Probability usig permutatios ad combiatios Multiplicatio priciple If A ca be doe i ways, ad B ca be doe i m ways, the A followed by B ca be doe i m ways. 1. A die ad a coi are throw together. How may results are possible? There are 6 outcomes for the die {1, 2, 3, 4, 5, 6} ad 2 outcomes for the coi {H, T} So there are 6 2 12 possible outcomes for the die ad the coi together. {1H, 2H, 3H, 4H, 5H, 6H, 1T, 2T, 3T, 4T, 5T, 6T} 2. Three 6-sided dice are throw together. Each dice has six possible outcomes, so the umber of possible outcomes for the three dice is: 6 6 6 216 Q. How may umber plates are possible if each umber plate is made up of 3 letters followed by a umber less tha 10 000 (excludig zero). A. The umber of plates is: 26 26 26 9 999 175 742 424 [for each letter there are 26 to choose from ad there are 9 999 umbers less tha 10 000 to choose from] Exercise A: The multiplicatio priciple 1. How may outfits (of shoes, skirt ad top) does a female Year 13 studet have for school if she has the choice of 2 pairs of shoes, 4 skirts ad 5 tops i her wardrobe? 2. A male studet wats to wear a 1960s outfit (trousers, shirt ad shoes) to the school ball. He goes to the local costume hire shop ad they have the followig items he thiks would be suitable: trousers (lime gree, fluorescet pik, vivid orage, purple) shirts (lurex, paisley, electric blue, black with rhiestoes, metallic gold) shoes (platform heels, wiklepickers, poited toes). 3. There are eight doors i a hall. I how may differet ways ca they be left either ope or closed? 4. I the UK, umber plates have 2 letters followed by 2 digits the 3 letters, e.g. SW43QPA. The first digit caot be 0. How may possible umber plates are there? 5. Number plates i Hog Kog are 2 letters followed by up to 4 digits, where the first digit caot be a zero. I Chia the umber plates are two letters followed by five digits or letters (first digit caot be zero). How may more possible umber plates are there i Chia tha i Hog Kog? How may possible choices of outfit does the studet have?
Factorials If is a whole umber, the factorial (writte ) is the product of all whole umbers from 1 to. For example, 5! 5 4 3 2 1 120 4! 4 3 2 1 24 [which is 5! 5 ] 3! 3 2 1 6 [which is 4! 4 ] 2! 2 1 2 [which is 3! 3 ] 1! 1 [which is 2! 2 ] Cotiuig this patter, 0! 1! 1 1 Note: There is a factorial key o scietific calculators. Usig factorials to cout arragemets The three objects A, B, C ca be arraged i 6 ways: ABC, ACB, BAC, BCA, CAB, CBA The umber of arragemets of 3 objects ca be worked out usig the followig reasoig. The first place ca be filled i 3 ways, the secod place i 2 ways, ad the fial place i 1 way. So there are 3! 3 2 1 6 possible arragemets [by the multiplicatio priciple] This result ca be geeralised: objects ca be arraged i a lie i ways where ( 1) ( 2) 3 2 1 Q. 8 people are waitig to go ito a room, oe at a time. I how may differet orders ca they go ito the room? A. 8! 40 320 ways Whe arragig objects i a circle, oe object is effectively held fixed ad the remaiig ( 1) objects arraged aroud it. objects ca be arraged i a circle i ( 1)! ways where ( 1)! ( 1) ( 2) 3 2 1 Q. I how may ways ca 8 guests be seated aroud a circular dier table? A. (8 1)! 7! 5 040 ways There may be various restrictios ivolved i a situatio, such as idetical objects i the group beig arraged (which reduces the umber of arragemets). Q. How may ways ca the letters i the word HARSH be arraged? A. If the H s are labelled H 1 ad H 2 (ad treated as beig differet) the there would be 5! 120 arragemets. Each pair of arragemets such as ASH 1 RH 2 ad ASH 2 RH 1 (i which the positios of H 1 ad H 2 are reversed) would the be couted as two differet arragemets. However, sice the H s are i fact the same (H 1 H 2 ) this pair of arragemets is really oly the sigle arragemet ASHRH. So 5! eeds to be divided by 2 (which is 2!, the umber of ways the two H s ca be arraged). So the umber of differet arragemets is 5! 