Permutations and CombinationsPermutations vs. CombinationsPermutationsSlide 4r-permutations examplePermutation formula proofPermutations vs. r-permutationsSample questionCombinationsCombinations exampleCombination formula proofSlide 13Slide 14Bit stringsSlide 16Corollary 1Corollary exampleCombinatorial proofCircular seatingsHorse racesA last note on combinations1Permutations and CombinationsCS/APMA 202Epp section 6.4Aaron Bloomfield2Permutations vs. Combinations•Both are ways to count the possibilities•The difference between them is whether order matters or not•Consider a poker hand:–A♦, 5♥, 7♣, 10♠, K♠•Is that the same hand as:–K♠, 10♠, 7♣, 5♥, A♦•Does the order the cards are handed out matter?–If yes, then we are dealing with permutations–If no, then we are dealing with combinations3Permutations•A permutation is an ordered arrangement of the elements of some set S–Let S = {a, b, c}–c, b, a is a permutation of S–b, c, a is a different permutation of S•An r-permutation is an ordered arrangement of r elements of the set–A♦, 5♥, 7♣, 10♠, K♠ is a 5-permutation of the set of cards•The notation for the number of r-permutations: P(n,r)–The poker hand is one of P(52,5) permutations4Permutations•Number of poker hands (5 cards):–P(52,5) = 52*51*50*49*48 = 311,875,200•Number of (initial) blackjack hands (2 cards):–P(52,2) = 52*51 = 2,652•r-permutation notation: P(n,r)–The poker hand is one of P(52,5) permutations)1)...(2)(1(),(  rnnnnrnP)!(!rnnnrnii15r-permutations example•How many ways are there for 5 people in this class to give presentations?•There are 27 students in the class–P(27,5) = 27*26*25*24*23 = 9,687,600–Note that the order they go in does matter in this example!6Permutation formula proof•There are n ways to choose the first element–n-1 ways to choose the second–n-2 ways to choose the third–…–n-r+1 ways to choose the rth element•By the product rule, that gives us:P(n,r) = n(n-1)(n-2)…(n-r+1)7Permutations vs. r-permutations•r-permutations: Choosing an ordered 5 card hand is P(52,5)–When people say “permutations”, they almost always mean r-permutations•But the name can refer to both•Permutations: Choosing an order for all 52 cards is P(52,52) = 52!–Thus, P(n,n) = n!8Sample question•How many permutations of {a, b, c, d, e, f, g} end with a?–Note that the set has 7 elements•The last character must be a–The rest can be in any order•Thus, we want a 6-permutation on the set {b, c, d, e, f, g} •P(6,6) = 6! = 720•Why is it not P(7,6)?10Combinations•What if order doesn’t matter?•In poker, the following two hands are equivalent:–A♦, 5♥, 7♣, 10♠, K♠–K♠, 10♠, 7♣, 5♥, A♦•The number of r-combinations of a set with n elements, where n is non-negative and 0≤r≤n is:)!(!!),(rnrnrnC11Combinations example•How many different poker hands are there (5 cards)?•How many different (initial) blackjack hands are there?2,598,960!47*1*2*3*4*5!47*48*49*50*51*52!47!5!52)!552(!5!52)5,52( C1,3261*251*52!50!2!52)!252(!2!52)2,52( C12Combination formula proof•Let C(52,5) be the number of ways to generate unordered poker hands•The number of ordered poker hands is P(52,5) = 311,875,200•The number of ways to order a single poker hand is P(5,5) = 5! = 120•The total number of unordered poker hands is the total number of ordered hands divided by the number of ways to order each hand•Thus, C(52,5) = P(52,5)/P(5,5)13Combination formula proof•Let C(n,r) be the number of ways to generate unordered combinations•The number of ordered combinations (i.e. r-permutations) is P(n,r)•The number of ways to order a single one of those r-permutations P(r,r) •The total number of unordered combinations is the total number of ordered combinations (i.e. r-permutations) divided by the number of ways to order each combination•Thus, C(n,r) = P(n,r)/P(r,r)14Combination formula proof)!(!!)!/(!)!/(!),(),(),(rnrnrrrrnnrrPrnPrnC15Bit strings•How many bit strings of length 10 contain:a) exactly four 1’s?Find the positions of the four 1’sDoes the order of these positions matter?•Nope!•Positions 2, 3, 5, 7 is the same as positions 7, 5, 3, 2Thus, the answer is C(10,4) = 210b) at most four 1’s?There can be 0, 1, 2, 3, or 4 occurrences of 1Thus, the answer is:•C(10,0) + C(10,1) + C(10,2) + C(10,3) + C(10,4)•= 1+10+45+120+210•= 38616Bit strings•How many bit strings of length 10 contain:c) at least four 1’s?There can be 4, 5, 6, 7, 8, 9, or 10 occurrences of 1Thus, the answer is:•C(10,4) + C(10,5) + C(10,6) + C(10,7) + C(10,8) + C(10,9) + C(10,10)•= 210+252+210+120+45+10+1•= 848Alternative answer: subtract from 210 the number of strings with 0, 1, 2, or 3 occurrences of 1d) an equal number of 1’s and 0’s?Thus, there must be five 0’s and five 1’sFind the positions of the five 1’sThus, the answer is C(10,5) = 25217 ! )()!(!),(rnnrnnrnnC)!(!!rnrn)!(!!),(rnrnrnCCorollary 1•Let n and r be non-negative integers with r ≤ n. Then C(n,r) = C(n,n-r)•Proof:18Corollary example•There are C(52,5) ways to pick a 5-card poker hand•There are C(52,47) ways to pick a 47-card hand•P(52,5) = 2,598,960 = P(52,47)•When dealing 47 cards, you are picking 5 cards to not deal–As opposed to picking 5 card to deal–Again, the order the cards are dealt in does matter19Combinatorial proof•A combinatorial proof is a proof that uses counting arguments to prove a theorem–Rather than some other method such as algebraic techniques•Essentially, show that both sides of the proof manage to count the same objects•Most of the questions in this section are phrased as, “find out how many possibilities there are if …”–Instead, we could phrase each question as a theorem:–“Prove there are x possibilities if …”–The same answer could be modified to be a combinatorial proof to the theorem20Circular seatings•How many ways are there to sit 6 people around a circular table, where seatings are considered to be the same if they can be obtained from each other by rotating the table?•First, place the first person in the north-most chair–Only one possibility•Then place