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    Discrete Structures
    GE-167
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    Topics
    1. Mathematical Reasoning: Propositional and Predicate Logic2. Propositional Logic: Logical Operators3. Translations Between Symbolic Expressions and Formal English Expression4. Logical Equivalences5. Predicate Logic: Quantifiers6. Nested Quantification7. Equivalences in Predicate Logic8. Translations Between Symbolic Forms and Formal English9. Rules of Inference: Proof Methods and Strategies10. Direct Proof11. Proof by Contraposition12. Proof by Induction13. Proof by Implication14. Existence Proof15. Uniqueness Proofs16. Trivial Proofs17. Vacuous Proofs18. Sets: Notations and Set Operations19. Venn Diagrams20. Countable and Uncountable Sets21. Relations: Equivalence Relations and Partitions22. Partial Orderings23. Recurrence Relations24. Functions: Injective, Surjective, Bijective25. Special Types of Functions26. Function Composition27. Inverse Functions28. Recursive Functions29. Compositions30. Number Theory: Sequences and Series31. Counting: Inclusion and Exclusion Principle32. Pigeonhole Principle33. Permutations and Combinations34. Integers and Divisibility: Division Theorem35. Modular Arithmetic36. LCM and GCD37. Euclidean and Extended Euclidean Method38. Finding Solutions to Congruence39. Primes: Fundamental Theorem of Arithmetic40. Characterizations of Primes41. Mersenne Primes42. Induction: Weak Induction43. Strong Induction44. Recursion and Recurrences: Formulation of Recurrences45. Closed Formulas46. Counting: Product Rule and Sum Rule47. Principle of Inclusion-Exclusion48. Binomial Coefficients49. Pascal's Identity and Pascal’s Triangle50. Binomial Theorem51. Relations: Reflexive, Symmetric, Transitive, and Antisymmetric52. Equivalence Relations and Equivalence Classes53. Partial Orders54. Graph Theory: Terminologies55. Elements of Graph Theory56. Planar Graphs57. Graph Coloring58. Euler Graph59. Hamiltonian Path60. Rooted Trees61. Graph Traversals62. Handshaking Lemma and Corollary63. Special Families of Graphs64. Graph Isomorphism65. Planarity in Graphs66. Eulerian and Hamiltonian Graphs67. Trees in Graph Theory
    GE-167›Binomial Coefficients
    Discrete StructuresTopic 48 of 67

    Binomial Coefficients

    9 minread
    1,587words
    Intermediatelevel

    Binomial Coefficients

    Binomial coefficients are fundamental in combinatorics and represent the number of ways to choose a subset of items from a larger set. They arise naturally in the expansion of powers of binomials, as in the binomial theorem, and are used in counting problems, probability theory, and algebra.

    The binomial coefficient is often denoted as:

    (nk)=n!k!(n−k)!\binom{n}{k} = \frac{n!}{k!(n-k)!}(kn​)=k!(n−k)!n!​

    Where:

    • nnn is the total number of elements.
    • kkk is the number of elements to choose.
    • n!n!n! (read "n factorial") is the product of all positive integers from 1 to nnn (i.e., n!=n×(n−1)×⋯×1n! = n \times (n-1) \times \cdots \times 1n!=n×(n−1)×⋯×1).

    Interpretation

    The binomial coefficient (nk)\binom{n}{k}(kn​) represents the number of ways to select kkk items from a set of nnn distinct items, without regard to the order in which the items are chosen. This is known as a combination. It answers the question:

    • How many ways can we choose kkk elements from a set of nnn elements?

    For example:

    • (52)\binom{5}{2}(25​) counts how many ways you can choose 2 items from 5 distinct items.

    Properties of Binomial Coefficients

    1. Symmetry: The binomial coefficient is symmetric, meaning that:

      (nk)=(nn−k)\binom{n}{k} = \binom{n}{n-k}(kn​)=(n−kn​)

      This is because choosing kkk items from a set of nnn items is the same as leaving out n−kn-kn−k items, so there are an equal number of ways to choose kkk items or n−kn-kn−k items.

