CBSENCERTSolutionsforClass11mathematicsChapter7

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Q.1. Howmanywords,withorwithoutmeaning,eachof2vowelsand3consonantscanbeformedfromthelettersofthewordDAUGHTER?

Solution: GivenwordiswordDAUGHTER.Thereare3vowelsnamely.A,U,andE,and5consonantsnamely,D,G,H,T,andR.

Weknowthat,robjectscanbeselectedfromnobjectsinCrnways.

So,numberofwaysofselecting2vowelsoutof3vowels=3C2=3!2!3-2!=3∵nCr=n!r!n-r!

And,numberofwaysofselecting3consonantsoutof5consonants

=5C3=5!3!5-3!=5×4×3!3!2!=10

Thus,numberofcombinationsof2vowelsand3consonants=3×10=30.

Here,numberofletterswegot=5.

Weknowthat,nobjectscanbearrangedamongthemselvesinn!ways.So,eachofthese30combinationsof2vowelsand3consonantscanbearrangedamongthemselvesin5!ways.Therefore,requirednumberofdifferentwords=30×5!=3600.

Q.2. Fromaclassof25students,10aretobechosenforanexcursionparty.Thereare3studentswhodecidethateitherallofthemwilljoinornoneofthemwilljoin.Inhowmanywayscantheexcursionpartybechosen?

Solution: Giventhatfromtheclassof25students,10aretobechosenforanexcursionparty.Thereare3studentswhodecidethateitherallofthemwilljoinornoneofthemwilljoin.

Therefore,therearetwocases.

Case1:Allthethreestudentsjoin.Ifall3studentsjoin,thentheremaining7studentscanbechosenfromtheremaining22students.Weknowthat,thenumberofwaysofselectingrobjectsfromnobjects=Crn=n!r!(n-r)!Therefore,Numberofwaysofselecting7studentsoutof22studentsis,22C7==22!7!(22-7)!=22!7!(15)!=170544

Case2:noneofthe3studentsjoin:

As,noneofthemarejoiningtheparty,10studentscanbeselectedfromtheremaining22students.

Thus,thenumberofwaysofselectionis,22C10=22!10!(22-10)!=22!10!12!=646646ways.Hence,totalnumberofwaysofselectingthestudentsfortheexcursionpartyisthesumofthewaysofselectingthestudentsinboththecases.Therefore,totalnumberofwaysinwhichtheexcursionpartycanbechosen=170544+646646=817190.

Q.3. InhowmanywayscanthelettersofthewordASSASSINATIONbearrangedsothatalltheS’saretogether?

Solution: GivenwordisASSASSINATION

TheletterAappears3times,Sappears4times,Iappears2times,Nappears2times,andalltheotherlettersappearonlyonce.Asallthewordshavetobearrangedinsuchawaythatallthe‘S’aretogether,SSSSistreatedasasingleobjectforthetimebeing.Thissingleobjecttogetherwiththeremaining9objectswillaccountfor10objects.

Weknowthat,thenumberofpermutationofnobjects,wherep1objectsareofonekind,p2soonuptoprobjectsareofsamekindis,n!p1!p2!p3!p4!......pr!.Therefore,numberofarrangementsof10objectsinwhichthereare3‘A’s,2‘I’sand2‘N’s.=10!3!2!2!=10×9×8×7×6×5×4×3!3!×2×1×2×1=151200Therefore,therequirednumberofwaysofarrangingthelettersofthegivenwordis151200.

Q.4. Howmanywords,withorwithoutmeaning,canbeformedusingallthelettersofthewordEQUATIONatatimesothatthevowelsandconsonantsoccurtogether?

Solution: GivenwordisEQUATION,thereare5vowels,namely.A,E,I,O,andU,and3consonants,namely,Q,T,andN.

Asallthevowelsandconsonantshavetooccurtogether,both(AEIOU)and(QTN)canbeassumedassingleobjects.

Andthepermutationsofthese2objectstakenallatatimearecounted.Thisnumberwouldbe2P2=2!

Correspondingtoeachofthesepermutations,thereare5!permutationsforthefivevowelstakenallatatimeand3!permutationsforthe3consonantstakenallatatime.

So,bymultiplicationprinciple,requirednumberofwords=2!×5!×3!=1440

Therefore,thenumberofwordsformedis1440.

Q.5. Acommitteeof7hastobeformedfrom9boysand4girls.Inhowmanywayscanthisbedonewhenthecommitteeconsistsof:exactly3girls?

Solution: Giventhatacommitteeof7hastobeformedfrom9boysand4girls.

Asexactly3girlsaretobethereinthecommittee,thecommitteemustconsistof7-3=4boys.

Here,Thenumberofwaysofformingthecommittee=Numberofwaysofselectingtheboys×NumberofwaysofselectingthegirlsWeknowthat,robjectscanbeselectedfromnobjectsinCrnways.

So,therequirednumberofways

=4C3×9C4=4!3!1!×9!4!5!∵nCr=n!r!n-r!

=4×3!3!×9×8×7×6×5!4!5!=4×126=504Therefore,thecommitteecanbeselectedin504ways.

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Q.6. Acommitteeof7hastobeformedfrom9boysand4girls.Inhowmanywayscanthisbedonewhenthecommitteeconsistsofatleast3girls?

