COLLEGE ALGEBRA LECTURES Copyrights and Author: Kevin Pinegar

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1 OLLEGE ALGEBRA LETURES opyrights ad Author: Kevi iegar 8.3 Advaced outig Techiques: ermutatios Ad ombiatios Factorial Notatio Before we ca discuss permutatio ad combiatio formulas we must itroduce factorial otatio. Whe you see a expressio like 5!, it is ot a excited umber. It is read 5 factorial ad it simply meas you multiply from 5 dow to 1. Example: 5*4*3*2*1=120 Factorial: Assume is ay whole umber greater tha or equal to 1.! = (-1)(-2)(-3)..(2)(1) 4!=4(3)(2)(1)=24 3!=3(2)(1)= 2!=2(1)=2 1!=1 0 Factorial: 0! is a bit disturbig to some studets but 0!=1 because mathematicias defied it that way. (It actually makes the math behid factorials work.) So just remember that, 0!=1 alculatig Factorials You ca calculate some factorials by had ad others usig a basic calculator usig multiplicatio. More advaced calculators have a factorial fuctio that does all the work for you. More o that later. We kow how to calculate 5!, 4! ad so o by had. You might ot wat to calculate 15! by had. Just eter ito your calculator 15*14*13*12*11*10*9*8*7**5*4*3*2*1 (you actually do t have to eter the 1 sice it will ot chage the aswer). With other factorial expressios there are some clever techiques we ca use. See below. 10! 9! We ca write 10! as 10*(9*8*7**5*4*3*2*1) or 10(9!) 10! 10(9!) 9! 9! Warig 3!4! 12! 10(9!) 9! 10 3!4! (3*2*1)(4*3*2*1) (24) ! 7! We ca take this cocept farther. 10! 10(9)(8)(7!) 7! 7! Warig 4! 2! 2! 4! 4*3*2*1 2! 2*1 10(9)(8) Usig A alculator To Fid Factorials Most scietific calculators will have a! key. You may have to locate it i the meu of your calculator. If you have o luck fidig it, tough! Just kiddig. Go do a google search or youtube search of somethig like calculatig factorials o the TI-84 calculator. Obviously replace TI-84 with your particular calculator model. You will fid plety of help. Oce you figure it out, test it o the earlier examples I gave you to see if you are doig it correctly {5!=120}. 1

2 Simple Factorial roblems A factorial is actually a multiplicatio priciple problem where you use all of the objects. I the previous sectio we asked how may 3 letter code words you could create from the first 9 letters of the alphabet without repetitio. What if I chage it to how may 9 letter code words you could create without repetitio? 9 choices for 1 st letter, 8 choices for 2 d letter, 7 choices for 3 rd letter ad so o. Aswer: 9(8)(7)()(5)(4)(3)(2)(1)=9! Here s aother example. Suppose 8 differet amio acids must lie up perfectly i a certai order for a bodily fuctio to work properly. How may ways could these amio acids lie up? Sice they are differet amio acids, we caot repeat. Aswer: 8(7)()(5)(4)(3)(2)(1)=8! Itroductio Of ermutatios To itroduce the cocept of permutatios we will use a example. Suppose we take the word math ad wat to kow how may 2-letter code words (without repetitio) we could create. We could do it the hard way by listig each possibility. (This is called the exhaustio method.) {(ma),(am),(mt),(tm),(mh),(hm),(at),(ta),(ah),(ha),(th),(ht)} I couted 12 code words. Each of these code words are a 2-letter permutatio of the 4 letters. A permutatio is simply a arragemet. I could have asked how may 3-letter code words or eve how may 4-letter code words. This leads to a defiitio. ermutatio A permutatio of a set of distict objects is a arragemet of the objects i a specific order without repetitio. osider the example above. math (ma) is a arragemet of two of the four objects. Repetitio was ot allowed ad (ma) is a specific order, because (am) would be a differet code word. Actually, each of the 12 code words is a differet permutatio. List all of the 3 object permutatios of the set: A={7,3,,4} (7,3,), (7,,3), (3,,7),(3,7,), (,3,7), (,7,3) (7,,4), (7,4,), (,4,7), (,7,4), (4,,7), (4,7,) (7,3,4), (7,4,3), (3,4,7), (3,7,4), (4,3,7), (4,7,3) (3,,4), (3,4,), (,4,3), (,3,4), (4,,3), (4,3,) There are 24 permutatios, but we kew this already from the multiplicatio priciple. There are 4 optios for 1 st umber, 3 optios for 2 d umber ad 2 optios for 3 rd umber. 