2 permutations and combinations lesson 8 hand out reference sheet and go over it
TRANSCRIPT
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Permutations and Permutations and CombinationsCombinations
Lesson 8Lesson 8
HAND OUT REFERENCE SHEET HAND OUT REFERENCE SHEET AND GO OVER ITAND GO OVER IT
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Permutations vs. CombinationsPermutations vs. Combinations
Both are ways to count the possibilitiesBoth are ways to count the possibilitiesThe difference between them is whether order The difference between them is whether order matters or notmatters or notConsider a poker hand:Consider a poker hand: AA♦, 5♥, 7♣, 10♠, K♠♦, 5♥, 7♣, 10♠, K♠
Is that the same hand as:Is that the same hand as: K♠, 10♠, 7♣, 5♥, K♠, 10♠, 7♣, 5♥, AA♦♦
Does the order the cards are handed out Does the order the cards are handed out matter?matter? If yes, then we are dealing with permutationsIf yes, then we are dealing with permutations If no, then we are dealing with combinationsIf no, then we are dealing with combinations
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PermutationsPermutations
A permutation is an ordered arrangement of the A permutation is an ordered arrangement of the elements of some set elements of some set SS Let Let SS = {a, b, c} = {a, b, c} c, b, a is a permutation of Sc, b, a is a permutation of S b, c, a is a b, c, a is a different different permutation of Spermutation of S
An An rr-permutation is an ordered arrangement of -permutation is an ordered arrangement of rr elements of the setelements of the set AA♦, 5♥, 7♣, 10♠, K♠ is a 5-permutation of the set of ♦, 5♥, 7♣, 10♠, K♠ is a 5-permutation of the set of
cardscards
The notation for the number of The notation for the number of rr-permutations: -permutations: PP((nn,,rr))
The poker hand is one of P(52,5) permutationsThe poker hand is one of P(52,5) permutations
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PermutationsPermutations
Number of poker hands (5 cards):Number of poker hands (5 cards): PP(52,5) = 52*51*50*49*48 = 311,875,200(52,5) = 52*51*50*49*48 = 311,875,200
Number of (initial) blackjack hands (2 cards):Number of (initial) blackjack hands (2 cards): PP(52,2) = 52*51 = 2,652(52,2) = 52*51 = 2,652
rr-permutation notation: -permutation notation: PP((nn,,rr)) The poker hand is one of P(52,5) permutationsThe poker hand is one of P(52,5) permutations
)1)...(2)(1(),( rnnnnrnP
)!(
!
rn
n
n
rni
i1
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rr-permutations example-permutations example
How many ways are there for 5 people in How many ways are there for 5 people in this class to give presentations?this class to give presentations?
There are 27 students in the classThere are 27 students in the class P(27,5) = 27*26*25*24*23 = 9,687,600P(27,5) = 27*26*25*24*23 = 9,687,600 Note that the order they go in does matter in Note that the order they go in does matter in
this example!this example!
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Permutation formula proofPermutation formula proof
There are There are nn ways to choose the first ways to choose the first elementelement nn-1 ways to choose the second-1 ways to choose the second nn-2 ways to choose the third-2 ways to choose the third …… nn--rr+1 ways to choose the +1 ways to choose the rrthth element element
By the product rule, that gives us:By the product rule, that gives us:
PP((nn,,rr) = ) = nn((nn-1)(-1)(nn-2)…(-2)…(nn--rr+1)+1)
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Permutations vs. Permutations vs. rr-permutations-permutations
rr-permutations: Choosing an ordered 5 -permutations: Choosing an ordered 5 card hand is card hand is PP(52,5)(52,5) When people say “permutations”, they almost When people say “permutations”, they almost
always mean always mean rr-permutations-permutationsBut the name can refer to bothBut the name can refer to both
Permutations: Choosing an order for all 52 Permutations: Choosing an order for all 52 cards is cards is PP(52,52) = 52!(52,52) = 52! Thus, Thus, PP((nn,,nn) = ) = nn!!
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Question 3Question 3
How many permutations of {a, b, c, d, e, f, g} How many permutations of {a, b, c, d, e, f, g} end with a?end with a? Note that the set has 7 elementsNote that the set has 7 elements
The last character must be aThe last character must be a The rest can be in any orderThe rest can be in any order
Thus, we want a 6-permutation on the set {b, c, Thus, we want a 6-permutation on the set {b, c, d, e, f, g} d, e, f, g} P(6,6) = 6! = 720P(6,6) = 6! = 720
Why is it not P(7,6)?Why is it not P(7,6)?
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CombinationsCombinations
What if order What if order doesn’tdoesn’t matter? matter?
In poker, the following two hands are equivalent:In poker, the following two hands are equivalent: AA♦, 5♥, 7♣, 10♠, K♠♦, 5♥, 7♣, 10♠, K♠ K♠, 10♠, 7♣, 5♥, K♠, 10♠, 7♣, 5♥, AA♦♦
The number of The number of rr-combinations of a set with -combinations of a set with nn elements, where elements, where nn is non-negative and 0≤ is non-negative and 0≤rr≤≤nn is: is:
)!(!
