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    MATH2113

    Discrete Mathematics

    Progress0 / 25 done
    Topics25
    1Sets, Subsets, Algebra of Sets2Propositions and Compound Statements3Basic Logical Operations4Propositional Logic and its Applications with Statement Problems5Propositions and Truth Tables6Tautologies and Contradictions7Conditional and Bi-conditional Statements8Arguments in Propositional Logic9Propositional Functions10Quantifiers and Negation of Quantified Statements11Relations and Equivalence Relations12Partial Ordering Relations13Functions and Recursively Defined Functions
    14Basics of Counting Methods15Combinations and Permutations16Pigeonhole Principle17Graphs and its Types18Graph Isomorphism19Trees in Graph Theory20Connectivity in Graphs21Eulerian and Hamiltonian Paths22Spanning Trees and Shortest Path Problem
    23Power, Floor, Increasing, Decreasing24Big O, Little O and Omega Notations25Orders of the Polynomial Functions
    9 min read
    Topic 15 of 25
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    15

    Combinations and Permutations

    Combinations and Permutations


    Combinations and Permutations


    Combinations and permutations are two fundamental concepts in combinatorics, dealing with the ways in which a set of objects can be arranged or selected. The key difference lies in whether the order of selection matters.


    1. Permutations

    A permutation refers to an arrangement of objects where the order of the objects matters. Essentially, it counts the number of ways to arrange rrr objects from a set of nnn distinct objects.

    Formula for Permutations:

    The number of ways to arrange rrr objects from nnn distinct objects is given by the permutation formula:

    P(n,r)=n!(n−r)!P(n, r) = \frac{n!}{(n - r)!}P(n,r)=(n−r)!n!​

    Where:

    • P(n,r)P(n, r)P(n,r) is the number of permutations of rrr objects chosen from nnn.
    • n!n!n! is the factorial of nnn, the total number of objects.
    • (n−r)!(n - r)!(n−r)! accounts for the number of arrangements of the remaining objects.

    Example 1:

    Suppose you have 5 distinct books and you want to arrange 3 of them on a shelf. The number of ways to do this is:

    P(5,3)=5!(5−3)!=5!2!=5×4×3×2!2!=60P(5, 3) = \frac{5!}{(5 - 3)!} = \frac{5!}{2!} = \frac{5 \times 4 \times 3 \times 2!}{2!} = 60P(5,3)=(5−3)!5!​=2!5!​=2!5×4×3×2!​=60

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

    Example 2:

    You have 7 people, and you need to select and arrange 4 of them for a team photo. The number of ways to do this is:

    P(7,4)=7!(7−4)!=7!3!=7×6×5×4×3!3!=840P(7, 4) = \frac{7!}{(7 - 4)!} = \frac{7!}{3!} = \frac{7 \times 6 \times 5 \times 4 \times 3!}{3!} = 840P(7,4)=(7−4)!7!​=3!7!​=3!7×6×5×4×3!​=840

    So, there are 840 ways to select and arrange 4 people from 7.


    2. Combinations

    A combination refers to a selection of objects where the order does not matter. Combinations are used when the arrangement of the selected items is irrelevant. It counts the number of ways to select rrr objects from a set of nnn distinct objects without regard to order.

    Formula for Combinations:

    The number of ways to select rrr objects from a set of nnn distinct objects is given by the combination formula:

    C(n,r)=(nr)=n!r!(n−r)!C(n, r) = \binom{n}{r} = \frac{n!}{r!(n - r)!}C(n,r)=(rn​)=r!(n−r)!n!​

    Where:

    • C(n,r)C(n, r)C(n,r) is the number of combinations of rrr objects chosen from nnn.
    • r!r!r! accounts for the fact that the order doesn't matter by dividing out the permutations of the rrr objects.