4(3)(2)=24 Note: If repetitio was allowed, I would have ra off the page: 4*4*4=4 possibilities, but they are t cosidered permutatios if repetitio is allowed. Suppose a police officer must lie up 5 people for a witess to idetify the prepetrator. How may ways ca the police officer lie them up? Solutio: Ask yourself some questios. 1) a the people be repeated? a I put oe suspect i two positios? No! So repetitio is ot allowed. 2) Is the order relevat? Of course it is. What if they always put the real suspect o the ed? That might get out. The lieup must be i some radom order. This is a permutatio questio but we ca solve permutatios questios with multiplicatio. 5 choices for 1 st spot,4 choices for 2 d spot,3 choices for 3 rd spot,2 choices for 2 d spot,1 choice for 5 th spot. 5(4)(3)(2)(1)=120 possible lieups. 2

3 Note: The key words are without repetitio ad specific order. First of all, if repetitio is allowed, you caot use the ermutatio (or the ombiatio formulas). You have to revert to the multiplicatio priciple. Also, if you cout the specific order of the sets (without repetitio), it is a ermutatio problem. Below gives the defiitio of ermutatio. A permutatio is simply a multiplicatio problem without repetitio where order is importat. For example if you had 12 players o a baseball team ad wated to kow how may battig orders you could submit, this would require a permutatio formula. For oe, you would ot list the same player battig 1 st ad 5 th. (o repetitio) Also, if you leadoff with Joh Doe battig first but ext time used the same 9 batters puttig Joh Doe as 9 th batter, that would be a totally differet battig order. Special ase: How may ways ca you arrage ALL of the objects. Suppose you have objects ad wish to arrage all of them i a specific order (like we did with the 8 amio acids). This is actually a special case of the permutatio formula. It is writte as follows, ; It ca also be writte as or (. r)!! I prefer this form:!, Example: How may ways ca you tur i a battig order for a baseball team if you oly have 9 players? Solutio: A battig order has 9 batters ad you have 9 players. So the umber of objects is 9 ad the umber we are arragig is 9. (Repetitio is ot allowed uless you pla o slippig old Joe i as your 1 st ad 5 th hitter. That s cheatig.) Obviously order is relevat. So we have the followig: 9 9 9! 32,280 Example: There are horses i a race. How may orders ca the horses fiish? Solutio: horses, fiishers. Repetitio makes o sese ad order is relevat.! 720 Example: The simplest protei molecule i biology is called vasopressi ad is composed of 8 amio acids that are chemically boud together i a particular order. The order i which these amio acids occur is of vital importace to the proper fuctioig of vasopressi. If these 8 amio acids were placed i a hat ad draw out radomly oe by oe, how may differet arragemets of these 8 amio acids are possible? 8 8 8! 40, 320 Geeral ermutatio Formula: May times the umber of selected objects (r) is less tha the total umber of objects (). I that case we use the stadard permutatio formula for how may ways you ca arrage r objects out of objects. r! Other Ways You Might See It:, (, r) ( r)! r Example: There are horses i a race. How may orders ca the top 4 horses fiish? Solutio: horses, 4 fiishers. Repetitio makes o sese ad order is relevat.! ( 4)!! 2!

