AMC AMS AMR ACS ACR ASR MSR MCR MCS CRS

Transcription

1 Sectio 6.5: Combiatios Example Recall our five frieds, Ala, Cassie, Maggie, Seth ad Roger from the example at the begiig of the previous sectio. The have wo tickets for a cocert i Chicago ad everybody would like to go. However, they caot afford two more tickets ad must choose a group of three people from the five to go to the cocert. Ala decides to make a fair decisio o who gets the tickets, by writig the iitials of each possible group of idividuals from the 5 frieds o a separate piece of paper ad puttig all of the pieces of paper i a hat. He will the draw oe piece of paper from the hat at radom to select the lucky group. The order i which the ames of the idividuals is writte is irrelevat. The complete list of the 0 possible groups of size three from the five frieds is show below: AMC AMS AMR ACS ACR ASR MSR MCR MCS CRS Note that for every group o this list, it appears 6! times o the list of possible photographs of three of the five frieds that we saw i the last lecture: AMC AMS AMR ACS ACR ACM ASM ARM ASC ARC CAM MAS MAR CAS CAR CMA MSA MRA CSA CRA M AC SAM RAM SAC RCA MCA SMA RMA SCA RAC ASR MSR MCR MCS CRS ARS MRS MRC MSC CSR SAR SMR RMC CMS RCS SRA SRM RCM CSM RSC RSA M RS CRM SM C SCR RAS M SR CM R SCM SRC Its easy to see that the reaso for this is because every group of three idividuals ca be permuted i! 6 ways to create! differet photographs. Whe selectig a group from the hat to receive the tickets, the order i which the ames of the idividuals appear o the slip of paper is irrelevat. The group AMC is the same as the group CAM, so oly oe of the! permutatios of this group should be i the hat for selectio. Thus the umber of ways to select a subset of three idividuals from the five frieds to receive the three tickets is 60! P (5, )! 5!!!. Ala has listed all Combiatios of the five frieds, take at a time to put i the hat. The umber of such combiatios, which is 0, is deoted by C(5, ). I terms of set theory he has listed all subsets of objects i the set of 5 objects {A, B, C, D, E}. Defiitio A Combiatio of objects take R at a time is a selectio (Sample, Team) of R objects take from amog the. The order i which the objects are chose does ot matter. The key characteristics of a combiatio are. A combiatio selects elemets from a sigle set.. Repetitios are ot allowed.

2 . The order i which the selected elemets are arraged is ot sigificat. The umber of such combiatios is deoted by the symbol C(, R) or ( R). We have C(, R) R ( ) ( R ) R (R ) (R ) Example Evaluate C(0, ). C(0, ) 0!! 7! P (, R) R!! R!( R)! Example(Choosig a team) How may ways are there to choose a team of 7 people from a class of 40 studets i order to make a team for Bookstore Basketball? C(40, 7) 40! !! , 64, 560 Example(Roud Robis) I a soccer touramet with 5 teams, each team must play each other team exactly oce. How may matches must be played? C(5, ) 5! !! Example(Poker Hads) A poker had cosists of five cards dealt at radom from a stadard deck of 5. How may differet poker hads are possible? C(5, 5) 5!, 598, 960 5! 47! Example A stadard deck of cards cosists of hearts, diamods, spades ad clubs. How may poker hads cosist etirely of clubs? C(, 5)!, 87 5! 8! Example How may poker hads cosist of red cards oly? There are 6 red cards so C(6, 5) 6! 65, 780 5!! Example How may poker hads cosist of kigs ad quees?

3 There are 4 kigs ad 4 quees. We ca select kigs i C(4, ) ways ad we ca select quees i C(4, ) ways. We ca distiguish kigs from quees so the aswer is C(4, ) C(4, ) Bous meditatio: Why is C(8, 5) C(4, 4) C(4, )C(4, ) C(4, )C(4, ) C(4, )C(4, ) C(4, 4)? Example (Quality Cotrol) A factory produces light bulbs ad ships them i boxes of 50 to their customers. A quality cotrol ispector checks a box by takig out a sample of size 5 ad checkig if ay of those 5 bulbs are defective. If at least oe defective bulb is foud the box is ot shipped, otherwise the box is shipped. How may differet samples of size five ca be take from a box of 50 bulbs? C(50, 5), 8, 760. Example If a box of 50 light bulbs cotais 0 defective light bulbs ad 0 o-defective light bulbs, how may samples of size 5 ca be draw from the box so that all of the light bulbs i the sample are good? C(0, 5) 4, 506.

