position: After choosing the first and second objects, you have 𝑛−2n−2 choices left for the third position.

    This process continues until you have filled all 𝑟r positions. Mathematically, this is expressed as:

    𝑛×(𝑛−1)×(𝑛−2)×…×(𝑛−𝑟+1)n×(n−1)×(n−2)×…×(n−r+1)

    This product can be simplified by using the factorial notation, leading to the formula:

    𝑃(𝑛,𝑟)=𝑛!(𝑛−𝑟)!P(n,r)=(n−r)!n!​

    Real-life Application

    Permutations are used in many real-life situations, such as:

    1. Seating Arrangements: Determining the number of ways people can sit in a row.
    2. Password Generation: Calculating the number of possible passwords from a set of characters.
    3. Scheduling: Finding the number of ways to schedule events or tasks.

    Step-by-Step Problem

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

    Solution:

    1. Identify the values:

      • 𝑛=5n=5
      • 𝑟=4r=4

    2. Apply the formula: 𝑃(5,4)=5!(5−4)!=5!1!=5!=5×4×3×2×1=120P(5,4)=(5−4)!5!​=1!5!​=5!=5×4×3×2×1=120

    So, there are 120 different 4-digit numbers that can be formed from the digits 1, 2, 3, 4, 5 with no repetition.

  5. 5.Permutations When All the Objects are Distinct


    Definition

    When dealing with permutations of distinct objects, we are arranging all or some of the objects in a specific order. If we have a set of 𝑛n distinct objects, and we want to find the number of ways to arrange 𝑟r of these objects, we use permutations.

    Factorial Notation

    Factorial notation is a shorthand way to represent the product of all positive integers up to a given number. It is denoted by an exclamation mark (!).

    Definition: 𝑛!=𝑛×(𝑛−1)×(𝑛−2)×…×1n!=n×(n−1)×(n−2)×…×1

    For example:

    • 5!=5×4×3×2×1=1205!=5×4×3×2×1=120
    • 3!=3×2×1=63!=3×2×1=6
    • By definition, 0!=10!=1

    Derivation of the Formula for 𝑛𝑃𝑟nPr​

    To derive the formula for the number of permutations of 𝑟r objects taken from 𝑛n distinct objects (𝑛𝑃𝑟nPr​), follow these steps:

    1. First Position: For the first position, you have 𝑛n choices because you can choose any of the 𝑛n objects.
    2. Second Position: After choosing one object for the first position, you have 𝑛−1n−1 choices left for the second position.
    3. Third Position: After choosing objects for the first and second positions, you have 𝑛−2n−2 choices left for the third position.

    This pattern continues until you have placed 𝑟r objects.

    Mathematically, this can be written as: 𝑛×(𝑛−1)×(𝑛−2)×…×(𝑛−𝑟+1)n×(n−1)×(n−2)×…×(n−r+1)

    To express this in factorial notation, note that the product above represents the first 𝑟r terms of 𝑛!n!, and the remaining terms are the factorial of 𝑛−𝑟n−r.

    𝑛𝑃𝑟=𝑛×(𝑛−1)×(𝑛−2)×…×(𝑛−𝑟+1)nPr​=n×(n−1)×(n−2)×…×(n−r+1) 𝑛𝑃𝑟=𝑛!(𝑛−𝑟)!nPr​=(n−r)!n!​

    Here, 𝑛!n! is the factorial of 𝑛n and (𝑛−𝑟)!(n−r)! is the factorial of 𝑛−𝑟n−r, which cancels out the terms after the first 𝑟r terms in 𝑛!n!.

    Example

    Let's use this formula in an example:

    Example: How many ways can you arrange 3 out of 5 distinct books?

    1. Identify the values:

      • 𝑛=5n=5 (total books)
      • 𝑟=3r=3 (books to arrange)

    2. Apply the formula: 𝑃(5,3)=5!(5−3)!=5!2!=5×4×3×2×12×1=1202=60P(5,3)=(5−3)!5!​=2!5!​=2×15×4×3×2×1​=2120​=60

    So, there are 60 different ways to arrange 3 out of 5 books.

