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- https://math.libretexts.org/Bookshelves/Mathematical_Logic_and_Proof/Book%3A_Mathematical_Reasoning__Writing_and_Proof_(Sundstrom)/04%3A_Mathematical_Induction/4.02%3A_Other_Forms_of_Mathematical_InductionSuppose that we want to prove that if P(k) is true, then P(k+1) is true. (This could be the inductive step in an induction proof.) To do this, we would be assuming that k!>2k and woul...Suppose that we want to prove that if P(k) is true, then P(k+1) is true. (This could be the inductive step in an induction proof.) To do this, we would be assuming that k!>2k and would need to prove that (k+1)!>2k+1. Hence, by the Second Principle of Mathematical Induction, we conclude that P(n) is true for all n∈N with n≥2, and this means that each natural number greater than 1 is either a prime number or is a product of prime numbers.
- https://math.libretexts.org/Bookshelves/Precalculus/Precalculus_(Stitz-Zeager)/09%3A_Sequences_and_the_Binomial_Theorem/9.04%3A_The_Binomial_TheoremSimply stated, the Binomial Theorem is a formula for the expansion of quantities for natural numbers.
- https://math.libretexts.org/Courses/Lorain_County_Community_College/Book%3A_Precalculus_(Stitz-Zeager)_-_Jen_Test_Copy/09%3A_Sequences_and_the_Binomial_Theorem/9.04%3A_The_Binomial_TheoremSimply stated, the Binomial Theorem is a formula for the expansion of quantities for natural numbers.
- https://math.libretexts.org/Bookshelves/Applied_Mathematics/Applied_Finite_Mathematics_(Sekhon_and_Bloom)/07%3A_Sets_and_Counting/7.03%3A_PermutationsThere are four choices for the first letter of our word, three choices for the second letter, and two choices for the third. Since the first letter must be a consonant, we have four choices for the fi...There are four choices for the first letter of our word, three choices for the second letter, and two choices for the third. Since the first letter must be a consonant, we have four choices for the first position, and once we use up a consonant, there are only three consonants left for the last spot. 7P2=7!5!=7⋅6⋅5⋅4⋅3⋅2⋅15⋅4⋅3⋅2⋅1=42
- https://math.libretexts.org/Courses/SUNY_Schenectady_County_Community_College/Discrete_Structures/04%3A_Mathematical_Induction/4.02%3A_Other_Forms_of_Mathematical_InductionSuppose that we want to prove that if P(k) is true, then P(k+1) is true. (This could be the inductive step in an induction proof.) To do this, we would be assuming that k!>2k and woul...Suppose that we want to prove that if P(k) is true, then P(k+1) is true. (This could be the inductive step in an induction proof.) To do this, we would be assuming that k!>2k and would need to prove that (k+1)!>2k+1. Hence, by the Second Principle of Mathematical Induction, we conclude that P(n) is true for all n∈N with n≥2, and this means that each natural number greater than 1 is either a prime number or is a product of prime numbers.
- https://math.libretexts.org/Courses/Lorain_County_Community_College/Book%3A_Precalculus_Jeffy_Edits_3.75/09%3A_Sequences_and_the_Binomial_Theorem/9.04%3A_The_Binomial_TheoremSimply stated, the Binomial Theorem is a formula for the expansion of quantities for natural numbers.
- https://math.libretexts.org/Courses/SUNY_Schenectady_County_Community_College/Discrete_Structures/11%3A_Counting/11.03%3A_Combinations_and_PermutationsA permutation is a (possible) rearrangement of objects.
- https://math.libretexts.org/Bookshelves/Algebra/Advanced_Algebra/09%3A_Sequences_Series_and_the_Binomial_Theorem/9.04%3A_Binomial_TheoremThe binomial theorem provides a method of expanding binomials raised to powers without directly multiplying each factor.
- https://math.libretexts.org/Courses/Chabot_College/Math_in_Society_(Zhang)/08%3A_Counting/8.01%3A_The_Fundamental_Counting_PrincipleIf there are n1 ways to of choosing the first item, n2 ways of choosing the second item after the first item is chosen, n3 ways of choosing the third item after the first two have been ...If there are n1 ways to of choosing the first item, n2 ways of choosing the second item after the first item is chosen, n3 ways of choosing the third item after the first two have been chosen, and so on until there are nk ways of choosing the last item after the earlier choices, then the total number of choices overall is given by
- https://math.libretexts.org/Courses/Fullerton_College/Math_100%3A_Liberal_Arts_Math_(Claassen_and_Ikeda)/06%3A_Probability/6.04%3A_Counting_MethodsSo far the problems we have looked at had rather small total number of outcomes. We could easily count the number of elements in the sample space. If there are a large number of elements in the sample...So far the problems we have looked at had rather small total number of outcomes. We could easily count the number of elements in the sample space. If there are a large number of elements in the sample space we can use counting techniques such as permutations or combinations to count them.
- https://math.libretexts.org/Courses/Las_Positas_College/Math_for_Liberal_Arts/12%3A_Counting/12.01%3A_The_Fundamental_Counting_PrincipleIf there are n1 ways to of choosing the first item, n2 ways of choosing the second item after the first item is chosen, n3 ways of choosing the third item after the first two have been ...If there are n1 ways to of choosing the first item, n2 ways of choosing the second item after the first item is chosen, n3 ways of choosing the third item after the first two have been chosen, and so on until there are nk ways of choosing the last item after the earlier choices, then the total number of choices overall is given by