2 60 Probabilities with factorials If p is the probability that a arragemet with a give restrictio occurs, the p Number of arragemets with restrictio Number of arragemets without restrictio Q. If 4 girls ad 1 boy are arraged radomly i a lie, fid the probability that 1. the boy is at oe ed of the lie. 2. the boy is i the middle A. 1. P(boy is at oe ed) Number of arragemets with boy at oe ed Number of arragemets without restrictio
There are 2 ways of placig the boy at oe ed of the lie (at the frot or at the back). For each placemet of the boy, there are 4! ways of arragig the girls. So the umber of arragemets with boy at oe ed 2 4! 48 [multiplicatio priciple] Number of arragemets without restrictio is 5! 120 P(boy is at oe ed) 48 120 2 5 2. P(boy i middle) 4 3 1 2 1 120 0.2 Exercise B: Factorials ad probability 1. Simplify eed to be i a circle. I how may differet ways could the circle of studets be arraged? a. c. 8! 7! 2 640! 2 639! b. d. 36! 34! ( + 1)! 5. There are 4 people i a family ad oe bathroom. I how may differet orders could they use the bathroom i the morig if: a. there is o restrictio? e. ( 5)! ( 6)! f. 5( + 3)! ( + 2)! b. oe particular family member always gets up ad uses the bathroom first? g. ( + 1)! h. ( 2)! 2. 11 people are liig up for the school catee. I how may ways could they lie up? 6. A ew school logo is beig developed ad the studets wat to have 7 differet symbols arraged i a circle. I how may ways could the symbols be arraged? 7. a. How may differet words ca be made from the word SURFIE? 3. A ew upholstery fabric is beig created that is made of 10 differetly coloured stripes. I how may differet ways ca the stripes be arraged? b. Fid the probability that a word begis ad eds with a vowel. 4. A primary school teacher has take a group of 10 studets outside to play a game i which they
8. a. How may differet words ca be made from the word SKATEBOARD? b. Fid the probability that a word has the two B s together ad the two I s together. b. Fid the probability that a word begis ad eds with a A. 10. The letters of the word TEAMS are arraged to make words. Fid the probability that the word : a. has alteratig cosoats ad vowels 9. a. How may differet words ca be made from the word PROBABILITY? b. has the two vowels together. Permutatios A permutatio is a arragemet i which order matters. For example, if two letters are selected from the word THE ad put i order, the there are 6 permutatios possible: TH, HT, TE, ET, HE, EH Note: The umber of arragemets ca be worked out usig the multiplicatio priciple: there are 3 ways of choosig the first letter ad 2 ways of choosig the secod letter, so altogether there are 3 2 6 ways of arragig two letters from three. The umber of permutatios (ordered arragemets) of 2 objects from 3 available objects is writte 3 P 2 So 3 P 2 6 The umber of permutatios of r objects selected from a group of differet objects is writte P r Q How may ways ca 4 letters of the word iphoe be arraged so that o letter is used more tha oce i each arragemet? A. There are 6 letters from which to choose 4, so the umber of permutatios is 6 P 4 There are 6 ways of choosig the first letter, 5 ways of choosig the secod letter (the first letter caot be used agai), 4 ways of choosig the third letter (the first two letters caot be used agai) ad 3 ways of choosig the fourth letter (the first three letters caot be used agai), so 6 P 4 6 5 4 3 360 Note: 6 P 4 6 5 4 3 which is 6 5 4 3 2 1 2 1 6! 2! 6! (6 4)! The formula for the umber of permutatios of r objects from objects is: P r ( r)! Scietific calculators have a P r P r key which ca be used for calculatig permutatios.