4 ermutatio ad ombiatio Formulas O A alculator Most scietific or graphig calculators have the ermutatio ad ombiatio fuctio i the meu. You ca search your calculator or search the iteret (google, youtube) o where it is located o your calculator. O the TI-83 ad TI-84, you hit the [MATH] key, the arrow left or right util you see [RB]. The scroll dow to [r] or [r]. Warig: You eed to eter the first umber () before pullig up the fuctio, the eter r followed by [ENTER] key. It will look like this 10r3=720 a t help you o others. Search! alculatig ermutatios By Had 7! (7 2)! 7()(5! ) 5! 7() Fid 72: ) How may ways ca the first 3 places be awarded i a race ivolvig 5 cotestats (excludig ties)? 5! 5! 5(4)(3)(2! ) 5 3 5(4)(3) 0 (5 3)! 2! 2! 2) How may ways ca the positios of presidet ad vice presidet be assiged from a group of 8 people? 8! (8 2)! 8!! 8(7)(! )! 8(7) 8 2 More Examples: 1) Number of ways to award the first three places i a cotest of 10 people ) Number of ways presidet, ad vice presidet ca be assiged from 8 people. 5 3) Number ways 4 people ca be arrage i 4 chairs Special ases of ermutatios(you do t eed to memorize these) Special ase I: ermutatio of Objects Take at a Time:! How may ways ca 4 people be seated i 4 chairs? 4 4=4!=24 Special ase II: ermutatio of objects Take (-1) at a Time:! How may ways ca 3 people be seated i 4 chairs? 4 3=4!=24 Special ase III: ermutatio of objects Take 1 at a time: 1 How may ways ca we award 1 st place i a race of 10 cotestats? 10 1=10 Special ase IV: ermutatio of objects Take 0 at a time: 0 1 How may ways ca we award o places i a race of 10 cotestats? 10 0=1 Next we will discuss what happes if the order does ot matter. Retur to the baseball team with 12 members. Suppose the coach is allowed to omiate 5 players for all coferece awards. How may groups could he select? First, lets radomly select 5 players. {Joe, Sam, Bob, Ray, Do} Whe the coach writes the ames dow, does the order matter? Would t {Sam, Joe, Do, Ray, Bob} be the same as the first group? Of course it would. Ad ay other arragemet of those 5 guys. Although the coach has other choices that would iclude the other 7 players. Basically, we just eed to kow how may subsets of 5 players ca be submitted. These are called combiatios ad we will discuss those ext. 1 4

5 ombiatios Itroductio Of ombiatios To itroduce the cocept of combiatios we will use a example. Suppose we have four sos {m,a,t,h} ear the same age ad we wi two tickets to the World Series. I am woderig how may ways I ca sed two of the four to the game. We wat to kow how may 2-so subsets are possible (without repetitio). We could do it with the exhaustio method. {(ma),(tm),(hm),(ta),(ha),(ht)} Note: I ca t give oe so both tickets o repetitio. I couted. Each of these is a 2-perso combiatio of the 4 letters. A combiatio is simply a subset. The order of the selectio is irrelevat. If I sed m ad a, what do they care about who I give the first ticket? If we had more tickets I could have asked how may 3-so subsets I could sed. This leads to a defiitio. ombiatio A combiatio of a set of distict objects (take r at a time without repetitio) is a r-elemet subset of the set of objects. The arragemet or order of the subset does ot matter. osider the example above. math (ma) is a subset of two of the four objects ad (am) is the same subset. Repetitio was ot allowed ad order does ot matter. Actually, each of the is a differet combiatio. List all of the 3 object combiatios of the set: A={7,3,,4} (7,3,), (7,,4), (7,3,4),(3,,4) There are oly 4 combiatios (subsets). Suppose a lawyer must select 4 jurors from a set of six cadidates? Solutio: Ask yourself some questios. 1) a the jurors be repeated? a I select oe juror to sit i two chairs? No! So repetitio is ot allowed. 2) Is the order of the selectio relevat? No. The jurors are just beig selected. They are ot beig assiged to partcular positio or order. This is a combiatio questio. We eed the combiatio formula to determie the umber of combiatios. Below is the formula for combiatios: r! May also be writte:, r (, r) ( r)! r! r For the juror selectio example, = ad r=4. The aswer to the questio is! ( 4)!4!! ()(5)(4!) 2!4! 2!4! ()(5)(4!) 2! O the TI-83 ad TI-84, you hit the [MATH] key, the arrow left or right util you see [RB]. The scroll dow to [r]. Warig: You eed to eter the first umber () before pullig up the fuctio, the eter r followed by [ENTER] key. It will look like this: r4=15 Example of Questios that lead to combiatio formulas: 1) Number of ways to select 3 people from a group of 10 people ) Number of groups of 5 products that ca be selected from a productio lie of ) Number of hugs possible i a family of 5 people (o repeats)