  6. 6.Permutations with Repetition Allowed

    Theorem 2

    Theorem: The number of permutations of 𝑛n different objects taken 𝑟r at a time, where repetition is allowed, is given by:

    𝑛𝑟nr

    Explanation

    When repetition is allowed, each of the 𝑟r positions can be filled by any of the 𝑛n objects. Hence, for each position, there are 𝑛n choices, resulting in a total of 𝑛𝑟nr permutations.

    Example

    Example: How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, 5, where repetition of digits is allowed?

    1. Identify the values:

      • 𝑛=5n=5 (total digits)
      • 𝑟=3r=3 (digits to be arranged)

    2. Apply the formula: 53=5×5×5=12553=5×5×5=125

    So, there are 125 different 3-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 with repetition allowed.

    Permutations When All Objects Are Not Distinct

    When the objects are not distinct, meaning some of the objects are repeated, the formula for permutations needs to account for these repetitions to avoid overcounting.

    Formula

    If we have a set of 𝑛n objects where 𝑛1n1​ objects are of one kind, 𝑛2n2​ objects are of another kind, and so on, the number of permutations is given by:

    𝑛!𝑛1!×𝑛2!×…×𝑛𝑘!n1​!×n2​!×…×nk​!n!​

    where 𝑛1+𝑛2+…+𝑛𝑘=𝑛n1​+n2​+…+nk​=n.

    Explanation

    This formula is derived by considering the total number of permutations of 𝑛n objects (which is 𝑛!n!) and then dividing by the factorial of the counts of each type of object (𝑛1!,𝑛2!,…,𝑛𝑘!n1​!,n2​!,…,nk​!) to correct for overcounting the indistinguishable arrangements.

    Example

    Example: How many distinct permutations can be made from the letters in the word "BALLOON"?

    1. Identify the values:

      • Total letters, 𝑛=7n=7 (B, A, L, L, O, O, N)
      • B occurs 1 time
      • A occurs 1 time
      • L occurs 2 times
      • O occurs 2 times
      • N occurs 1 time

    2. Apply the formula: 7!1!×1!×2!×2!×1!=50401×1×2×2×1=50404=12601!×1!×2!×2!×1!7!​=1×1×2×2×15040​=45040​=1260

    So, there are 1260 distinct permutations of the letters in the word "BALLOON".

  7. 7.Theorem 3 & Theorem 4

    Theorem 3: Permutations with Repeated Objects

    Theorem 3: The number of permutations of 𝑛n objects, where 𝑝p objects are of the same kind and the rest are all different, is given by:

    𝑛!𝑝!p!n!​

    Explanation

    This theorem accounts for the fact that when some objects are identical, the identical objects can be arranged among themselves in 𝑝!p! ways which do not create a new unique permutation. Thus, the total number of distinct permutations is reduced by a factor of 𝑝!p!.

    Example

    Example: How many distinct permutations can be made from the letters in the word "LEVEL"?

    1. Identify the values:

      • Total letters, 𝑛=5n=5 (L, E, V, E, L)
      • E occurs 2 times
      • L occurs 2 times

    2. Apply the formula: 5!2!×2!=1204=302!×2!5!​=4120​=30

    So, there are 30 distinct permutations of the letters in the word "LEVEL".

    Theorem 4: General Permutations with Multiple Sets of Repeated Objects

    Theorem 4: The number of permutations of 𝑛n objects, where 𝑝1p1​ objects are of one kind, 𝑝2p2​ are of a second kind,..., 𝑝𝑘pk​ are of 𝑘k-th kind, and the rest, if any, are of different kinds, is given by:

    𝑛!𝑝1!×𝑝2!×…×𝑝𝑘!p1​!×p2​!×…×pk​!n!​

    Explanation

    This theorem generalizes the idea that when multiple sets of objects are identical, we need to account for the permutations within each set of identical objects. This is done by dividing the total permutations 𝑛!n! by the factorial of each set size 𝑝1!,𝑝2!,…,𝑝𝑘!p1​!,p2​!,…,pk​!.