    Example 1:

    Suppose you have 5 distinct books and you want to select 3 books to read. The number of ways to do this is:

    C(5,3)=5!3!(5−3)!=5×4×3!3!×2!=10C(5, 3) = \frac{5!}{3!(5 - 3)!} = \frac{5 \times 4 \times 3!}{3! \times 2!} = 10C(5,3)=3!(5−3)!5!​=3!×2!5×4×3!​=10

    So, there are 10 ways to select 3 books out of 5, where the order of selection doesn’t matter.

    Example 2:

    You have 7 people, and you need to select 4 of them to form a committee. The number of ways to do this is:

    C(7,4)=7!4!(7−4)!=7×6×5×4!4!×3!=35C(7, 4) = \frac{7!}{4!(7 - 4)!} = \frac{7 \times 6 \times 5 \times 4!}{4! \times 3!} = 35C(7,4)=4!(7−4)!7!​=4!×3!7×6×5×4!​=35

    So, there are 35 ways to select 4 people from 7 without regard to the order in which they are selected.


    3. Key Differences Between Permutations and Combinations

    • Order Matters:

      • Permutations: The order of the selected objects matters. For example, arranging books on a shelf (where the position of each book matters) involves permutations.
      • Combinations: The order of the selected objects does not matter. For example, selecting books from a shelf to read (where the selection itself, not the order of selection, matters) involves combinations.
    • Formulae:

      • Permutations: P(n,r)=n!(n−r)!P(n, r) = \frac{n!}{(n - r)!}P(n,r)=(n−r)!n!​
      • Combinations: C(n,r)=n!r!(n−r)!C(n, r) = \frac{n!}{r!(n - r)!}C(n,r)=r!(n−r)!n!​

    4. Examples of Applications of Permutations and Combinations

    Permutations in Real Life:

    1. Arranging people in a line: If you have 10 people and you want to arrange 3 of them in a specific order (e.g., for a photo or seating arrangement), you would use permutations.
    2. Assigning tasks: If you are assigning 3 tasks to 3 people out of 10 people, the order in which tasks are assigned matters, so you use permutations.

    Combinations in Real Life:

    1. Choosing a team: If you have 10 players and need to select 5 to form a team, the order of selection does not matter, so combinations are used.
    2. Choosing a subset of books: If you have 10 books and need to choose 3 books to read, the order doesn’t matter, so you use combinations.

    5. Multinomial Coefficients (Generalization of Combinations)

    When you need to select items from more than two groups (i.e., if you have multiple categories), you use multinomial coefficients. A multinomial coefficient generalizes combinations to the case of selecting multiple subsets from a set of objects divided into categories.

    For example, the number of ways to divide a set of nnn objects into kkk groups with sizes n1,n2,…,nkn_1, n_2, \dots, n_kn1​,n2​,…,nk​ is given by:

    (nn1,n2,…,nk)=n!n1!n2!…nk!\binom{n}{n_1, n_2, \dots, n_k} = \frac{n!}{n_1! n_2! \dots n_k!}(n1​,n2​,…,nk​n​)=n1​!n2​!…nk​!n!​

    Example:

    Suppose you have 10 objects, and you want to divide them into 3 groups: 3 objects in the first group, 4 objects in the second group, and 3 objects in the third group. The number of ways to do this is:

    (103,4,3)=10!3!4!3!\binom{10}{3, 4, 3} = \frac{10!}{3!4!3!}(3,4,310​)=3!4!3!10!​

    Summary:

    • Permutations: Arrangements where order matters. The formula is P(n,r)=n!(n−r)!P(n, r) = \frac{n!}{(n - r)!}P(n,r)=(n−r)!n!​.
    • Combinations: Selections where order does not matter. The formula is C(n,r)=n!r!(n−r)!C(n, r) = \frac{n!}{r!(n - r)!}C(n,r)=r!(n−r)!n!​.
    • Key Difference: Permutations are used when order is important; combinations are used when order does not matter.
    • Multinomial Coefficients: A generalization of combinations used when dealing with multiple groups.

    These counting methods form the core of combinatorics and are essential tools in solving problems related to arrangements, selections, and probability.

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    Combinatorics: Basics of Counting Methods
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    Pigeonhole Principle

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