21 Permutations and Combinations

8/2/2019 21 Permutations and Combinations

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Permutations andPermutations and

CombinationsCombinationsCS/APMA 202CS/APMA 202

Rosen section 4.3Rosen section 4.3Aaron BloomfieldAaron Bloomfield


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Permutations vs. CombinationsPermutations vs. Combinations

BothBoth areare waysways toto countcount thethe possibilitiespossibilities

TheThe differencedifference betweenbetween themthem isis whetherwhether orderordermattersmatters oror notnot

ConsiderConsider aa pokerpoker handhand:: AA,, 55,, 77,, 1010,, KK

IsIs thatthat thethe samesame handhand asas:: KK,, 1010,, 77,, 55,, AA

DoesDoes thethe orderorder thethe cardscards areare handedhanded outoutmatter?matter? IfIf yes,yes, thenthen wewe areare dealingdealing withwith permutationspermutations

IfIf no,no, thenthen wewe areare dealingdealing withwith combinationscombinations


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PermutationsPermutations

AA permutationpermutation isis anan orderedordered arrangementarrangement ofof thetheelementselements ofof somesome setset SS LetLet SS== {a,{a, b,b, c}c}

c,c, b,b, aa isis aa permutationpermutation ofof SS b,b, c,c, aa isis aa differentdifferent permutationpermutation ofof SS

AnAn rr--permutationpermutation isis anan orderedordered arrangementarrangement ofofrrelementselements ofof thethe setset

AA,, 55,, 77,, 1010,, KK isis aa 55--permutationpermutation ofof thethe setset ofofcardscards

TheThe notationnotation forfor thethe numbernumber ofof rr--permutationspermutations::PP((nn,,rr)) TheThe pokerpoker handhand isis oneone ofof P(P(5252,,55)) permutationspermutations


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PermutationsPermutations

NumberNumber ofof pokerpoker handshands ((55 cards)cards)::

PP((5252,,55)) == 5252**5151**5050**4949**4848 == 311311,,875875,,200200

NumberNumber ofof (initial)(initial) blackjackblackjack handshands ((22 cards)cards)::

PP((5252,,22)) == 5252**5151 == 22,,652652rr--permutationpermutation notationnotation:: PP((nn,,rr))

TheThe pokerpoker handhand isis oneone ofof P(P(5252,,55)) permutationspermutations

)1)...(2)(1(),( ! rnnnnrnP

)!(

!

rn

n

!

!

!

n

rni

i

1


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rr--permutations examplepermutations example

HowHow manymany waysways areare therethere forfor55 peoplepeople inin

thisthis classclass toto givegive presentations?presentations?

ThereThere areare 2727 studentsstudents inin thethe classclass

P(P(2727,,55)) == 2727**2626**2525**2424**2323 == 99,,687687,,600600

NoteNote thatthat thethe orderorder theythey gogo inin doesdoes mattermatter ininthisthis example!example!


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Permutation formula proofPermutation formula proof

ThereThere areare nn waysways toto choosechoose thethe firstfirst

elementelement

nn--11 waysways toto choosechoose thethe secondsecond nn--22 waysways toto choosechoose thethe thirdthird

nn--rr++11 waysways toto choosechoose thethe rrthth elementelement

ByBy thethe productproduct rule,rule, thatthat givesgives usus::

PP((nn,,rr)) == nn((nn--11)()(nn--22))((nn--rr++11))


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Permutations vs.Permutations vs. rr--permutationspermutations

rr--permutationspermutations:: ChoosingChoosing anan orderedordered 55

cardcard handhand isis PP((5252,,55))

When

When peoplepeople saysay permutations,permutations, theythey almostalmostalwaysalways meanmean rr--permutationspermutations

ButBut thethe namename cancan referrefer toto bothboth

PermutationsPermutations:: ChoosingChoosing anan orderorder forfor allall 5252

cardscards isis PP((5252,,5252)) == 5252!!

Thus,Thus, PP((nn,,nn)) == nn!!


