Uiversity of Waterloo Faculty of Mathematics Cetre for Educatio i Mathematics ad Computig Itermediate Math Circles November 4, 009 Coutig II Last time, after lookig at the product rule ad sum rule, we looked at coutig permutatios. I particular, at how may ways i which we could choose k object from distict objects where the order we picked them mattered. Today, we will look at combiatios: the umber of ways i which we ca choose k objects from distict objects, whe order does ot matter. Example : A math studet is give a list of 5 math problems ad is asked to do ay of the problems. How may differet choices ca the studet make? Solutio: Let us ame the problems,,, 4, ad 5. The, the possible choices are: {,, } {,, 4} {, 4, 5} {,, 5} {,, 4} {,, 5} {,, 4} {,, 5} {, 4, 5} {, 4, 5} Observe that sice it does ot matter which order the studet does the problems, the choice {,, } would be the same as the choice {,, } of {,, }, etc. We wat to figure out how to cout these mathematically. Oe way we ca thik about this is we wat to cout all the permutatios of the objects take k at a time, ad the remove all the permutatios which are the same. I the example above, what are the total umber of permutatios? 5 4 = 60. For ay particular permutatio, how may permutatios are equal to it? If we cosider the permutatio {(,, }, the ay permutatio cotaiig just these umbers are the same... how may of these are there? This is just all the permutatios o objects so! = 6 Hece, we have 60 total permutatios, but we ca orgaize these ito groups of 6, which are all equivalet if we are sayig order does ot matter. Hece, the total umber of possible choices is 60 objects divided by 6 groups = 0, which matches what we did above. Thus, the umber of ways i which we ca choose k objects from distict objects whe order does ot matter is ( ( (k k!
The formula does look very ice i this form, so we will try to make it look prettier. Observe that 5 4 = 5 4, i particular = 5!! ( ( (k =! ( k! Thus, we ca write the umber of permutatios o objects take k at a time as! ( k!. So, the umber of ways i which we ca choose k objects from distict objects is! ( k!k!. We deote this with the symbol ( k ad say choose k. (! = k ( k!k! Example : At a cafeteria, a studet is allowed to pick 4 items from the followig list: { pop, juice, milk, water, burger, hotdog, vegetable soup, baaa, orage, apple pie }. a. How may differet choices does the studet have? Solutio: ( 0 = 0! 4 6!4! = 0 9 8 7 4 = 0 7 = 0. b How may differet choices does the studet have, if they do t like apple pie? Solutio: ( 9 = 9! 4 6!4! = 9 8 7 6 4 = 7 6 = 6. c How may differet choice does the studet have, if they must pick oe, ad oly oe drik? Solutio: The studet has 4 choices to which drik they could pick. The, the studet ca pick ay of the remaiig 6 items so they have ( 6 ways of pickig those. Hece, by the product rule, the studet has 4 ( 6 = 4 6 5 4 = 4 0 = 80, choices
Exercises. Evaluate the followig: a. ( 7 b. ( 0 c. ( 6 4 ( 6. For a quest, the kight, Sir Cumferece, eeds to pick 4 out of his 0 fellow kights. a. I how may ways ca he do this? b. I how may ways ca he do this, if he must pick oe particular kight, Sir Kull. c. I how may ways ca he do this, if he ca t pick Sir Kull?. A studet takig a math test ad is told to aswer ay 7 of the 0 questios. I how may ways ca the studet do this? 4. A high school of 50 boys ad 480 girls must select 5 studets to represet them at a competitio. Aswers: a. I how may ways ca they do this? b. I how may ways ca they do this if boys are ot allowed to be selected? c. I how may ways ca they do this if the group must have boys ad girls? d. I how may ways ca they do this if the group must have more boys tha girls?. a 5, b 45, c. a ( ( 0 4 = 0, b 9 ( = 84, c 9 4 = 6. ( 0 7 4. a ( 000 5, b 5, c ( 50, d ( 50 5 + ( 50 4 + ( 580 Pascal s Triagle If oe looks at values of ( k as we chage ad k, oe quickly otices a patter. Let us do this ad fid the patter. We have ( 0 0... hmmm, this is the umber of ways we ca pick 0 objects from 0 objects... we are ot really doig much the, so we will say there is just way of doig this. So ( 0 0 =. What is ( 0? How may ways ca we pick 0 objects from object? way... we do t pick ay objects. So ( 0 =. Notice the that we have! =... so we defie 0! =. 0!! What is (? How may ways ca we pick object from object? way... we pick the object. So =. (
4 We the calculate that ( 0 =! =, (!0! =! =, (!! =! =. 0!! Lets start makig these ito a table i the form of a triagle ( 0 ( 0 ( ( 0 ( ( 0 Substitutig i the values gives Exercise: Fill i the ext rows. Try to fid the patter i the triagle ad use it to fill i more rows. Aswer: Observe that the eds of each row are always ad the middle umbers are just the sum of the two umbers above it. This gives the triagle which is ow called Pascal s triagle. 4 6 4 Pascal s triagle has may iterestig properties ad uses. Oe immediate use is that it suggests some properties of ( ( k. I particular, otice that ( k = k. But, Pascal s triagle is ot a proof of this fact... so let us prove it. Propositio: For ay o-egative itegers ad k with k we have ( ( k = k. Proof: We have ( k =! =! = ( ( ( k!( k! ( k!k! k. The triagle also suggests aother property (which is much harder to prove, amely that ( ( ( + = +. k + k k +
5 Compare this ow with Exercise above. Observe that we i fact had ( ( ( 0 9 9 = +. 4 4 Which follows from the sum rule... the umber of ways of choosig the kights is the umber of ways of choosig a group with Sir Kull plus the umber of ways of choosig a group without Sir Kull. Exercises. a. Mr E. Lipps has i frot of him a circle, a square, ad a rectagle. I how may ways ca he select some of the shapes? (he may pick ay umber of shapes b. If he has 4 shapes istead of three, i how may ways ca he select some of the shapes? What if he had shapes?. How may games are held i a roud-robi sigles teis touramet ivolvig players?. Determie ( k ( k. Aswers:. a 7 =, b 5 = 4,.. ( = (.. k+. k
6 Problem Set. If 50 differet studets try out for a team of 0 players, i how may differet ways ca the coach choose the team?. How may groups ca be formed from 8 me ad 5 wome if a. the group must have wome ad me? b. the group ca be ay size, but must have a equal umber of me ad wome?. How may subsets of size 7 of the set {,,..., 0} have a. 9 as the largest elemet? b. 9 as the middle elemet? c. the differece betwee the largest ad smallest elemets equal to 4? 4. With a stadard deck of 5 cards, a subset of 5 cards is called a had. a. How may hads are there? b. How may hads cotai oe pair? ( of a kid ad differet cards c. How may hads have 4 of a kid? 5. How may permutatios of the umbers,,..., 0 take 7 at a time a. cotai odd ad 4 eve umbers? b. cotai sigle-digit umbers ad 5 two-digit umbers? c. have the 7 umbers arraged i icreasig order? 6. Evaluate the followig without the use of a calculator. ( 7 ( ( 8 a ( 7 4 b ( 9 4 c (,. Aswers to last week s Problem Set. a 8, b 5;. 500. 6 4 0 4. 9 5. 70 6. a 5!, b 9 4!, c 6 4! 7. 4 6 5 4, b 6 5 4