    2. Recurrence Relation: Binomial coefficients satisfy the recurrence relation:

      (nk)=(n−1k−1)+(n−1k)\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}(kn​)=(k−1n−1​)+(kn−1​)

      This recurrence shows that the number of ways to choose kkk elements from nnn elements can be broken into two parts:

      • The number of ways to choose k−1k-1k−1 elements from n−1n-1n−1 elements (if one particular element is included).
      • The number of ways to choose kkk elements from n−1n-1n−1 elements (if that particular element is not included).
    3. Boundary Conditions:

      • (n0)=1\binom{n}{0} = 1(0n​)=1 for all nnn, because there is exactly one way to choose 0 items from nnn items (the empty set).
      • (nn)=1\binom{n}{n} = 1(nn​)=1 for all nnn, because there is exactly one way to choose all nnn items from nnn.
      • (nk)=0\binom{n}{k} = 0(kn​)=0 if k>nk > nk>n, because it is impossible to choose more items than are available in the set.
    4. Binomial Theorem: The binomial coefficient plays a key role in the binomial theorem, which describes the expansion of the binomial expression (x+y)n(x + y)^n(x+y)n. It states:

      (x+y)n=∑k=0n(nk)xn−kyk(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k(x+y)n=k=0∑n​(kn​)xn−kyk

      This expansion expresses (x+y)n(x + y)^n(x+y)n as a sum of terms, where each term is a product of a binomial coefficient, powers of xxx, and powers of yyy.

    Examples of Binomial Coefficients

    1. (52)\binom{5}{2}(25​): To choose 2 elements from a set of 5, you can calculate:

      (52)=5!2!(5−2)!=5×4×3!2×1×3!=5×42×1=10\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4 \times 3!}{2 \times 1 \times 3!} = \frac{5 \times 4}{2 \times 1} = 10(25​)=2!(5−2)!5!​=2×1×3!5×4×3!​=2×15×4​=10

      So, there are 10 ways to choose 2 elements from a set of 5.

    2. (63)\binom{6}{3}(36​): To choose 3 elements from a set of 6:

      (63)=6!3!(6−3)!=6×5×4×3!3×2×1×3!=6×5×43×2×1=20\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4 \times 3!}{3 \times 2 \times 1 \times 3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20(36​)=3!(6−3)!6!​=3×2×1×3!6×5×4×3!​=3×2×16×5×4​=20

      So, there are 20 ways to choose 3 elements from a set of 6.

    3. (100)\binom{10}{0}(010​): Choosing 0 elements from a set of 10:

      (100)=1\binom{10}{0} = 1(010​)=1

      There is exactly one way to choose 0 elements, which is to choose the empty set.

    4. (77)\binom{7}{7}(77​): Choosing all 7 elements from a set of 7:

      (77)=1\binom{7}{7} = 1(77​)=1

      There is exactly one way to choose all the elements from the set.

    Applications of Binomial Coefficients

    1. Counting Subsets: Binomial coefficients are often used in problems involving the counting of subsets. For example, given a set of size nnn, the number of ways to choose kkk elements (a subset of size kkk) is given by (nk)\binom{n}{k}(kn​).

    2. Combinatorics: Binomial coefficients are used in combinatorics for problems such as:

      • Counting the number of ways to arrange objects.
      • Counting the number of possible selections, partitions, or groupings of items.
    3. Probability: In probability theory, binomial coefficients are used to compute probabilities in the binomial distribution. For example, the probability of getting exactly kkk successes in nnn independent trials with probability ppp of success on each trial is given by:

      P(X=k)=(nk)pk(1−p)n−kP(X = k) = \binom{n}{k} p^k (1 - p)^{n-k}P(X=k)=(kn​)pk(1−p)n−k
    4. Pascal’s Triangle: The binomial coefficients are arranged in Pascal’s Triangle, where each number is the sum of the two numbers directly above it. This triangular arrangement visually represents the recurrence relation for binomial coefficients.

      Here's how Pascal’s Triangle looks for the first few rows:

      11112113311464115101051\begin{matrix} 1 \\ 1 & 1 \\ 1 & 2 & 1 \\ 1 & 3 & 3 & 1 \\ 1 & 4 & 6 & 4 & 1 \\ 1 & 5 & 10 & 10 & 5 & 1 \\ \end{matrix}111111​12345​13610​1410​15​1​

      The numbers in each row are the binomial coefficients for successive values of nnn and kkk, where each entry (nk)\binom{n}{k}(kn​) appears in the nnn-th row and kkk-th column of the triangle.

    Conclusion

    Binomial coefficients are essential in combinatorics, probability, and algebra, as they provide a way to count combinations and play a key role in binomial expansions. Understanding their properties and applications is crucial for solving counting problems and analyzing various mathematical models involving selections and arrangements.

    Previous topic 47
    Principle of Inclusion-Exclusion
    Next topic 49
    Pascal's Identity and Pascal’s Triangle

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