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Rosen, section 4.3, question 3Rosen, section 4.3, question 3

HowHow manymany permutationspermutations ofof {a,{a, b,b, c,c, d,d, e,e, f,f, g}g}endend withwith a?a? NoteNote thatthat thethe setset hashas 77 elementselements

TheThe lastlast charactercharacter mustmust bebe aa TheThe restrest cancan bebe inin anyany orderorder

Thus,Thus, wewe wantwant aa 66--permutationpermutation onon thethe setset {b,{b, c,c,

d,d, e,e, f,f, g}g}P(P(66,,66)) == 66!! == 720720

WhyWhy isis itit notnot P(P(77,,66)?)?


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CombinationsCombinations

WhatWhat ifif orderorderdoesntdoesnt matter?matter?

InIn poker,poker, thethe followingfollowing twotwo handshands areare equivalentequivalent::

AA,, 55,, 77,, 1010,, KK

KK,, 1010,, 77,, 55,, AA

TheThe numbernumber ofof rr--combinationscombinations ofof aa setset withwith nn

elements,elements, wherewhere nn isis nonnon--negativenegative andand 00rrnn isis::

)!(!

!),(

rnr

nrnC

!


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Combinations exampleCombinations example

HowHow manymany differentdifferent pokerpoker handshands areare therethere

((55 cards)?cards)?

HowHow manymany differentdifferent (initial)(initial) blackjackblackjack

handshands areare there?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

LetLet CC((5252,,55)) bebe thethe numbernumber ofof waysways toto generategenerate

unorderedunordered pokerpoker handshands

TheThe numbernumber ofof orderedordered pokerpoker handshands isis PP((5252,,55)) ==

311311,,875875,,200200

TheThe numbernumber ofof waysways toto orderorder aa singlesingle pokerpoker

handhand isis PP((55,,55)) == 55!! == 120120

TheThe totaltotal numbernumber ofof unorderedunordered pokerpoker handshands isisthethe totaltotal numbernumber ofof orderedordered handshands divideddivided byby thethe

numbernumber ofof waysways toto orderorder eacheach handhand

Thus,Thus, CC((5252,,55)) == PP((5252,,55)/)/PP((55,,55))


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LetLet CC((nn,,rr)) bebe thethe numbernumber ofof waysways toto generategenerateunorderedunordered combinationscombinations

TheThe numbernumber ofof orderedordered combinationscombinations (i(i..ee.. rr--

permutations)permutations) isis PP((nn,,rr))TheThe numbernumber ofof waysways toto orderorder aa singlesingle oneone ofofthosethose rr--permutationspermutations PP((r,rr,r))

TheThe totaltotal numbernumber ofof unorderedunordered combinationscombinations isis

thethe totaltotal numbernumber ofof orderedordered combinationscombinations (i(i..ee.. rr--permutations)permutations) divideddivided byby thethe numbernumber ofof waysways totoorderorder eacheach combinationcombination

Thus,Thus, CC((n,rn,r)) == PP((n,rn,r)/)/PP((r,rr,r))


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NoteNote thatthat thethe textbooktextbook explainsexplains itit slightlyslightlydifferently,differently, butbut itit isis samesame proofproof

)!(!

!

)!/(!

)!/(!

),(

),(),(

rnr

n

rrr

rnn

rrP

rnPrnC

!

!!


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Computer bugsComputer bugs


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HowHow manymany bitbit stringsstrings ofof lengthlength 1010 containcontain::

a)a) exactlyexactly fourfour 11s?s?FindFind thethe positionspositions ofof thethe fourfour 11ss

DoesDoes thethe orderorder ofof thesethese positionspositions matter?matter?Nope!Nope!PositionsPositions 22,, 33,, 55,, 77 isis thethe samesame asas positionspositions 77,, 55,, 33,, 22

Thus,Thus, thethe answeranswer isis CC((1010,,44)) == 210210

b)b) atat mostmost fourfour 11s?s?