Q. There are 220 Year 13 studets eligible to be the chairperso, treasurer ad secretary of the ball committee. How may possible ways ca these offices be filled? A. 220 P 3 10 503 240 possible ways [from 220 studets, select ad arrage 3 studets] 220! Note: By the formula this is (220 3)! which is too large to be worked out directly o a calculator. Simplifyig gives: 220! 220 219 218 217! (220 3)! 217! 220 219 218 10 503 240 Probability with permutatios Probability problems may deal with restrictios o the permutatios of objects. 5-letter words are beig formed from the word COMPUTER. Fid the probability that a word begis with C ad eds with M. Solutio: P(word begis with C ad eds with M) Number of words begiig with C ad edig with M Number of 4-letter words without restrictio The word is of the form C _ M The umber of ways of placig the C ad the M is 1 [C must be first ad M last] This leaves 6 letters from which to choose ad arrage the remaiig 3 letters. This ca be doe i 6 P 3 120 ways. So the umber of words begiig with C ad edig with M is 1 120 120 The umber of 5-letter words (without restrictio) is 8 P 5 6 720 P(word begis with C ad eds with M) is 120 6 720 1 56 Exercise C: Permutatios ad probability 1. Studets i a secodary school are asked to fill out a questioaire rakig the five thigs they are most satisfied with about their school. The choices they have are: teachers, workig eviromet, facilities, computer access, fellow studets, uiform, disciplie, buildigs, sports facilities, activities ad catee. The rakig is to be show by the umbers 1, 2, 3, 4, 5 where 1 is the thig they are most satisfied with ad 2 the ext most satisfied. How may differet ways ca a studet complete the questioaire? 3. A school has 224 Year 13 studets. The pricipal of the school wats to select a group of three Year 13 studets to work together o a committee to fudraise for World Visio. Oe studet will be the leader of the group, aother will take charge of the moey ad the third will be the publicist. I how may ways could the pricipal select this committee: a. without restrictio? b. if the leader of the group must be Bethay, the Head Girl? 2. A combiatio lock has 10 differet digits (0, 1, 2,, 9) ad i order to ope the lock a combiatio of three differet digits must be selected i the correct order. How may differet combiatios could be selected for this lock? 4. a. How may differet 4-letter words ca be made from the word TRIUMPH?
b. What is the probability that a 4-letter word made from the word TRIUMPH starts with a P? 5. Fid the probability that a 5-letter word made from the word PROBLEMS has first letter P ad last letter S. Combiatios A combiatio is a selectio i which order does t matter. For example, if two letters are to be selected from the four letters A, B, C, D, without order of selectio matterig ( A the B is the same selectio as B the A ), the there are six combiatios possible: A,B A,C A,D B,C B,D C,D The umber of combiatios (uordered selectios) of 2 objects from 4 available objects is writte 4 C 2 So 4 C 2 6 Q. A school selects 3 girls from a group of 8 girls to play teis agaist aother school. How may combiatios of team members are possible? A. The umber of permutatios would be 8 P 3 336 But sice the order does t matter there will be fewer combiatios If the three girls chose are called Aroha (A), Beatrice (B) ad Carissa (C) the there would be 3! 6 permutatios possible for this selectio (ABC, ACB, BAC, BCA, CAB, CBA). All six permutatios correspod to a sigle combiatio {A, B, C}. So there are 6 times as may permutatios as there would be combiatios, so the umber of permutatios eeds to be divided by 6: Number of possible teis teams is: 8 P 3 3! 336 6 56 The formula for the umber of combiatios of r objects from objects is: C r P r r! C r ca also be writte as r ( r)!r! Scietific calculators have a C r C r key which ca be used for calculatig combiatios. Read the the problem carefully to determie if permutatios or combiatios are required. Q. From a club of 12 members, how may ways are there of selectig: 1. a presidet, vice-presidet ad a treasurer? 2. a committee of 3 people? A. 1. The order matters so there are 12 P 3 1 320 ways 2. Order does t matter so there are 12 C 3 220 ways Probabilities with combiatios Remember to multiply for AND ad add for OR. Q. 1. I how may ways ca a committee of 5 be chose from 3 wome ad 7 me if the committee must cotai at least 2 wome? 2. If a committee of 5 is chose radomly from 3 wome ad 7 me, what is the probability the committee has: a. at least 2 wome? b. o wome?