6 alculatig A ombiatio By Had 1) How may ways ca 3 ruers be selected for the Olympics from a field of 5 cotestats? 5! 5! 5(4)(3! ) 5(4) !(5 3)! 3!2! 3!2! 2! 2 2) How may ways ca two me be selected from a group of 8 me to be o a committee? 8! 8! 8(7)(! ) 8(7) !(8 2)! 2!! 2!! 2! 2 Special ases of ombiatios (you do t eed to memorize these) Special ase I: ombiatio of Objects Take at a Time: How may 4-people committees ca be formed from a group of 4 people? 44=1 Special ase II: ombiatio of objects Take (-1) at a Time: 1 How may 3-people committees ca be formed from a group of 4 people? 43=4 Special ase III: ombiatio of objects Take 1 at a time: How may ways ca 1 ma be selected from a group of 8 me to be o a committee? 81=8 Special ase IV: ombiatio of objects Take 0 at a time: How may ways ca o me be selected from a group of 8 me to be o a committee? 80= Note about Special ases: Although the special cases might be iterestig, it is ot ecessary to memorize them. Ay permutatio or combiatio ca be calculated by either usig the formulas or a calculator. See a couple of examples below. You ca show the others i like fashio if you wish. From earlier we kow! From above we kow 1 By the permutatio formula: By the combiatio formula:!!!! ( )! 0! 1 1! ( 1)! ( 1)!1! ( 1)!1 1 Studet Questio: How do I kow whe to use a permutatio or a combiatio? Aswer: First, oly use permutatios or combiatio if repetitio is ot allowed. To determie whether to use or, try the followig: Take a subset from the set. Rearrage the subset. If you get somethig ew, that tells you that the order matters ad therefore is a permutatio. If you do t get somethig ew, order does ot matter ad therefore you have a combiatio problem. See examples to follow. Example: You have 9 families you would like to ivite to a weddig. Ufortuately, you ca oly ivite families. How may differet sets of ivitatios could you write? Solutio: Obviously you will ot sed two ivitatios to the same family. So repetitio is ot allowed. To determie if this is a permutatio or combiatio, select 7 families. Let s call the families {a,b,c,d,e,f,g,h,i} ad select of these families. {a,b,d,e,g,h} Now rearrage the families i way you like. {a,c,b,g,e,d} Does rearragig the families give you somethig ew? NO! It is still the same set of ivitatios. This is a combiatio problem with =9 ad r=. 9! 9(8)(7)(!) 9(8)(7) 504 Aswer: 9 84 sets (9 )!! 3!! 3!

7 ermutatio ad ombiatios Examples Side By Side It helps to see similar problems side by side. Suppose a seior class at a small high school cosists of 40 studets. a) How may ways could we assig the positios of presidet, vice presidet, secretary ad treasurer? Solutio: Repetitio is ot allowed because we will ot assig the same perso to two positios. Select 4 studets: {a,b,c,d}. Rearrage them: {c,b,a,d}. Selectio {a,b,c,d} might mea that a=presidet, b=v.p., c=sec., d=treas. Selectio {c,b,a,d} might mea that c=presidet, b=v.p., a=sec., d=treas. Order matters (is relevat) ermutatio 2,193,30 (from calculator) 40 4 b) How may ways could we select a group of 4 to serve o the seior trip committee? Solutio: Repetitio is ot allowed because we will ot assig the same perso to two positios. Select 4 studets: {a,b,c,d}. Rearrage them: {c,b,a,d}. Selectio {a,b,c,d} ad selectio {c,b,a,d} represet the same committee. Order is ot relevat) ombiatio 91,390 (from calculator) 40 4 outig Examples: It would ot be fair to oly give permutatio ad combiatios here. I will mix i some additio ad