ThereThere cancan bebe 00,, 11,, 22,, 33,, oror 44 occurrencesoccurrences ofof 11Thus,Thus, thethe answeranswer isis::

CC((1010,,00)) ++ CC((1010,,11)) ++ CC((1010,,22)) ++ CC((1010,,33)) ++ CC((1010,,44))

== 11++1010++4545++120120++210210

== 386386


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End of lecture on 30 March 2005End of lecture on 30 March 2005


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HowHow manymany bitbit stringsstrings ofof lengthlength 1010 containcontain::

c)c) atat leastleast fourfour 11s?s?ThereThere cancan bebe 44,, 55,, 66,, 77,, 88,, 99,, oror 1010 occurrencesoccurrences ofof 11

Thus,Thus, thethe answeranswer isis::CC((1010,,44)) ++ CC((1010,,55)) ++ CC((1010,,66)) ++ CC((1010,,77)) ++ CC((1010,,88)) ++ CC((1010,,99))++ CC((1010,,1010))

== 210210++252252++210210++120120++4545++1010++11

== 848848

AlternativeAlternative answeranswer:: subtractsubtract fromfrom 221010 thethe numbernumber ofof

stringsstrings withwith 00,, 11,, 22,, oror 33 occurrencesoccurrences ofof 11d)d) anan equalequal numbernumber ofof 11ss andand 00s?s?

Thus,Thus, therethere mustmust bebe fivefive 00ss andand fivefive 11ss

FindFind thethe positionspositions ofof thethe fivefive 11ss

Thus,Thus, thethe answeranswer isis CC((1010,,55)) == 252252


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Corollary 1Corollary 1

LetLet nn andand rr bebe nonnon--negativenegative integersintegers withwith

rr nn.. ThenThen CC((nn,,rr)) == CC((nn,,nn--rr))

ProofProof::

)!(!

!),(

rnr

nrnC

!

? A!)()!(

!),(

rnnrn

nrnnC

!

)!(!

!

rnr

n

!


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Corollary exampleCorollary example

ThereThere areare C(C(5252,,55)) waysways toto pickpick aa 55--cardcard pokerpoker

handhand

ThereThere areare C(C(5252,,4747)) waysways toto pickpick aa 4747--cardcard handhand

P(P(5252,,55)) == 22,,598598,,960960 == P(P(5252,,4747))

WhenWhen dealingdealing 4747 cards,cards, youyou areare pickingpicking 55 cardscards

toto notnot dealdeal AsAs opposedopposed toto pickingpicking 55 cardcard toto dealdeal

Again,Again, thethe orderorder thethe cardscards areare dealtdealt inin doesdoes mattermatter


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Combinatorial proofCombinatorial proof

AA combinatorialcombinatorial proofproof isis aa proofproof thatthat usesuses countingcounting argumentsarguments totoproveprove aa theoremtheorem RatherRather thanthan somesome otherother methodmethod suchsuch asas algebraicalgebraic techniquestechniques

Essentially,Essentially, showshow thatthat bothboth sidessides ofof thethe proofproof managemanage toto countcount thethe

samesame objectsobjects

InIn aa typicaltypical RosenRosen example,example, hehe doesdoes notnot dodo muchmuch withwith thisthis proofproofmethodmethod inin thisthis sectionsection WeWe willwill seesee moremore inin thethe nextnext sectionssections

MostMost ofof thethe questionsquestions inin thisthis sectionsection areare phrasedphrased as,as, findfind outout howhowmanymany possibilitiespossibilities therethere areare ifif Instead,Instead, wewe couldcould phrasephrase eacheach questionquestion asas aa theoremtheorem::

ProveProve therethere areare xxpossibilitiespossibilities ifif

TheThe samesame answeranswer couldcould bebe modifiedmodified toto bebe aa combinatorialcombinatorial proofproof toto thethetheoremtheorem


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HowHow manymany waysways areare therethere toto sitsit 66 peoplepeople aroundaround aa circularcircular table,table,

wherewhere seatingsseatings areare consideredconsidered toto bebe thethe samesame ifif theythey cancan bebe

obtainedobtained fromfrom eacheach otherother byby rotatingrotating thethe table?table?