    Example

    Example: How many distinct permutations can be made from the letters in the word "MISSISSIPPI"?

    1. Identify the values:

      • Total letters, 𝑛=11n=11 (M, I, S, S, I, S, S, I, P, P, I)
      • M occurs 1 time
      • I occurs 4 times
      • S occurs 4 times
      • P occurs 2 times

    2. Apply the formula: 11!1!×4!×4!×2!=399168001×24×24×2=399168001152=345601!×4!×4!×2!11!​=1×24×24×239916800​=115239916800​=34560

    So, there are 34,560 distinct permutations of the letters in the word "MISSISSIPPI".

  8. 8.Exercise Questions

    Question 1

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

    Solution:

    • For the first digit, we have 9 choices (1 to 9).
    • For the second digit, we have 8 choices (since one digit is already used).
    • For the third digit, we have 7 choices (since two digits are already used).

    So, the total number of 3-digit numbers is: 9×8×7=5049×8×7=504

    Question 2

    How many 4-digit numbers are there with no digit repeated?

    Solution:

    • For the first digit, we have 9 choices (1 to 9, excluding 0).
    • For the second digit, we have 9 choices (including 0 but excluding the first digit).
    • For the third digit, we have 8 choices (excluding the first two digits).
    • For the fourth digit, we have 7 choices (excluding the first three digits).

    So, the total number of 4-digit numbers is: 9×9×8×7=45369×9×8×7=4536

    Question 3

    How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 if no digit is repeated?

    Solution:

    • The last digit must be even. The choices are 2, 4, 6, 8 (4 choices).
    • For the first digit, we have 8 choices (since one digit is used for the last digit and 0 is not included).
    • For the second digit, we have 7 choices (excluding the two digits used).

    So, the total number of 3-digit even numbers is: 4×8×7=2244×8×7=224

    Question 4

    Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?

    Solution: Total 4-digit numbers:

    • First digit: 5 choices (1 to 5)
    • Second digit: 4 choices
    • Third digit: 3 choices
    • Fourth digit: 2 choices

    Total 4-digit numbers: 5×4×3×2=1205×4×3×2=120

    Even 4-digit numbers:

    • The last digit must be even. The choices are 2 and 4 (2 choices).
    • First digit: 4 choices
    • Second digit: 3 choices
    • Third digit: 2 choices

    Total even 4-digit numbers: 2×4×3×2=482×4×3×2=48

    Question 5

    From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person can not hold more than one position?

    Solution:

    • For the chairman position, we have 8 choices.
    • For the vice chairman position, we have 7 choices (since one person is already chosen as chairman).

    Total ways to choose: 8×7=568×7=56

    Question 6

    Find 𝑛n if 𝑛−1𝑃3=6×𝑛𝑃3n−1P3​=6×nP3​.

    Solution: We have: (𝑛−1)!(𝑛−4)!=6×𝑛!(𝑛−3)!(n−4)!(n−1)!​=6×(n−3)!n!​

    Simplify: (𝑛−1)!(𝑛−4)!=6×𝑛×(𝑛−1)!(𝑛−3)!(n−4)!(n−1)!​=6×(n−3)!n×(n−1)!​ (𝑛−1)!(𝑛−4)!=6𝑛×(𝑛−3)(n−4)!(n−1)!​=6n×(n−3) (𝑛−1)(𝑛−2)(𝑛−3)=6𝑛(𝑛−3)(n−1)(n−2)(n−3)=6n(n−3) (𝑛−1)(𝑛−2)=6𝑛(n−1)(n−2)=6n 𝑛2−3𝑛+2=6𝑛n2−3n+2=6n 𝑛2−9𝑛+2=0n2−9n+2=0

    Using the quadratic formula 𝑛=−𝑏±𝑏2−4𝑎𝑐2𝑎n=2a−b±b2−4ac​​ where 𝑎=1a=1, 𝑏=−9b=−9, and 𝑐=2c=2: 𝑛=9±81−82n=29±81−8​​ 𝑛=9±732n=29±73​​

    Since 𝑛n must be a positive integer, we discard non-integer solutions.