Solution: Giventhatacommitteeof7hastobeformedfrom9boysand4girls.

Asatleast3girlsaretobethereineverycommittee,thecommitteecanconsistof

3girlsand4boysor4girlsand3boys.Here,Thenumberofwaysofformingthecommittee=Numberofwaysofselectingtheboys×NumberofwaysofselectingthegirlsWeknowthat,robjectscanbeselectedfromnobjectsinCrnways.So,3girlsand4boyscanbeselectedin4C3×9C4ways.And,4girlsand3boyscanbeselectedin4C4×9C3ways.

Thus,inthiscase,requirednumberofways

=4C3×9C4+4C4×9C3

=4!3!1!×9!4!5!+4!4!×9!3!6!∵Crn=n!r!(n-r)!=504+84=588Therefore,thenumberofwaysofselectingthecommitteeis588.

Q.7. Acommitteeof7hastobeformedfrom9boysand4girls.Inhowmanywayscanthisbedonewhenthecommitteeconsistsofatmost3girls?

Solution: Giventhatacommitteeof7hastobeformedfrom9boysand4girls.

Atmost3girlsaretobethereineverycommittee.

So,thecommitteecanconsistof,(A)3girlsand4boys(B)2girlsand5boys(C)1girland6boys(D)Nogirland7boys

Weknowthat,robjectscanbeselectedfromnobjectsinCrnways.

Here,Thenumberofwaysofformingthecommittee=Numberofwaysofselectingtheboys×Numberofwaysofselectingthegirls

So,(A)3girlsand4boyscanbeselectedin4C3×9C4ways.(B)2girlsand5boyscanbeselectedin4C2×9C5ways.(C)1girland6boyscanbeselectedin4C1×9C6ways.(D)Nogirland7boyscanbeselectedin4C0×9C7ways.

Thus,inthiscase,requirednumberofways

=4C3×9C4+4C2×9C5+4C1×9C6+4C0×9C7

=4!3!1!×9!4!5!+4!2!2!×9!5!4!+4!1!3!×9!6!3!+4!0!4!×9!7!2!∵nCr=n!r!n-r!=504+756+336+36=1632Therefore,thenumberofwaysofformingthecommitteeis1632.

Q.8. IfthedifferentpermutationsofalltheletterofthewordEXAMINATIONarelistedasinadictionary,howmanywordsarethereinthislistbeforethefirstwordstartingwithE?

Solution: InthegivenwordEXAMINATION,thereare11lettersoutofwhich,A,I,andNappear2timesandalltheotherlettersappearonlyonce.

Here,AistheonlyletterinthegivenwordthatcomesbeforeEaspertheenglishalphabets.

ThewordsthatwillbelistedbeforethewordsstartingwithEinadictionarywillbethewordsthatstartwithAonly.

Thus,togetthenumberofwordsstartingwithA,theletterAisfixedattheextremeleftposition,andthentheremaining 10letterstakenallatatimearerearranged.

Weknowthat,innlettersifoneletterisrepeatingxtimesandanotherletterisrepeatingytimes,thethenumberofpossiblewordsformed=n!x!·y!.

Sincethereare2‘I’and2‘N’intheremaining10letters,NumberofwordsstartingwithA=10×9×8×7×6×5×4×3×2!2!×2!=907200Therefore,therequirednumbersofwordsis907200.

Q.9. Howmany6-digitnumberscanbeformedfromthedigits0,1,3,5,7and9whicharedivisibleby10andnodigitisrepeated?

Solution: Anumberisdivisibleby10ifitsunitsdigitsis0.

Therefore,0isfixedattheunitsplace.

Therefore,Therewillbeasmanywaysastherearewaysoffilling5vacantplacesinsuccessionbytheremaining5digits(i.e.,1,3,5,7and9).Weknowthat,nobjectscanbearrangedinnplacesinn!ways.So,the5vacantplacescanbefilledin5!ways.Hence,requirednumberof6-digitnumbers=5!=120

Q.10. TheEnglishalphabethas5vowelsand21consonants.Howmanywordswithtwodifferentvowelsand2differentconsonantscanbeformedfromthealphabet?

Solution: Giventhatthereare2differentvowelsand2differentconsonantsaretobeselectedfromtheEnglishalphabet.

Weknowthat,thereare5vowelsintheEnglishalphabet.

Thenumberofwaystoselectrobjectsfromnobjects=Crn=n!r!(n-r)!.Thus,Thenumberofwaysofselecting2differentvowelsfromthealphabet=5C2=5!2!3!=5×4×3!2×1×3!=10

Similarly,asthereare21consonantsintheEnglishalphabet,numberofwaysofselecting2differentconsonantsfromthealphabet

=21C2=21!2!19!=21×20×19!2×1×19!=210

Therefore,numberofcombinationsof2differentvowelsand2differentconsonants=10×210=2100Eachofthese2100combinationshas4letters,whichcanbearrangedamongthemselvesin4!ways.Therefore,requirednumberofwords=2100×4!=50400.

Q.11. Inanexamination,aquestionpaperconsistsof12questionsdividedintotwopartsi.e.,PartIandPartII,containing5and7questions,respectively.Astudentisrequiredtoattempt8questionsinall.Selectingatleast3fromeachpart.Inhowmanywayscanastudentselectthequestions?