multiplicatio priciple problems to keep you hoest. roblem Solutio 1. Suppose we have to select 5 maagers from Let the maagers be called (a,b,c,d,e,f,g,h,i,j) a list of 10. How may ways ca this be doe? We caot select the same maager twice, therefore there is o repetitio. (a,b,c,d,e) ad (a,c,b,e,d) represet the same set of maagers. (order is irrelevat) ombiatio roblem: 105= Suppose we have to select a maager, assistat maager, ad ight maager from a list of 10 people. How may ways ca this be doe? 3. How may ways ca a 3-card had be selected from a stadard 52-card deck? 4. Three cards are selected radomly ad give to 3 players. How may possibilities exist? (a,b,c) implies a=maager, b= asst. maager, c=ight maager (c,a,b) implies c=maager, a= asst. maager, b= ight maager (order is relevat) ermutatio roblem: 103=720 No repetitio because the same perso caot get 3-clubs twice. (3-clubs, 9-diamods, 8-spades) is the same set as (3-clubs, 8-spades, 9-diamods) (order is irrelevat) ombiatio roblem: 523=22100 (3-clubs, 9-diamods, 8-spades) implies player A gets 3-clubs, player B gets 9-diamods ad player gets 8-spades; while (3-clubs, 8- spades, 9-diamods) implies player A gets 3- clubs, player B gets 8-spades ad player gets 9-diamods. (order is relevat) ermutatio roblem: 523=

8 5. A card is select from a stadard deck of cards, the put back ad the deck is shuffled. This is doe 3 times. How may 3-card hads ca you receive?. At a Fiat dealership a total of 3 cars of a particular model must be trasported to aother dealership. If there are 25 cars of this type, how may choices are available for trasport? 7. At a Fiat dealership a total of 3 cars of a particular model must be trasported to aother dealership. If there are 25 cars of this type, how may ways ca they be loaded oto the truck for trasport? 8. At a Fiat dealership there are 25 cars of a certai model. Fiftee have automatic trasmissio. Twelve have leather seats. Te cars have both automatic trasmissio ad leather seats. a) How may have either automatic trasmissio or leather seats. b) How may have either automatic trasmissio or leather seats. (You ca set up a Ve diagram to make this easier.) 9. Social Security umbers cosist of 9 digits (0-9). If there are o restrictios, how may differet social security umbers are possible? 10. Suppose licese plates i oe state have 4 letters followed by 2 digits. a) How may licese plates ca be created if there are o other restrictios? b) What if oly letters caot be repeated? c) What if oly digits caot be repeated? Note: Obviously the actual umber of possibilities would have to be adjusted based o commo sese. Thik about it. Would you wat a licese plate like : TURD-01 or BUTT-45 I kept it G, but you get my poit. Repetitio is possible sice the card is replaced before the ext card is selected. You could get the Ace of hearts all 3 times. Therefore we caot use a permutatio or combiatio formula. We must resort to the multiplicatio priciple. There are 52 selectios per trial 52(52)(52)=52 3 =140,08 Repetitio does ot make sese ad the order i which they are selected is irrelevat. (c1,c15,c20) ad (c15,c20,c1) represet the same set of 3 cars ombiatio roblem: 253=2300 Repetitio does ot make sese ad the order i which they are selected is relevat. (c1,c15,c20) ad (c15,c20,c1) represet the differet ways to load the cars. ermutatio roblem: 253=13,800 This is a additio priciple problem. You should idetify this by the way it is stated. Let A be those with automatic trasmissio. Let B be those with leather seats. By the additio priciple, those that have either automatic trasmissio or leather seats is represeted by AB. a) (AB)=(A)+(B)-(AB)= =17 To fid how may have either, we subtract 17 from 25. b) (A B )=25-17=8 Repetitio is possible so we must use multiplicatio priciple. 10*10*10*10*10*10*10*10*10=10 9 =1,000,000,000 Do you thik a billio is eough? By the way, this icludes all possibilities. Eve { }. Solutio a) We actually have two operatios. Operatio 1 is selectig the letters. Operatio 2 is selectig the digits. Sice repetitio is allowed, O1=2(2)(2)(2)=2 4 =45,97 O2=10(10)=10 2 =100 Fial Aswer: 4597(100)=45,97,00 Solutio b) O1=2(25)(24)(23)=358,800 Fial Aswer: (100)=35,880,000 Solutio c) O1=45,97; O2=10(9)=90 Fial Aswer: 4597(90)= 41,127,840 8