    Question 7

    Find 𝑟r if (i) 9𝑃𝑟=2×8𝑃𝑟9Pr​=2×8Pr​ (ii) 9𝑃𝑟=6×9𝑃𝑟−19Pr​=6×9Pr−1​

    Solution: (i) 9𝑃𝑟=2×8𝑃𝑟9Pr​=2×8Pr​: 9!(9−𝑟)!=2×8!(8−𝑟)!(9−r)!9!​=2×(8−r)!8!​ 9!(9−𝑟)!=2×8×8!(8−𝑟)!(9−r)!9!​=2×8×(8−r)!8!​ 9×8!(9−𝑟)!=16×8!(8−𝑟)!(9−r)!9×8!​=16×(8−r)!8!​ 99−𝑟=169−r9​=16 9=16×(9−𝑟)9=16×(9−r) 9=144−16𝑟9=144−16r 16𝑟=13516r=135 𝑟=13516r=16135​

    (ii) 9𝑃𝑟=6×9𝑃𝑟−19Pr​=6×9Pr−1​: 9!(9−𝑟)!=6×9!(9−𝑟+1)!(9−r)!9!​=6×(9−r+1)!9!​ (9−𝑟+1)=6(9−r+1)=6 9−𝑟+1=69−r+1=6 10−𝑟=610−r=6 𝑟=4r=4

    Question 8

    How many words, with or without meaning, can be formed using all the letters of the word "EQUATION," using each letter exactly once?

    Solution: "EQUATION" has 8 distinct letters.

    So, the total number of permutations is: 8!=403208!=40320

  9. 9.Exercise Questions

    Question 1

    If 𝐶8=𝐶2C8​=C2​, find 𝐶𝑟Cr​.

    Solution:

    Using the symmetry property of combinations: 𝐶(𝑛,𝑟)=𝐶(𝑛,𝑛−𝑟)C(n,r)=C(n,n−r)

    Given 𝐶8=𝐶2C8​=C2​: 𝐶(𝑛,8)=𝐶(𝑛,2)C(n,8)=C(n,2)

    From the property, we know: 8=𝑛−28=n−2 𝑛=10n=10

    So, 𝐶𝑟=𝐶(10,𝑟)Cr​=C(10,r).

    Question 2

    Determine 𝑛n if (i) 2𝑛𝐶3:𝐶3=12:12nC3​:C3​=12:1 (ii) 2𝑛𝐶3:𝐶3=11:12nC3​:C3​=11:1

    Solution: (i) 2𝑛𝐶3:𝐶3=12:12nC3​:C3​=12:1:

    2𝑛𝐶3𝐶3=12C3​2nC3​​=12 2𝑛=122n=12 𝑛=log⁡212n=log2​12 𝑛=3.58496n=3.58496

    Since 𝑛n must be an integer, this case has no valid 𝑛n.

    (ii) 2𝑛𝐶3:𝐶3=11:12nC3​:C3​=11:1:

    2𝑛𝐶3𝐶3=11C3​2nC3​​=11 2𝑛=112n=11 𝑛=log⁡211n=log2​11 𝑛=3.45943n=3.45943

    Since 𝑛n must be an integer, this case also has no valid 𝑛n.

    Question 3

    How many chords can be drawn through 21 points on a circle?

    Solution:

    The number of chords that can be drawn through 𝑛n points on a circle is given by the number of combinations of 2 points from 𝑛n:

    𝐶(𝑛,2)=𝑛(𝑛−1)2C(n,2)=2n(n−1)​ 𝐶(21,2)=21×202=210C(21,2)=221×20​=210

    So, 210 chords can be drawn through 21 points on a circle.