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Solution: Giventhatthequestionpaperconsistsof12questionsdividedintotwoparts-PartIandPartII,containing5and7questions,respectively.

Astudenthastoattempt8questions,selectingatleast3fromeachpart.Thiscanbedoneasfollows,

(A)3questionsfrompartIand5questionsfrompartII(B)4questionsfrompartIand4questionsfrompartII(C)5questionsfrompartIand3questionsfrompartII

Here,Thenumberofwaysofselectingthequestions=NumberofwaysofselectingquestionsfrompartI×NumberofwaysofselectingquestionsfrompartII

Weknowthat,robjectscanbeselectedfromnobjectsinCrnways.

So,3questionsfrompartIand5questionsfrompartIIcanbeselectedin5C3×7C5ways.4questionsfrompartIand4questionsfrompartIIcanbeselectedin5C4×7C4ways.5questionsfrompartIand3questionsfrompartIIcanbeselectedin5C5×7C3ways.

Hence,therequirednumberofwaysofselectingquestions,

=5C3×7C5+5C4×7C4+5C5×7C3

=5!3!2!×7!5!2!+5!4!1!×7!4!3!+5!5!0!×7!3!4!∵Crn=n!r!(n-r)!=210+175+35=420Therefore,thenumberofwaysinwhichastudentcanselectthequestionsis420.

Q.12. Determinethenumberof5cardcombinationsoutofadeckof52cardsifeachselectionof5cardshasexactlyoneking.

Solution: Weknowthatinadeckof52cards,thereare4kings.Acombinationof5cardshavetobemadeinwhichthereisexactlyoneking.

Weknowthat,robjectscanbeselectedfromnobjectsinCrnways.

So,1kingcanbeselectedoutof4kingsin4C1waysandTheremaining4cardscanbeselectedoutofthe48cardsin48C4ways.

Here,Thenumberofwaysofselectingthecards=Numberofwaysofselectingaking×Numberofwaysofselectingtheothercards

Hence,bymultiplicationprinciple,requirednumberof5cardcombinations,

=4C1×48C4=4!1!4-1!×48!4!48-4!∵Crn=n!r!(n-r)!=4×3!3!×48×47×46×45×44!4!44!=4×194580=778320Therefore,thenumberofwaysofselectingthecardsis778320.

Q.13. Itisrequiredtoseat5menand4womeninarowsothatthewomenoccupytheevenplaces.Howmanysucharrangementsarepossible?

Solution: Weknowthat5menand4womenaretobeseatedinarowsuchthatthewomenoccupytheevenplaces.

Herethetotalnumberofplaces=9inwhich,the5oddplaceswillbeoccupiedbythemenandthe4evenplaceswillbeoccupiedbythewomenasshownbelow.

MWMWMWMWMWeknowthatnobjectscanbearrangedinnplacesinn!ways.Hence,the5mencanbeseatedintheoddplacesin5!ways.And,thewomencanbeseatedintheoddplacesin4!ways.Weknowthat,Totalnumberofarrangements=Numberofwaystoarrangethemen×NumberofwaystoarrangethewomenTherefore,possiblenumberofarrangements=4!×5!=24×120=2880

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Exercise7.1

Q.1. Howmany3-digitnumberscanbeformedfromthedigits1,2,3,4and5assumingthatrepetitionofthedigitsisallowed?

Solution: Giventhatrepetitionisallowed,

Letthe3vacantplacesbe□□□.

Inthiscase,repetitionofdigitsisallowed.Thus,theunitsplacetensplaceandhundredsplacecanbefilledinbyanyofthegivenfivedigits. So,eachplacewecanfillin5differentways.Therefore,bythemultiplicationprinciple,Thetotalnumberof3-digitnumbersare5×5×5=125.Hence,numberof3-digitnumbersformedis125.

Q.2. Howmany3-digitnumberscanbeformedfromthedigits1,2,3,4and5assumingthatrepetitionofthedigitsisnotallowed?

Solution: Giventhat,repetitionofdigitsisnotallowed.

Ifunitsplaceisfilledinfirst,thenitcanbefilledbyanyofthegivenfivedigits.

Thus,thenumberofwaysoffillingtheunitsplaceofthethree-digitnumberis5andthetensplacecanbefilledwithanyoftheremaining4digitsandthehundredsplacecanbefilledwithanyoftheremaining3digits.Therefore,bythemultiplicationprinciple,Thetotalnumberof3-digitnumbersformedusingthegivendigitsare5×4×3=60Hence,numberof3-digitnumbersformedis60.

Q.3. Howmany3-digitevennumberscanbeformedfromthedigits1,2,3,4,5,6ifthedigitscanberepeated?

Solution: Letthe3vacantplacesforthethree-digitnumberbe□□□.

Giventhatthenumbersthatcanbeformedshouldbeeven.

Therefore,theunitsplaceshouldbefilledwithevennumbersonlyso,itcanbefilledby2or4or6onlyi.e.,theunitsplacecanbefilledin3ways,i.e.□□3Asthedigitscanberepeated,thetensplacecanbefilledbyanyofthe6digitsin6differentways,i.e.□63.Also,thehundredsplacecanbefilledbyanyofthe6digitsin6differentwaysi.e.663.Thus,bythemultiplicationprinciple,therequirednumberofthreedigitevennumbersare3×6×6=108.Hence,Numberofthree-digitnumbersthatcanbeformedis108.