4 Problems usig a mixture of coutig priciples Example How may poker hads have at least two kigs? There are C(4, ) ways to get kigs ad C(48, ) ways to fill out the had. Hece there are C(4, ) C(48, ) hads with exactly kigs. There are C(4, ) C(48, ) hads with exactly kigs ad there are C(4, 4) C(48, ) hads with exactly 4 kigs. Hece there are C(4, ) C(48, ) C(4, ) C(48, ) C(4, 4) C(48, ) hads with at least two kigs. The umber is 6 7, 96 4, , 6. Example I the Notre Dame Jugglig club, there are 5 graduate studets ad 7 udergraduate studets. All would like to atted a jugglig performace i Chicago. However, they oly have fudig from Studet Activities for 5 people to atted. The fudig will oly apply if at least three of those attedig are udergraduates. I how may ways ca 5 people be chose to go to the performace so that the fudig will be grated? Agai it is useful to break the problem up, i this case by umber-of-udergraduates. We eed to work out how may ways we ca get udergraduates, how may ways we ca get 4 udergraduates, how may ways we ca get 5 udergraduates, ad the add these umbers. Three udergraduates: C(7, ) C(5, ); Four udergraduates: C(7, 4) C(5, ); Five udergraduates: C(7, 5) C(5, 0). The umber is cot 546. Example Gio s Pizza Parlor offers three types of crust, types of cheese, 4 vegetable toppigs ad meat toppigs. Pat always chooses oe type of crust, oe type of cheese, vegetable toppigs ad two meat toppigs. how may differet pizzas ca Pat create? Pat s choices are idepedet so C(, ) C(, ) C(4, ) C(, ) Example How may subsets of a set of size 5 have at least 4 elemets? C(5, 4) C(5, 5). Special Cases ad Formulas 4

5 It is immediate from our formula C(, R)! R!( R)! that C(, R) C(, R). Thus C(0, ) C(0, 7) ad C(00, 98) C(00, ) etc... By defiitio (for coveiece) 0! ad C(, 0), which makes sese sice there is exactly oe subset with zero elemets i every set. C(, )! ( )!, which makes sese sice there are exactly subsets with oe elemet i a set with elemets. C(, ) C(, 0) which makes sese because... (fiish the setece). The Biomial Theorem The Biomial theorem says that for ay positive iteger ad two umbers x ad y, we have (x y) x 0 x y For example if 4, the the theorem says that ( ) ( ) 4 4 (x y) 4 x 4 x y 0 ( ) ( ) x y xy x y xy y 4. 4 y Example (a) Check that the above equatio is true for 4, x, y. (b) Check that the above equatio is true for 4, x, y. Thigs to Note 5

6 4 4 (X Y) 4 ( 4 0)x 4 ( 4 )x y ( 4 )x y ( 4 )x y ( 4 4)y 4 Symmetry C(4, ) C(4, ) Powers of x ad y are switched Note how the powers relate to the lower umbers i the coefficiets. Note how the powers decrease o the x s ad icrease o the y s as we move from left to right. Note how the powers i each term of the expressio add to. Note the symmetry i the expasio. ( 4 ) x y is a Term of the expressio ad ( 4 ) is the coefficiet of that term. Note that the sum of the coefficiets of the terms of the expasio of (x y) 4 is equal to the total umber of subsets oe ca make usig a set of size 4. Settig x ad y i the above equatio, we get the formula: 4 0 Applyig the same proof to the geeral case, we get the formula: 0 4 C(, 0) C(, ) C(, ) C(, ) C(, ) for ay coutig umber. Thus the umber of subsets of a set with elemets is. Example A set has te elemets. How may of its subsets have at least two elemets? C(0, )C(0, )C(0, 4)C(0, 5)C(0, 6)C(0, 7)C(0, 8)C(0, 9)C(0, 0). To actually compute this umber it is easier to compute 0 ( C(0, 0) C(0, ) ) 04 ( 0) 0 6

7 Example How may tips could you leave at a restaurat, if you have a half-dollar, a oe dollar coi, a two dollar ote ad a five dollar ote? You ca leave ay subset of your moey. You have 4 items so there are 4 6 possibilities. Extra: Taxi Cab Geometry revisited. Recall that the umber of taxi cab routes (always travelig south or east) from A to B is the umber of differet rearragemets of the sequece SSSSEEEEE which is 9! 4!5! C(9, 4) C(9, 5). The sequece SSSSEEEEE is show i red ad the sequece ESSEEESES i blue. A Ca you thik of ay reaso why the umber of routes should equal the umber of ways to choose 4 objects from a set of 9 objects? A route is specified by a strig of S s ad E s with exactly 4 S s ad 5 E s ad ay such strig specifies a route. I other words you eed a ie letter word made up of 4 S s ad 5 E s. If I give you a subset of 4 elemets of the set {,,, 9} ad you put S s i those positios, the you ca fill i the rest of the word with E s. As a example, give {,, 7, 6} start with S S SS ad get S E S E E S S E. B 7