    Question 4

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

    Solution:

    Select 3 boys from 5: 𝐶(5,3)=5!3!(5−3)!=10C(5,3)=3!(5−3)!5!​=10

    Select 3 girls from 4: 𝐶(4,3)=4!3!(4−3)!=4C(4,3)=3!(4−3)!4!​=4

    Total ways to form the team: 𝐶(5,3)×𝐶(4,3)=10×4=40C(5,3)×C(4,3)=10×4=40

    So, there are 40 ways to form a team of 3 boys and 3 girls from 5 boys and 4 girls.

    Question 5

    Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls, and 5 blue balls if each selection consists of 3 balls of each color.

    Solution:

    Select 3 red balls from 6: 𝐶(6,3)=6!3!(6−3)!=20C(6,3)=3!(6−3)!6!​=20

    Select 3 white balls from 5: 𝐶(5,3)=5!3!(5−3)!=10C(5,3)=3!(5−3)!5!​=10

    Select 3 blue balls from 5: 𝐶(5,3)=5!3!(5−3)!=10C(5,3)=3!(5−3)!5!​=10

    Total ways to select 9 balls: 𝐶(6,3)×𝐶(5,3)×𝐶(5,3)=20×10×10=2000C(6,3)×C(5,3)×C(5,3)=20×10×10=2000

    So, there are 2000 ways to select 9 balls with 3 balls of each color.

    Question 6

    Determine the number of 5-card combinations out of a deck of 52 cards if there is exactly one ace in each combination.

    Solution:

    Select 1 ace from 4 aces: 𝐶(4,1)=4!1!(4−1)!=4C(4,1)=1!(4−1)!4!​=4

    Select 4 non-aces from 48 cards: 𝐶(48,4)=48!4!(48−4)!C(48,4)=4!(48−4)!48!​ 𝐶(48,4)=48×47×46×454×3×2×1=194580C(48,4)=4×3×2×148×47×46×45​=194580

    Total ways to select 5 cards with exactly one ace: 𝐶(4,1)×𝐶(48,4)=4×194580=778320C(4,1)×C(48,4)=4×194580=778320

    So, there are 778320 ways to select 5 cards with exactly one ace.

    Question 7

    In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?

    Solution:

    Select 4 bowlers from 5: 𝐶(5,4)=5!4!(5−4)!=5C(5,4)=4!(5−4)!5!​=5

    Select 7 non-bowlers from the remaining 12 players: 𝐶(12,7)=12!7!(12−7)!C(12,7)=7!(12−7)!12!​ 𝐶(12,7)=12×11×10×9×85×4×3×2×1=792C(12,7)=5×4×3×2×112×11×10×9×8​=792

    Total ways to form the cricket team: 𝐶(5,4)×𝐶(12,7)=5×792=3960C(5,4)×C(12,7)=5×792=3960

    So, there are 3960 ways to select a cricket team of 11 with exactly 4 bowlers from 17 players.

    Question 8

    A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.

    Solution:

    Select 2 black balls from 5: 𝐶(5,2)=5!2!(5−2)!=10C(5,2)=2!(5−2)!5!​=10

    Select 3 red balls from 6: 𝐶(6,3)=6!3!(6−3)!=20C(6,3)=3!(6−3)!6!​=20

    Total ways to select 2 black and 3 red balls: 𝐶(5,2)×𝐶(6,3)=10×20=200C(5,2)×C(6,3)=10×20=200

    So, there are 200 ways to select 2 black and 3 red balls from the bag.

    Question 9

    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?

    Solution:

    Out of the 5 courses, 2 are compulsory. Therefore, we need to choose 3 more courses from the remaining 7 courses.

    Select 3 courses from 7: 𝐶(7,3)=7!3!(7−3)!=7×6×53×2×1=35C(7,3)=3!(7−3)!7!​=3×2×17×6×5​=35

    So, there are 35 ways for a student to choose a program of 5 courses from 9 available courses with 2 compulsory course

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