Q.4. Howmany4-lettercodescanbeformedusingthefirst10lettersoftheEnglishalphabet,ifnolettercanberepeated?

Solution: Letthe4vacantplacesforthecodebe□□□□.

Numberofletterstobeusedisfirst10lettersofEnglish.

Therefore,thefirstplacecanbefilledin10differentwaysbyanyofthefirst10lettersoftheEnglishalphabetfollowingwhichthesecondplacecanbefilledinbyanyoftheremaininglettersin9differentways.

Thethirdplacecanbefilledinbyanyoftheremaining8lettersin8differentwaysandthefourthplacecanbefilledinbyanyoftheremaining7lettersin7differentways.

Thus,bymultiplicationprincipletotalnumberofcodesthatcanbeformedis,10×9×8×7=5040

Therefore,5040four-lettercodescanbeformedusingthefirst10lettersoftheEnglishalphabet,ifnoletterisrepeated.

Q.5. Howmany5-digittelephonenumberscanbeconstructedusingthedigits0to9ifeachnumberstartswith67andnodigitappearsmorethanonce?

Solution: Giventhat,Thedigitsfrom0to9istobeusedtoform5-digittelephonenumberssothateachnumberalwaysstartwith67.

Here,thenumberofdigits=10andthenumberofplaces=5.

Letthearrangementisgivenby,67.So,therewillbeasmanyphonenumbersastherearewaysoffilling3vacantplacesbythedigits0,1,2,3,4,5,8,9astheothertwodigitsarealreadyused.Thus,theonesplacecanbefilledbyanyofthedigitsfrom0to9,exceptdigits6and7.

Thus,theunitsplacecanbefilledin8differentwaysfollowingwhich,thetensplacecanbefilledinbyanyoftheremaining7digitsin7differentways,andthehundredsplacecanbefilledinbyanyoftheremaining6digitsin6differentways.

Therefore,bymultiplicationprinciple,wehave8×7×6=336

Hence,336telephonenumberscanbeconstructed.

Q.6. Acoinistossed3timesandtheoutcomesarerecorded.Howmanypossibleoutcomesarethere?

Solution: Weknowthatwegettwooutcomeswhenacoinistossed(Headandtail)

LetheadisrepresentedbyHandtailisrepresentedbyT.

Ineachtosstheoutcomecanbeeitherheadortail.Thetotalnumberofoutcomescanberepresentedbytheset:HHH,HHT,HTH,THH,TTH,THT,HTT,TTTHerewecanseethatnumberofpossibleoutcomesis8.

Alternativelywecanalsocalculatebyusingthemultiplicationprincipleofcounting.

Herethecoinistossedthriceandineachtosstotalnumberofpossibleoutcomesis2

Therefore,therequirednumberofpossibleoutcomesis2×2×2=8.Hence,thereare8possibleoutcomes.

Q.7. Given5flagsofdifferentcolours,howmanydifferentsignalscanbegeneratedifeachsignalrequirestheuseof2flags,onebelowtheother?

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Solution: Giventhat,

Thenumberofflags=5.

Numberofflagsrequiredforeachsignal=2.Tofind:Thenumberofsignalsthatcanbegenerated.Here,tofindthenumberofsignalsthatcanbegenerated,wecanfindthenumberofwaysinwhichfiveflagscanbearrangedintwoplaces.Lettheplacesare:.Forthefirstplaceanyofthe5flagscanbeused,sowehave5waysofusingaflagforthefirstplace. Forthesecondplace,anyofthe4remainingflagscanbeused,i.e.54.Therefore,bymultiplicationprinciplewehavetotalnumberofarrangementsis5×4=20.Hence,20differentsignalscanbegenerated.

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Exercise7.2

Q.1. Evaluate8!

Solution: Thefactorialofanumberistheproductofalltheintegersfrom1tothatnumber.

Factorialisnotdefinedfornegativenumbersandthefactorialofzeroisone.

So,n!=nn-1n-2.........1Weneedtoevaluate8!⇒8×7×6×5×4×3×2×1=40320Therefore,valueof8!is40320.

Q.2. Evaluate4!-3!.

Solution: Thefactorialofanumberistheproductofalltheintegersfrom1tothatnumber.

Factorialisnotdefinedfornegativenumbersandthefactorialofzeroisone.

So,n!=nn-1n-2.........1Weneedtoevaluate4!-3!⇒4!=4×3×2×1=24,3!=3×2×1=6Therefore,4!-3!=24-6=18

Q.3. Is3!+4!=7!?

Solution: Thefactorialofanumberistheproductofalltheintegersfrom1tothatnumber.

Factorialisnotdefinedfornegativenumbersandthefactorialofzeroisone.

So,n!=nn-1n-2.........1.Therefore,3!=3×2×1=6and4!=4×3×2×1=24.Hence,3!+4!=6+24=30Whereas,7!=7×6×5×4×3×2×1=5040.Therefore,3!+4!≠7!

