CBSE NCERT Solutions for Class 11 mathematics Chapter 7
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CBSE NCERT Solutions for Class 11 mathematics Chapter 7 Miscellaneous exercise on chapter 7 Q.1. How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER? Solution: Given word is word DAUGHTER. There are 3 vowels namely. A, U, and E, and 5 consonants namely, D, G, H, T, and R. We know that, r objects can be selected from n objects in Crn ways. So, number of ways of selecting 2 vowels out of 3 vowels =3C2=3!2!3-2!=3 ∵nCr=n!r!n-r! And, number of ways of selecting 3 consonants out of 5 consonants =5C3=5!3!5-3!=5×4×3!3!2!=10 Thus, number of combinations of 2 vowels and 3 consonants =3×10=30. Here, number of letters we got =5. We know that, n objects can be arranged among themselves in n! ways. So, each of these 30 combinations of 2 vowels and 3 consonants can be arranged among themselves in 5! ways. Therefore, required number of different words =30×5!=3600. Q.2. From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen? Solution: Given that from the class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. Therefore, there are two cases. Case 1: All the three students join. If all 3 students join,then the remaining 7 students can be chosen from the remaining 22 students. We know that, the number of ways of selecting r objects from n objects =Crn=n!r!(n-r)! Therefore, Number of ways of selecting 7 students out of 22 students is, 22C7==22!7!(22-7)!=22!7!(15)!=170544 Case 2: none of the 3 students join: As, none of them are joining the party, 10 students can be selected from the remaining 22 students. Thus, the number of ways of selection is, 22C10=22!10!(22-10)!=22!10!12!=646646 ways. Hence, total number of ways of selecting the students for the excursion party is the sum of the ways of selecting the students in both the cases. Therefore, total number of ways in which the excursion party can be chosen =170544+646646=817190. Q.3. In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together? Solution: Given word is ASSASSINATION The letter A appears 3 times, S appears 4 times, I appears 2 times, N appears 2 times, and all the other letters appear only once. As all the words have to be arranged in such a way that all the ‘S’ are together, SSSS is treated as a single object for the time being. This single object together with the remaining 9 objects will account for 10 objects. We know that, the number of permutation of n objects, where p1 objects are of one kind, p2so on up to pr objects are of same kind is, n!p1! p2!p3!p4!......pr!. Therefore, number of arrangements of 10 objects in which there are 3‘A’s, 2 ‘I’ s and 2‘N’s. =10!3!2!2!=10×9×8×7×6×5×4×3!3!×2×1×2×1=151200 Therefore, the required number of ways of arranging the letters of the given word is 151200. Q.4. How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together? Solution: Given word is EQUATION, there are 5 vowels, namely. A, E, I, O, and U, and 3 consonants, namely, Q, T, and N. As all the vowels and consonants have to occur together, both (AEIOU) and (QTN) can be assumed as single objects. And the permutations of these 2 objects taken all at a time are counted. This number would be 2P2=2! Corresponding to each of these permutations, there are 5! permutations for the five vowels taken all at a time and 3! permutations for the 3 consonants taken all at a time. So, by multiplication principle, required number of words =2!×5!×3!=1440 Therefore, the number of words formed is 1440. Q.5. A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of: exactly 3 girls? Solution: Given that a committee of 7 has to be formed from 9 boys and 4 girls. As exactly 3 girls are to be there in the committee, the committee must consist of 7-3=4 boys. Here, The number of ways of forming the committee=Number of ways of selecting the boys×Number of ways of selecting the girls We know that, r objects can be selected from n objects in Crn ways. So, the required number of ways =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 =504 Therefore, the committee can be selected in 504 ways. Mathematics Textbook for Class 11 Chapter 7 Permutations and combinations Practice more on Permutations and combinations Page 1 www.embibe.com
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Transcript of CBSE NCERT Solutions for Class 11 mathematics Chapter 7
CBSENCERTSolutionsforClass11mathematicsChapter7
Miscellaneousexerciseonchapter7
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?
MathematicsTextbookforClass11 Chapter7Permutationsandcombinations
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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
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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
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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
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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
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