!),(
rnr
nrnC
1111
Combinations exampleCombinations example
How many different poker hands are there How many different poker hands are there (5 cards)?(5 cards)?
How many different (initial) blackjack How many different (initial) blackjack hands are there?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(
C
1,3261*2
51*52
!50!2
!52
)!252(!2
!52)2,52(
C
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Combination formula proofCombination formula proof
Let Let CC(52,5) be the number of ways to generate (52,5) be the number of ways to generate unordered poker handsunordered poker hands
The number of ordered poker hands is The number of ordered poker hands is PP(52,5) = (52,5) = 311,875,200311,875,200
The number of ways to order a single poker The number of ways to order a single poker hand is hand is PP(5,5) = 5! = 120(5,5) = 5! = 120
The total number of unordered poker hands is The total number of unordered poker hands is the total number of ordered hands divided by the the total number of ordered hands divided by the number of ways to order each handnumber of ways to order each hand
Thus, Thus, CC(52,5) = (52,5) = PP(52,5)/(52,5)/PP(5,5)(5,5)
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Combination formula proofCombination formula proof
Let Let CC((nn,,rr) be the number of ways to generate ) be the number of ways to generate unordered combinationsunordered combinationsThe number of ordered combinations (i.e. The number of ordered combinations (i.e. rr--permutations) is permutations) is PP((nn,,rr))The number of ways to order a single one of The number of ways to order a single one of those those rr-permutations -permutations PP((r,rr,r) ) The total number of unordered combinations is The total number of unordered combinations is the total number of ordered combinations (i.e. the total number of ordered combinations (i.e. rr--permutations) divided by the number of ways to permutations) divided by the number of ways to order each combinationorder each combinationThus, Thus, CC((n,rn,r) = ) = PP((n,rn,r)/)/PP((r,rr,r))
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Combination formula proofCombination formula proof
Note that the textbook explains it slightly Note that the textbook explains it slightly differently, but it is same proofdifferently, but it is same proof
)!(!
!
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)!/(!
),(
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rnr
n
rrr
rnn
rrP
rnPrnC
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Question 11Question 11
How many bit strings of length 10 contain:How many bit strings of length 10 contain:a)a) exactly four 1’s?exactly four 1’s?
Find the positions of the four 1’sFind the positions of the four 1’sDoes the order of these positions matter?Does the order of these positions matter?
Nope!Nope!Positions 2, 3, 5, 7 is the same as positions 7, 5, 3, 2Positions 2, 3, 5, 7 is the same as positions 7, 5, 3, 2
Thus, the answer is Thus, the answer is CC(10,4) = 210(10,4) = 210
b)b) at most four 1’s?at most four 1’s?There can be 0, 1, 2, 3, or 4 occurrences of 1There can be 0, 1, 2, 3, or 4 occurrences of 1Thus, the answer is:Thus, the answer is:
CC(10,0) + (10,0) + CC(10,1) + (10,1) + CC(10,2) + (10,2) + CC(10,3) + (10,3) + CC(10,4)(10,4)= 1+10+45+120+210= 1+10+45+120+210= 386= 386
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Question 11Question 11
How many bit strings of length 10 contain:How many bit strings of length 10 contain:c)c) at least four 1’s?at least four 1’s?
There can be 4, 5, 6, 7, 8, 9, or 10 occurrences of 1There can be 4, 5, 6, 7, 8, 9, or 10 occurrences of 1Thus, the answer is:Thus, the answer is:
c)c) CC(10,4) + (10,4) + CC(10,5) + (10,5) + CC(10,6) + (10,6) + CC(10,7) + (10,7) + CC(10,8) + (10,8) + CC(10,9) (10,9) + + CC(10,10)(10,10)= 210+252+210+120+45+10+1= 210+252+210+120+45+10+1= 848= 848
Alternative answer: subtract from 2Alternative answer: subtract from 21010 the number of the number of strings with 0, 1, 2, or 3 occurrences of 1strings with 0, 1, 2, or 3 occurrences of 1
an equal number of 1’s and 0’s?an equal number of 1’s and 0’s?Thus, there must be five 0’s and five 1’sThus, there must be five 0’s and five 1’sFind the positions of the five 1’sFind the positions of the five 1’sThus, the answer is Thus, the answer is CC(10,5) = 252(10,5) = 252
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Corollary 1Corollary 1
Let Let nn and and rr be non-negative integers with be non-negative integers with rr ≤ ≤ nn. Then . Then CC((nn,,rr) = ) = CC((nn,,n-rn-r))
Proof:Proof:
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!),(
rnr
nrnC
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!),(
rnnrn
nrnnC
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rnr
n
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Corollary exampleCorollary example
There are C(52,5) ways to pick a 5-card poker There are C(52,5) ways to pick a 5-card poker handhand
There are C(52,47) ways to pick a 47-card handThere are C(52,47) ways to pick a 47-card hand
P(52,5) = 2,598,960 = P(52,47)P(52,5) = 2,598,960 = P(52,47)
When dealing 47 cards, you are picking 5 cards When dealing 47 cards, you are picking 5 cards to not dealto not deal As opposed to picking 5 card to dealAs opposed to picking 5 card to deal Again, the order the cards are dealt in does matterAgain, the order the cards are dealt in does matter
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Combinatorial proofCombinatorial proof
A A combinatorial proofcombinatorial proof is a proof that uses counting arguments to is a proof that uses counting arguments to prove a theoremprove a theorem
Rather than some other method such as algebraic techniquesRather than some other method such as algebraic techniques
Essentially, show that both sides of the proof manage to count the Essentially, show that both sides of the proof manage to count the same objectssame objects
In a typical Rosen example, he does not do much with this proof In a typical Rosen example, he does not do much with this proof method in this sectionmethod in this section
We will see more in the next sectionsWe will see more in the next sections
Most of the questions in this section are phrased as, “find out how Most of the questions in this section are phrased as, “find out how many possibilities there are if …”many possibilities there are if …”
Instead, we could phrase each question as a theorem:Instead, we could phrase each question as a theorem: ““Prove there are Prove there are xx possibilities if …” possibilities if …” The same answer could be modified to be a combinatorial proof to the The same answer could be modified to be a combinatorial proof to the
theoremtheorem
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Question 40Question 40
How many ways are there to sit 6 people around a circular table, 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 where seatings are considered to be the same if they can be obtained from each other by rotating the table?obtained from each other by rotating the table?