Q.4. Compute8!6!×2!

Solution: Thefactorialofanumberistheproductofalltheintegersfrom1tothatnumber.

Factorialisnotdefinedfornegativenumbersandthefactorialofzeroisone.

n!=nn-1n-2.........1Therefore,8!6!×2!=8×7×6!6!×2×1=8×72=28

Q.5. If16!+17!=x8!,findx.

Solution: Given,16!+17!=x8!

⇒16!×77+17!=x8!∵LCM of6!and7!is7!as7!=7·6!

⇒77!+17!=x8!⇒7+17!=x8!⇒87!×8!=x⇒87!×8×7!=x⇒x=64Therefore,thevalueofxis64.

Q.6. Evaluaten!n-r!,whenn=6,r=2

Solution: Tofind:n!(n-r)!,forn=6,r=2

Therefore,

n!n-r!=6!6-2!=6!4!Byusingtheformula,n!=n(n-1)(n-2)!weget,6!4!=6×5×4!4!=30Therefore,thevalueofthegivenexpressionis30.

Q.7. Evaluaten!n-r!,whenn=9,r=5.

Solution: Tofind:n!(n-r)!,forn=9,r=5

Therefore,

n!n-r!=9!9-5!=9!4!Byusingtheformula,n!=n(n-1)(n-2)(n-3)(n-4)(n-5)!weget,9!4!=9×8×7×6×5×4!4!=15120Therefore,thevalueofthegivenexpressionis15120.

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Exercise7.3

Q.1. Howmany3-digitnumberscanbeformedbyusingthedigits1to9ifnodigitisrepeated?

Solution: Weneedtoform3-digitnumberswithoutrepetitionusingthedigits1to9.

Weknowthattherewillbeasmany3-digitnumbersastherearepermutationsof9differentdigitstaken3atatime.

Therefore,therequirednumberof3-digitnumbers=9P3=9!9-3!∵Prn=n!n-r!

=9!6!=9×8×7×6!6!

=9×8×7=504Therefore,wecanform504numbersbyusingthedigits1to9.

Q.2. InhowmanyofthedistinctpermutationsofthelettersinMISSISSIPPIdothefourI’snotcometogether?

Solution: GivenwordMISSISSIPPI,Iappear4times,Sappears4times,Pappears2times,andMappearsjustonce.Thus,numberofdistinctpermutationsofthelettersinthegivenword⇒11!4!4!2!=11×10×9×8×7×6×5×4!4!×4×3×2×1×2×1=11×10×9×8×7×6×54×3×2×1×2×1=34650

Weknowthathereare4‘I’sinthegivenword.Whentheyoccurtogether,theyaretreatedasasingleobjectforthetimebeing.Thissingleobject‘IIII’togetherwiththeremaining7objectswillaccountfor8objects.Therefore,these8objectsinwhichthereare4‘S’and2‘P’canbearrangedin8!4!2!=8×7×6×5×4!4!×2×1=840ways.Numberofarrangementswhereall‘I’occurtogether=840Therefore,numberofdistinctpermutationsofthelettersinMISSISSIPPIinwhichfour‘I’donotcometogether=34650-840=33810

Q.3. InhowmanywayscanthelettersofthewordPERMUTATIONSbearrangedifthewordsstartwithPandendwithS,

Solution: GivenwordisPERMUTATIONS,thereare2‘T’andalltheotherlettersappearonlyonce.IfPandSarefixedattheextremeends(PattheleftendandSattherightend),then10lettersareleft.

Therefore,inthiscase,requirednumberofarrangements10!2!=10×9×8×7×6×5×4×3×2!2!=1814400

Q.4. InhowmanywayscanthelettersofthewordPERMUTATIONSbearrangedifthevowelsarealltogether,

Solution: GivenwordisPERMUTATIONS,thereare2‘T’andalltheotherlettersappearonlyonce.Inthegivenwordthereare5vowels,eachappearingonlyonce.Astheyhavetoalwaysoccurtogether,theyaretreatedasasingleobjectforthetimebeing.Thissingleobjecttogetherwiththeremaining7objectswillaccountfor8objects.These8objectsinwhichthereare2‘T’canbearrangedin8!2!.

Therefore,correspondingtoeachofthesearrangements,the5differentvowelscanbearrangedin5!ways.Thus,therequirednumberofarrangements⇒8!2!×5!=8×7×6×5×4×3×2!2!×120=2419200

Q.5. InhowmanywayscanthelettersofthewordPERMUTATIONSbearrangedifthetherearealways4lettersbetweenPandS?

Solution: GivenwordisPERMUTATIONS,thereare2‘T’andalltheotherlettersappearonlyonce.Weshouldarrangethelettersinsuchawaythattherearealways4lettersbetweenPandS.Thus,inaway,theplacesofPandSarefixedandtheremaining10lettersinwhichthereare2‘T’canbearrangedin10!2!ways.AndthelettersPandScanbeplacedsuchthatthereare4lettersbetweenthemin2×7=14ways.

Thus,bymultiplicationprinciple,requirednumberofarrangementsinthiscase⇒10!2!×14=10×9×8×7×6×5×4×3×2!2!×14=25401600Therefore,wecanarrangein25401600ways.

