Permutations • Definition ○ A permutation of a set of distinct objects is an ordered arrangement of these objects. ○ An ordered arrangement of r elements of a set is called an r-permutation. • Example ○ Let S = {1,2,3}. ○ The ordered arrangement 3,1,2 is a permutation of S. ○ The ordered arrangement 3,2 is a 2-permutation of S. ○ The number of r-permutations of a set with n elements is denoted by P(n,r). ○ The 2-permutations of S = {1,2,3} are 1,2; 1,3; 2,1; 2,3; 3,1; and 3,2 ○ Hence, P(3,2) = 6. A Formula for the Number of Permutations • Theorem 1 ○ If n is a positive integer and r is an integer with 1 ≤ r ≤ n, then there are ○ P(n,r)=n(n−1)(n−2)⋯(n−r+1) ○ r-permutations of a set with n distinct elements. • Proof ○ Use the product rule. ○ The first element can be chosen in n ways. ○ The second in n − 1 ways ○ And so on until there are (n−(r−1)) ways to choose the last element. ○ Note that P(n,0) = 1, since there is only one way to order zero elements. • Corollary 1 ○ If n and r are integers with 1 ≤ r ≤ n, then ○ P(n,r)=n!/(n−r)! • Example ○ How many ways are there to select a first-prize winner, a second prize winner, and a third-prize winner from 100 different people who have entered a contest? ○ P(100,3) = 100 ∙ 99 ∙ 98 = 970,200 Example ○ Suppose that a saleswoman has to visit eight different cities. ○ She must begin her trip in a specified city ○ But she can visit the other seven cities in any order she wishes. ○ How many possible orders can the saleswoman use when visiting these cities? ○ The first city is chosen, and the rest are ordered arbitrarily. ○ Hence the orders are: 7! = 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 5040 ○ If she wants to find the tour with the shortest path that visits all the cities, ○ she must consider 5040 paths! Example ○ How many permutations of the letters ABCDEFGH contain the string ABC ? ○ We solve this problem by counting the permutations of six objects, ABC, D, E, F, G, H. ○ 6! = 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 720 Combinations • Definition ○ An r-combination of elements of a set is an unordered selection of r elements from the set. ○ Thus, an r-combination is simply a subset of the set with r elements. ○ The number of r-combinations of a set with n distinct elements is denoted by C(n, r). ○ The notation (█(n@r)) is also used and is called a binomial coefficient. ○ (We will see the notation again in the binomial theorem in Section 6.4.) • Example: ○ Let S be the set {a, b, c, d}. ○ Then {a, c, d} is a 3-combination from S. ○ It is the same as {d, c, a} since the order listed does not matter. ○ C(4,2) = 6 because the 2-combinations of {a, b, c, d} are the six subsets ○ {a, b}, {a, c}, {a, d}, {b, c}, {b, d}, and {c, d}. • Theorem 2 ○ The number of r-combinations of a set with n elements, where n≥r ≥ 0, equals ○ C(n,r)=n!/(n−r)!r! • Proof ○ By the product rule P(n, r)=C(n,r)⋅P(r,r). Therefore, ○ C(n,r)=P(n,r)/P(r,r) =(n!∕(n−r)!)/(r!∕(r−r)!)=n!/(n−r)!r! • Example ○ How many poker hands of five cards can be dealt from a standard deck of 52 cards? ○ Also, how many ways are there to select 47 cards from a deck of 52 cards? ○ Since the order in which the cards are dealt does not matter ○ the number of five card hands is: ○ C(52,5)=52!/5!47!=(52⋅51⋅50⋅49⋅48)/(5⋅4⋅3⋅2⋅1)=26⋅17⋅10⋅49⋅12=2,598,960 ○ The different ways to select 47 cards from 52 is ○ C(52,47)=52!/47!5!=C(52,5)=2,598,960 • Corollary 2 ○ Let n and r be nonnegative integers with r≤n. Then C(n, r) = C(n, n−r). • Proof ○ From Theorem 2, it follows that ○ C(n,n−r)=n!/(n−r)!(n−(n−r))!=n!/(n−r)!r!=C(n,r) ○ Hence, C(n, r)=C(n, n−r) Combinatorial Proofs • Definition ○ A combinatorial proof of an identity is a proof that uses one of the following methods. ○ Double Counting Proof § A double counting proof uses counting arguments to prove that § both sides of an identity count the same objects, but in different ways. ○ Bijective Proof § A bijective proof shows that there is a bijection § between the sets of objects counted by the two sides of the identity. • Example ○ Here are two combinatorial proofs that C(n, r)=C(n, n−r) ○ Bijective Proof § Suppose that S is a set with n elements. § The function that maps a subset A of S to A ̅ is a bijection between § the subsets of S with r elements and the subsets with n−r elements. § Since there is a bijection between the two sets § They must have the same number of elements. ○ Double Counting Proof § By definition the number of subsets of S with r elements is C(n, r). § Each subset A of S can also be described by § specifying which elements are not in A, i.e., those which are in A ̅. § Since the complement of a subset of S with r elements has n−r elements § There are also C(n,n−r) subsets of S with r elements. More Examples • How many words can you formed by rearranging the letters in the word: ○ Combine § 7! ○ Permutation § Pick where t s go § Arrange remaining 9 letters § (█(11@2))⋅9!=11!/2! ○ Rearrange § Pick where r s go § Pick where e s go § Pick where a s go § Arrange remaining 2 letters § (█(9@3))⋅(█(6@2))⋅(█(4@2))⋅2!=9!/3!2!2! • In a game of cards a hand consists of 13 cards. How many possible hands are there with ○ Exactly one ace § Pick which ace § Pick the rest § 4⋅(█(52−4@12)) ○ At least one ace § 1 Ace + 2 Aces + 3 Aces + 4 Aces § 4⋅(█(52−4@12))+(█(4@2))⋅(█(52−4@11))+(█(4@3))⋅(█(52−4@10))+(█(4@4))⋅(█(52−4@9)) ○ Exactly one ace and two diamonds § Case 1: We pick one diamond and a diamond ace □ 1⋅12⋅(█(52−4−12@11)) § Case 2: We pick two diamonds and a different ace □ 3⋅(█(12@2))⋅(█(52−4−12@10)) § So the total number of hands is 1⋅12⋅(█(52−4−12@11))+3⋅(█(12@2))⋅(█(52−4−12@10))
