21 Permutations and Combinations
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Transcript of 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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Combination formula proofCombination formula proof
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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Combination formula proofCombination formula proof
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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Rosen, section 4.3, question 11Rosen, section 4.3, question 11
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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Rosen, section 4.3, question 11Rosen, section 4.3, question 11
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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Rosen, section 4.3, question 40Rosen, section 4.3, question 40
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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Rosen, section 4.3, question 42Rosen, section 4.3, question 42
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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Quick surveyQuick survey
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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Quick surveyQuick survey
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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