Q.6. Howmany4-digitnumbersaretherewithnodigitrepeated?

Solution: Weneedtofindthetotalnumberoffourdigitsnumbers.

Here,weneedtomakethe4-digitnumbersusingthedigits0,1,2,3,4,5,6,7,8,9.

Ifthenumberstartingwithzerolike0457,0248,....whichareactually3-digitnumbers.Therefore,Requirednumberoffourdigitnumbers=TotalNumberoffourdigitnumbers-NumberoffourdigitnumbersstartingwithzeroNumberofdigitfrom0to9=10So,thetotalnumberof4-digitnumbersareP410.Here,P410=10!(10-4)!=10!6!=10×9×8×7=5040

Thethousandsplacearefilledwithzeroandthedigitscannotberepeatedinthe 4-digitnumbers.

Thehundreds,tensandunitplacesaretobefilledbytheremaining9digits.

Weknowthattherewillbeasmanysuchdifferentwaysastherearepermutationsof9differentdigitstaken3atatime.Therefore,thenumberofwaystofillremaining3places=9P3=9!9-3!=9!6!=9×8×7×6!6!=9×8×7=504Requirednumber=5040-504=4536.Therefore,numberof4-digitnumbersthatcanbeformedis4536.

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Q.7. Howmany3-digitevennumberscanbemadeusingthedigits1,2,3,4,6,7ifnodigitisrepeated?

Solution: Weneedtofindthetotalnumberofeven3-digitnumbersformedusingthedigits1,2,3,4,6,7withoutrepetition.

So,unitsdigitscanbefilledin3waysbyanyofthedigits,2,4or6.

Asthedigitscannotberepeatedinthe3-digitnumbersandunitsplaceisalreadyoccupiedwithadigit(whichiseven),thehundredsandtensplaceistobefilledbytheremaining5digits.

Weknowthat,numberofwaysofarrangingnobjectsinrplacesisPrn=n!(n-r)!.

Where,n>randn,r>0.

Therefore,numberofevennumbersif2isattheunitsplace,=5P2=5!5-2!=5!3!=5×4×3!3!=5×4=20.Similarly,Thenumberofevennumbersif4isataunitsplace=20Thenumberofevennumbersif6isataunitsplace=20Therefore,therequirednumbersof3-digitnumbersis20+20+20=60.

Q.8. Findthenumberof4-digitnumbersthatcanbeformedusingthedigits1,2,3,4,5ifnodigitisrepeated.Howmanyofthesewillbeeven?

Solution: Givendigits:1,2,3,4,and5.

Weneedtoform4-digitnumbersusingthedigits.

Therefore,numberofplacestobefilledisfourandnumberofdigitsisfive.Weknowthat,numberofwaysofarrangingndigitsinrplacesisgivenby,Prn=n!(n-r)!Where,n>randn,r>0Thus,therequirednumberof4-digitnumbers=5P4=5!5-4!=5!1!=5×4×3×2×11=120

Amongthe4-digitnumbersformedbyusingthedigits,1,2,3,4,5,evennumbersendwitheither2or4.

Therefore,thenumberofwaysinwhichtheunitplacecanbefilled=P12=2!(2-1)!=2.

As,theunitplaceisfilledbyonenumber,restoftheplacescanbefilledbyusingfour-digitsasfollows. Numberofwaysoffillingtheremainingplaces,=4P3=4!4-3!=4!1!=4×3×2×11=24Therefore,Numberofevennumbers=Numberofwaysoffillingtheunitplace×Numberofwaysoffillingtheremainingplaces=24×2=48So,outofallthenumbersformed,thereare48evennumbers.

Q.9. Fromacommitteeof8persons,inhowmanywayscanwechooseachairmanandavicechairmanassumingonepersoncannotholdmorethanoneposition?

Solution: Thereisacommitteeof8personsfromwhichachairmanandavicechairmanaretobechoseninsuchawaythatonepersoncannotholdmorethanoneposition.Now,thenumberofwaysofchoosingachairmanandavicechairmanisthepermutationof8differentobjectstaken2atatime.

Thus,therequirednumberofways,=8P2=8!8-2!=8!6!=8×7×6!6!=8×7=56

Therefore,thenumberofwaysinwhichachairmanandavicechairmancanbeselectedis56.

Q.10. Findn,ifn-1P3:nP4=1:9.

Solution: Given,n-1P3:nP4=1:9⇒n-1P3nP4=19⇒(n-1)!n-1-3!n!n-4!=19∵nPr=n!n-r!⇒n-1!n!=19⇒n-1!n×n-1!=19⇒1n=19⇒n=9Therefore,thevalueofnis9.

Q.11. Findrif5Pr=26Pr-1

Solution: Given,5Pr=26Pr-1⇒5!5-r!=2×6!6-r+1!∵nPr=n!n-r!⇒7-r!5-r!=2×6!5!

⇒7-r6-r5-r!5-r!=2×6×5!5!⇒7-r6-r=12⇒42-13r+r2=12

⇒r2-13r+30=0⇒r2-10r-3r+30=0⇒rr-10-3r-10=0⇒r-10r-3=0⇒r=3orr=10Weknowthat,0≤r≤n,heren=5therefore,r≠10.Therefore,r=3.