A. 1. The committee must have either 2 wome (ad 3 me) OR 3 wome (ad 2 me) The umber of ways of formig a committee with 2 wome AND 3 me is: 3 C 2 7 C 3 105 [multiplicatio priciple] The umber of ways of formig a committee with 3 wome AND 2 me is: 3 C 3 7 C 2 21 [multiplicatio priciple] So the total umber of committees possible with at least 2 wome is 105 + 21 126 2. a. P(committee has at least 2 wome) Number of committees with at least 2 wome Number of committees without restrictio There are 7 + 3 10 people available from which to select a committee of size 5. The umber of committees that ca be formed without restrictio is 10 C 5 252 P(committee has at least 2 wome) is 126 252 1 2 b. P(o wome) 3 C 0 7 C 5 10 C 5 1 12 Exercise D: Combiatios ad probability 1. A caterig studet wats to make a fruit salad with 8 fruits so he goes to the supermarket ad fids there are 23 differet fruits to choose from. How may differet possible varieties of fruit salad could he make? b. i. If the class decides it wats to sed 2 girls ad 2 boys, how may differet ways ca the four represetatives be chose? ii. If the class decides to sed at least 1 girl, how may differet ways ca the four represetatives be chose? 2. At the local ice-cream parlour there are 8 differet flavours to choose from. If Maree wats to buy a sudae with oe scoop of each of three differet flavours of ice-cream, how may possible sudaes could she choose? 3. A school has 140 Year 13 studets. The pricipal of the school wats to select a group of 6 studets to work together o a committee to fudraise for World Visio. I how may ways could the Pricipal select this committee? c. Calculate the the probability that a radomly selected group of four studets from 13Pmm is made up of: i. 2 girls ad 2 boys ii. at least 1 girl. 4. There are 15 girls ad 13 boys i 13Pmm. The school wat to sed four studets from the class to represet the school at the Azac Day service. a. How may differet ways are there to select the four represetatives? 5. Oe of the teachers at school likes to give studets Fruitbursts sweets if they have bee workig very hard. Misbah always works hard so the teacher decided to give her two Fruitbursts. The teacher
held oe of each colour i his had (purple, red, gree, yellow ad orage) ad asked her to choose two. a. How may differet combiatios of colour could Misbah choose? b. What is the probability her selectio does ot iclude a purple Fruitburst? a. I how may ways could the coach select this group to iterview? The coach makes a selectio ad the group is made up completely of boys. b. Do you cosider it likely that this group of six was radomly selected? Justify your aswer with suitable probability calculatios. 6. A class of 20 studets icludes a pair of twis. a. A group of 4 studets is selected from the class. What is the probability that the group comprises: i. both twis? ii. either twi? iii. exactly oe of the twis? b. Let the radom variable X the umber of twis i a selectio of 4 studets from this class of 20 studets. Draw up a probability distributio table for X. 8. A group of 5 Year 13 mathematics studets is studyig simple radom samplig. Each studet chooses a sample of 4 jelly sakes from a populatio of 12 differetly coloured jelly sakes. The studets otice that all 5 samples are differet ad they woder what the chaces are of two studets choosig exactly the same colours i their sample of 4 sakes. They also woder how the situatio would chage if the group of studets was a differet size, e.g. a whole class of 30 studets performig this samplig process? Discuss this situatio. Iclude some appropriate probability calculatios with your aswer. 7. A coach has a squad of 15 football players, of whom 9 are boys ad 6 are girls. The coach decides to iterview a radom group of 6 of these players, to fid out their opiio o traiig schedules.