9 11. A shipmet of 20 disk drives were received by a computer store. Four of the drives are defective. A sample of 2 are selected radomly. a) How may differet samples ca be selected? b) How may of the samples cotai 2 defective drives? c) Suppose oe of the samples is tested ad oe sample is sold. How may ways ca this be doe? 12. Suppose a 5-card had is selected from a stadard deck of cards. How may ways ca the followig be doe? a) Select 3 Kigs ad 2 Aces b) Select exactly 3 fours. c) At least 4 hearts 13. Suppose there are 3 sales forces i your compay (A,B,). A has 8 members, B has 7 members ad has 5 members. You must assig a salesperso ad a alterate from each group to atted a coferece. How may ways ca this doe? 14. Suppose we have a office of 5 wome ad me ad eed to select a 4 perso committee. How may ways ca we select a. 2 me ad 2 wome? b. 3 me ad 1 woma? c. All wome? 15. A lottery cosists of 54 umbers. To purchase a ticket, you select umbers from 54 without repetitio. How may selectios are possible? (I lotteries, the order is geerally ot relevat.) Solutio a) (d5,d19) ad (d19,d5) are same set ad repetitio makes o sese. ombiatio roblem: 202=190 Solutio b) You must select 2 out of the 4 defective drives. ombiatio roblem: 42= Solutio c) (d5,d19) implies drive 5 is tested ad drive 19 is sold; whereas (d19,d5) implies the reverse. ermutatio roblem: 202=380 Solutio a) O1 is selectig 3 Kigs from 4. O2 is selectig 2 Aces from 4. N1=43=4, N2=42= Aswer: (43)(42)=4()=24 Solutio b) O1 is selectig 3 fours from 4. O2 is selectig 2 other cards from 48 (ofours) N1=43=4, N2=482=1128 Aswer: (43)(482)=4(1128)=4512 Solutio c) I this example we must fid the umber of hads that have exactly 4 hearts ad add the aswer to the umber of hads that have exactly 5 hearts. (At least 4 hearts meas all hads that have 4 or 5 hearts.) 4 hearts: 134(391)=715(39)=27,885 {4 hearts from 13 ad 1 oheart from 39} 5 hearts: 135(390)=1287(1)=1287 {5 hearts from 13 ad 0 oheart from 39} Aswer: =29,172 Solutio: Order is relevat sice we assig two positios to each selectio. Group A: 82, Group B: 72, Group : 52 Aswer: =5*42*20=47040 Solutio a) Solutio b) 0 Solutio c) ,827,15 Just for fu, what if you had to get the umbers i the order selected? 54 18,595,558,800 Iterestig Website: 9

10 ractice roblems 1. A total of 37,451 hd s were eared i There were 114 hd s i busiess ad 292 of those were wome. How may me eared hd s i somethig other tha busiess? 2. Out of 30 applicats, 11 are female, 17 are college graduates, 7 are biligual, 3 are female graduates, 2 are biligual wome, are biligual graduates ad 2 are biligual female graduates. Fid the umber of female graduates that are ot biligual. 3. There are 30 members of a club o campus. How may ways ca a presidet, vice presidet ad secretary be elected. 4. How may 3 letter code words ca be selected if there are o restrictios? How may 3 letter code words ca be selected if repetitio is ot allowed? 5. A shirt compay has 4 desigs, 5 patters, colors ad 3 styles. How may differet shirts ca be created?. How may 5-card hads ca be dealt from a stadard 52-card deck? How may of these hads cotai 2 hearts ad 3 clubs? 7. Te horses are etered i a race. How may ways ca the horses cross the fiish lie (excludig ties)? 8. Seve cois are tossed. How may differet ways ca they lad? 9. A test has 3 T/F questios ad 7 multiple choice questios that have four choices each. How may differet aswer sheets ca be submitted? 10. There are 7 wome ad 5 me i a class. The istructor must select 5 to be o a committee. How may ways ca the istructor select, a group of 3 wome ad 2 me? a group of 2 wome ad 3 me? a group of all wome? a group of all me? Aswer: 22,797 Aswer: 1 Aswer: 24,30 Aswer: 17,57 Aswer: 15,00 Aswer: 30 Aswer: 2,598,90 Aswer: 22,308 Aswer: 3,28,800 Aswer: 128 Aswer: 131,072 Aswer: 350 Aswer: 210 Aswer: 21 Aswer: 1 I the ext sectio we will itroduce probability. 10