Q.12. Findrif5Pr=6Pr-1

MathematicsTextbookforClass11 Chapter7Permutationsandcombinations

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Solution: Given,5Pr=6Pr-1⇒5!5-r!=6!6-r+1!∵nPr=n!n-r!⇒5!5-r!=6×5!7-r!⇒15-r!=67-r6-r5-r!⇒1=67-r6-r

⇒42-13r+r2=6⇒r2-13r+36=0⇒r2-9r-4r+36=0⇒rr-9-4r-9=0⇒r-4r-9=0⇒r=4orr=9Weknowthat,0≤r≤n,heren=5or6therefore,r≠9.Therefore,r=4.

Q.13. Howmanywords,withorwithoutmeaning,canbeformedusingallthelettersofthewordEQUATION,usingeachletterexactlyonce?

Solution: NumberoflettersinthewordEQUATIONis8.

Thus,thenumberofwordsthatcanbeformedusingallthelettersofthewordEQUATION,usingeachletterexactlyonce,isthenumberofpermutationsof8differentobjectstaken8atatime.

Weknowthat,nletterscanbearrangedinnplacesinPnnwaysandPrn=n!(n-r)!.Therefore,Pnn=n!(n-n)!=n!∵0!=1Therefore,requirednumberofwordsthatcanbeformedis,8P8=8!=8×7×6×5×4×3×2×1=40320Hence,40320wordscanbeformed.

Q.14. Howmanywords,withorwithoutmeaningcanbemadefromthelettersofthewordMONDAY,assumingthatnoletterisrepeated,if4lettersareusedatatime,

Solution: Numberof4-letterwordsthatcanbeformedfromthelettersofthewordMONDAY,withoutrepetitionofletters,isthenumberofpermutationsof6differentobjectstaken4atatime.

Weknowthat,nletterscanbearrangedinrplacesinPrnwaysandPrn=n!(n-r)!.

Therefore,therequirednumberofwordsthatcanbeformedusing4lettersatatimeisgivenby6P4=6!6-4!=6!2!=6×5×4×3×2!2!=6×5×4×3=360Therefore,thenumberofwordsformedis360.

Q.15. Howmanywords,withorwithoutmeaningcanbemadefromthelettersofthewordMONDAY,assumingthatnoletterisrepeated,ifalllettersareusedatatime.

Solution: NumberofwordsthatcanbeformedbyusingallthelettersofthewordMONDAYatatimeisthenumberofpermutationsof6differentobjectstaken6atatime.

Weknowthat,nletterscanbearrangedinnplacesinPnnwaysandPrn=n!(n-r)!.

Therefore,Pnn=n!(n-n)!=n!∵0!=1Therefore,therequirednumberofwordsthatcanbeformedwhenalllettersareusedatatimeis,P66=6!=6×5×4×3×2×1=720.Hence,720wordscanbeformed.

Q.16. Howmanywords,withorwithoutmeaningcanbemadefromthelettersofthewordMONDAY,assumingthatnoletterisrepeated,ifalllettersareusedbutfirstletterisavowel?

Solution: Givenword:MONDAY.

Numberofdistinctvowelsinthegivenword=2

Giventhat:Thefirstletterisavowel.Therefore,theleftmostplaceistobefilledwithoneofthevowels.Thiscanbedonein2ways.Astheletterscannotberepeatedandtheleft-mostplaceisalreadyoccupiedwithaletter(whichisavowel),theremainingfiveplacesaretobefilledbytheremaining5letters.Thiscanbedonein5P5=5!ways.Here,Numberofwordsformed=Numberofwaysoffillingthefirstplacebyavowel×NumberofwaysoffillingtheremainingplacesTherefore,inthiscase,requirednumberofwordsthatcanbeformedis5!×2=120×2=240

MathematicsTextbookforClass11 Chapter7Permutationsandcombinations

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Exercise7.4

Q.1. IfnC8=nC2,findnC2.

Solution: Given,nC8=nC2.ItisknownthatifnCa=nCb,thena=bora+b=n.Therefore,nC8=nC2⇒n=8+2⇒n=10Thus,nC2=10C2

Now,C2n=10!2!10-2!=10!2!8!∵Crn=n!r!(n-r)!=10×9×8!2×1×8!=5×9=45

Q.2. Determinenif2nC3:nC3=12:1

Solution: Given,2nC3:nC3=12:1⇒2nC3nC3=121⇒(2n)!3!2n-3!n!3!n-3!=121nCr=n!r!n-r!

⇒2n×2n-1×2n-2×2n-3!2n-3!n×n-1×n-2×n-3!n-3!=121⇒2×2n-1×2n-2n-1×n-2=121⇒2×2n-1×2n-1n-1×n-2=121⇒42n-1n-2=121⇒8n-4=12n-24⇒4n=20⇒n=5

Q.3. Determinenif2nC3:nC3=11:1

Solution: Given,2nC3:nC3=11:1⇒2nC3nC3=111⇒2n!3!2n-3!n!3!n-3!=111∵nCr=n!r!n-r!⇒2n×2n-1×2n-2×2n-3!2n-3!n×n-1×n-2×n-3!n-3!=111

⇒2×2n-1×2n-2n-1×n-2=111⇒2×2n-1×2n-1n-1×n-2=111⇒42n-1n-2=111⇒8n-4=11n-22⇒3n=18Therefore,n=6

Q.4. Howmanychordscanbedrawnthrough21pointsonacircle?