First,First, placeplace thethe firstfirst personperson inin thethe northnorth--mostmost chairchair OnlyOnly oneone possibilitypossibility

ThenThen placeplace thethe otherother55 peoplepeople

ThereThere areare P(P(55,,55)) == 55!! == 120120 waysways toto dodo thatthat

ByBy thethe productproduct rule,rule, wewe getget 11**120120 ==120120

AlternativeAlternative meansmeans toto answeranswer thisthis::

ThereThere areare P(P(66,,66)=)=720720 waysways toto seatseat thethe 66 peoplepeople aroundaround thethe tabletable

ForFor eacheach seating,seating, therethere areare 66 rotationsrotations ofof thethe seatingseating

Thus,Thus, thethe finalfinal answeranswer isis 720720//66 == 120120


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HowHow manymany waysways areare therethere forfor 44 horseshorses toto finishfinish ifif tiesties areare allowed?allowed? NoteNote thatthat orderorder doesdoes matter!matter!

SolutionSolution byby casescases NoNo tiesties

TheThe numbernumber ofof permutationspermutations isis P(P(44,,44)) == 44!! == 2424

TwoTwo horseshorses tietieThereThere areare CC((44,,22)) == 66 waysways toto choosechoose thethe twotwo horseshorses thatthat tietie

ThereThere areare PP((33,,33)) == 66 waysways forfor thethe groupsgroups toto finishfinish AA groupgroup isis eithereither aa singlesingle horsehorse oror thethe twotwo tyingtying horseshorses

ByBy thethe productproduct rule,rule, therethere areare 66**66 == 3636 possibilitiespossibilities forfor thisthis casecase

TwoTwo groupsgroups ofof twotwo horseshorses tietieThereThere areare CC((44,,22)) == 66 waysways toto choosechoose thethe twotwo winningwinning horseshorses

TheThe otherother twotwo horseshorses tietie forfor secondsecond placeplace ThreeThree horseshorses tietie withwith eacheach otherother

ThereThere areare CC((44,,33)) == 44 waysways toto choosechoose thethe twotwo horseshorses thatthat tietie

ThereThere areare PP((22,,22)) == 22 waysways forfor thethe groupsgroups toto finishfinish

ByBy thethe productproduct rule,rule, therethere areare 44**22 == 88 possibilitiespossibilities forfor thisthis casecase

AllAll fourfour horseshorses tietieThereThere isis onlyonly oneone combinationcombination forfor thisthis

ByBy thethe sumsum rule,rule, thethe totaltotal isis 2424++3636++66++88++11 == 7575


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A last note on combinationsA last note on combinations

AnAn alternativealternative (and(and moremore common)common) wayway toto

denotedenote anan rr--combinationcombination::

IllIll useuse C(C(nn,,rr)) wheneverwhenever possible,possible, asas itit isiseasiereasier toto writewrite inin PowerPointPowerPoint

!

r

nrnC ),(


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Quick surveyQuick survey

I felt I understood the material in this slide setI felt I understood the material in this slide set

a)a) Very wellVery well

b)b) With some review, Ill be goodWith some review, Ill be goodc)c) Not reallyNot really

d)d) Not at allNot at all


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The pace of the lecture for this slide set wasThe pace of the lecture for this slide set was

a)a) FastFast

b)b) About rightAbout rightc)c) A little slowA little slow

d)d) Too slowToo slow


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How interesting was the material in this slideHow interesting was the material in this slide

set? Be honest!set? Be honest!

a)a) Wow!T

hat was SOOOOOO cool!Wow!T

hat was SOOOOOO cool!b)b) Somewhat interestingSomewhat interesting

c)c) Rather bortingRather borting

d)d) ZzzzzzzzzzzZzzzzzzzzzz


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21 Permutations and Combinations