First, place the first person in the north-most chairFirst, place the first person in the north-most chair Only one possibilityOnly one possibility
Then place the other 5 peopleThen place the other 5 people There are P(5,5) = 5! = 120 ways to do thatThere are P(5,5) = 5! = 120 ways to do that
By the product rule, we get 1*120 =120By the product rule, we get 1*120 =120
Alternative means to answer this:Alternative means to answer this:
There are P(6,6)=720 ways to seat the 6 people around the tableThere are P(6,6)=720 ways to seat the 6 people around the table
For each seating, there are 6 “rotations” of the seatingFor each seating, there are 6 “rotations” of the seating
Thus, the final answer is 720/6 = 120Thus, the final answer is 720/6 = 120
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Question 42Question 42
How many ways are there for 4 horses to finish if ties are allowed?How many ways are there for 4 horses to finish if ties are allowed? Note that order does matter!Note that order does matter!
Solution by casesSolution by cases No tiesNo ties
The number of permutations is P(4,4) = 4! = 24The number of permutations is P(4,4) = 4! = 24 Two horses tieTwo horses tie
There are There are CC(4,2) = 6 ways to choose the two horses that tie(4,2) = 6 ways to choose the two horses that tieThere are There are PP(3,3) = 6 ways for the “groups” to finish(3,3) = 6 ways for the “groups” to finish
A “group” is either a single horse or the two tying horsesA “group” is either a single horse or the two tying horses
By the product rule, there are 6*6 = 36 possibilities for this caseBy the product rule, there are 6*6 = 36 possibilities for this case Two groups of two horses tieTwo groups of two horses tie
There are There are CC(4,2) = 6 ways to choose the two winning horses(4,2) = 6 ways to choose the two winning horsesThe other two horses tie for second placeThe other two horses tie for second place
Three horses tie with each otherThree horses tie with each otherThere are There are CC(4,3) = 4 ways to choose the two horses that tie(4,3) = 4 ways to choose the two horses that tieThere are There are PP(2,2) = 2 ways for the “groups” to finish(2,2) = 2 ways for the “groups” to finishBy the product rule, there are 4*2 = 8 possibilities for this caseBy the product rule, there are 4*2 = 8 possibilities for this case
All four horses tieAll four horses tieThere is only one combination for thisThere is only one combination for this
By the sum rule, the total is 24+36+6+8+1 = 75By the sum rule, the total is 24+36+6+8+1 = 75
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A last note on combinationsA last note on combinations
An alternative (and more common) way to An alternative (and more common) way to denote an denote an rr-combination:-combination:
I’ll use C(I’ll use C(nn,,rr) whenever possible, as it is ) whenever possible, as it is easier to write in PowerPointeasier to write in PowerPoint
r
nrnC ),(
EXAMPLE 1EXAMPLE 1
A home security system had buttons A home security system had buttons numbered 0 through 9. The system can numbered 0 through 9. The system can be disarmed by pressing 3 different be disarmed by pressing 3 different buttons in the correct sequence. How buttons in the correct sequence. How man different 3-button sequences are man different 3-button sequences are possible?possible?
PermutationPermutation
n = 10, r = 3n = 10, r = 32323
Example 2Example 2
Janie has 12 CDs. Her CD player holds 3 Janie has 12 CDs. Her CD player holds 3 CDs at a time. How many ways can she CDs at a time. How many ways can she choose 3 CDs to put into her CD player?choose 3 CDs to put into her CD player?
CombinationCombination
n = 12n = 12
r = 3r = 3
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Example 3Example 3
A committee of 3 students is to be chosen A committee of 3 students is to be chosen from a group of 5 students. Jason, Lily, from a group of 5 students. Jason, Lily, and Mariene are students in the group. and Mariene are students in the group. What is the probability that all 3 of them What is the probability that all 3 of them will be chosen for the committee?will be chosen for the committee?
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