Solution: Weknowthatfordrawingonechordonacircle,only2pointsarerequired.Weneedtofindthenumberofchordsthatcanbedrawnthroughthegiven21pointsonacircle,thenumberofcombinationshavetobecounted.Thus,therewillbeasmanychordsastherearecombinationsof21pointstaken2atatime.

Therefore,therequirednumberofchords21C2=21!2!21-2!∵nCr=n!r!n-r!=21!2!19!=21×20×19!2×1×19!=21×10=210Thus,210chordscanbedrawnthrough21pointsonacircle.

Q.5. Inhowmanywayscanateamof3boysand3girlsbeselectedfrom5boysand4girls?

Solution: Weshouldselectateamof3boysand3girlsfrom5boysand4girls.So,3boyscanbeselectedfrom5boysin5C3ways.Also,3girlscanbeselectedfrom4girlsin4C3ways.

Thus,bymultiplicationprinciple,numberofwaysinwhichateamof3boysand3girlscanbeselected=5C3×4C3=5!3!5-3!×4!3!4-3!∵nCr=n!r!n-r!=5×4×3!3!2!×4×3!3!1!=10×4=40

Q.6. Findthenumberofwaysofselecting9ballsfrom6redballs,5whiteballsand5blueballsifeachselectionconsistsof3ballsofeachcolour.

Solution: Giventhatthereareatotalof6redballs,5whiteballs,and5blueballs.Weneedtoselect9ballsinsuchawaythateachselectionconsistsof3ballsofeachcolour.So,3ballscanbeselectedfrom6redballsin6C3ways.3ballscanbeselectedfrom5whiteballsin5C3ways.3ballscanbeselectedfrom5blueballsin5C3ways.

Therefore,bymultiplicationprinciple,requirednumberofwaysofselecting9balls⇒6C3×5C3×5C3=6!3!6-3!×5!3!5-3!×5!3!5-3!∵nCr=n!r!n-r!=6×5×4×3!3!3!×5×4×3!3!2!×5×4×3!3!2!=5×4×5×2×5×2=20×10×10=2000

Q.7. Determinethenumberof5cardcombinationsoutofadeckof52cardsifthereisexactlyoneaceineachcombination

MathematicsTextbookforClass11 Chapter7Permutationsandcombinations

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Solution: Weknowthatinadeckof52cards,thereare4aces.Acombinationof5cardshavetobemadeinwhichthereisexactlyoneace.So,1acecanbeselectedoutof4acesin4C1waysandTheleft4cardscanbeselectedoutofthe48cardsin48C4ways.

Therefore,bymultiplicationprinciple,requirednumberof5cardcombinations=4C1×48C4=4!1!4-1!×48!4!48-4!∵nCr=n!r!n-r!=4×3!3!×48×47×46×45×44!4!44!=4×194580=778320

Q.8. Inhowmanywayscanoneselectacricketteamofelevenfrom17playersinwhichonly5playerscanbowlifeachcricketteamof11mustincludeexactly4bowlers?

Solution: Giventhatoutof17players,5playersarebowlers.Acricketteamof11playersistobeselectedinsuchawaythatthereareexactly4bowlers.And,4bowlerscanbeselectedoutof5bowlersin5C4waysandTheremaining7playerscanbeselectedoutofthe12playersin12C7ways.

Therefore,bymultiplicationprinciple,requirednumberofwaysofselectingcricketteam.=5C4×12C7=5!4!5-4!×12!7!12-7!∵nCr=n!r!n-r!=5×4!4!1!×12×11×10×9×8×7!7!5!=5×792=3960

Q.9. Abagcontains5blackand6redballs.Determinethenumberofwaysinwhich2blackand3redballscanbeselected.

Solution: Giventhatthereare5blackand6redballsinthebag.So,2blackballscanbeselectedoutof5blackballsin5C2waysand3redballscanbeselectedoutof6redballsin6C3ways.

Therefore,bymultiplicationprinciple,requirednumberofwaysofselecting2blackand3red.=5C2×6C3=5!2!5-2!×6!3!6-3!∵nCr=n!r!n-r!=5×4×3!2!3!×6×5×4×3!3!3!=10×20=200

Q.10. Inhowmanywayscanastudentchooseaprogramof5coursesif9coursesareavailableand2specificcoursesarecompulsoryforeverystudent?

Solution: Giventhatthereare9coursesavailableoutofwhich,2specificcoursesarecompulsoryforeverystudentandhastochoose3coursesoutoftheremaining7courses.So,2coursescanbeselectedoutof2specifiedcoursesin2C2waysand3coursescanbeselectedoutoftheremaining7coursesin7C3ways.

Therefore,requirednumberofwaysofchoosingtheprogramme⇒2C2×7C3=2!2!2-2!×7!3!7-3!∵nCr=n!r!n-r!=2!2!0!×7×6×5×4!3!4!=1×35=35

MathematicsTextbookforClass11 Chapter7Permutationsandcombinations

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MathematicsTextbookforClass11 Chapter7Permutationsandcombinations

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