COMBINATORICS 2. Recall, in the previous lesson, we looked at Taxicabs machines, which always took the shortest path home

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1 COMBINATORICS BEGINNER CIRCLE 1/0/ ADVANCE TAXICABS Recall, i the previous lesso, we looked at Taxicabs machies, which always took the shortest path home taxipath We couted the umber of ways that a taxicab could drive a passeger back home. Problem 1. O the city grid below, label each itersectio with the umber of ways a taxicab ca drive to it. (I ve doe the first few for you!) Problem. Last time, we also looked at the umber of ways to choose a small umber of objects from a big group of objects. We said that the umber of ways to choose k objects from a set of objects was ( ) k. Compute the followig values. Jeff Hicks/Los Ageles Math Circle/UCLA Departmet of Mathematics. 1

2 LAMC hadout 1/0/13 (a) ( ) 1 ways (b) ( ) 3 ways as well (c) ( ) 5 10 ways (d) ( ) ways as well (e) I full seteces, explai how the ( is related to the taxicab problem triagle. The umber of paths to the itersectio that is itersectios away ad sittig i the k itersectio has ( paths to it. This is because the path is itersectios away, ad you must choose to go right k times whe you reach a itersectio, so the umber of paths you ca make is equal to the umber of ways you ca choose to make k rights out of itersectios. Problem 3. What is the total umber of ways to all the itersectios that are blocks away. (Hit: at every itersectio, the taxicab ca make a choice of goig right

3 LAMC hadout 1/0/13 3 or dow) There are differet paths to take i this situatio Problem. What is the total umber of paths to itersectios that are blocks away, ad southeast of the driver s curret positio? Explai your reasoig i full seteces. As before, there are differet paths. At each itersectio, the driver has the choice of goig either right or goig dow. Problem 5 (Summed Choice, First time Aroud). Usig the relatioship betwee taxicabs ad choice, explai why ( ) 0 ( ) 1 ( ) ( ) ( ) ( ) = Use full seteces. From the previous problem, we kow that the umber of paths to the itersectios that are blocks away is. However, the umber of paths to each specific itersectio is (. Therefore, the sum of the ( is. 3

4 LAMC hadout 1/0/13 Problem 6. Whe Derek goes home, he first wats to stop by Isaac s house to drop off some math circle papers. How may ways ca he do that? There are ( ( ) ways from the car to Isaac s house, ad ) ways from Isaac s house to Derek s. This gives a total of ( ( ) ( ) = ) ways to travel to Derek s house via Isaac s house. Problem 7. Every sigle path to Derek s house passes through oe of his 5 frieds houses. Usig this fact, ad the previous problem, explai why the umber of paths to Derek s house is ( ) ( ) ( ) ( ) ( ) Explai your aswer i full seteces. By the previous problem, we kow that the umber of ways to Derek s house through each of the labeled itersectios is ( ) k

5 LAMC hadout 1/0/13 5 Because each of the paths to Derek s house must pass through oe of the 5 houses, the total umber of paths from the car to Derek s house is equal to the sum of the umber of paths passig through each oe of the 5 houses. Problem 8. Jeff ad Joatha decide to meet up i the middle of the week to discuss math circle problems. Jeff starts at his house, ad Joatha starts at his house, ad they decide to stop i the middle. How may ways ca they do this? You may leave your solutio as a equatio. (Hit: Compute the umber of ways that they could meet at oe of the meetig spots, the add up across all of the meetigs spots) Suppose that Jeff ad Joatha decide to meet up at the itersectio i the k row that is a equal distace from both of them. As Jeff has ( paths to a specified meetig spot, ad Joatha has ( ( paths to the same meetig spot, they have ( paths to the meetig spot. We sum this solutio across all the possible meetig poits, to get ( ) ( ) ( ) ( )

6 LAMC hadout 1/0/13 6 Problem 9 (Bous). Usig the above two problems for ispiratio, show that ( ) ( ) ( ) ( ) ( ) ( )... = The total umber of paths to a city itersectio that is blocks away, ad sittig i the colum is ( ). O the other had, each o of these paths passes through a midpoit, ad the total umber of paths from oe side to the other that passes through a midpoit i the k colum is ( ) (. Summig across these two values gives the required result. 6

7 LAMC hadout 1/0/13 7. REARRANGEMENTS Aother combiatorial thig we may look at is rearragemets. The rearragemet umber is the total ways that oe ca take a few objects ad rearrage them. For example, there are 6 ways to rearrage 3 objects abc acb bac bca cab cba Oe way to make sure that you got all of the possible rearragemets is to list them i alphabetical order Problem 10. List all the arragemets for these four letters i alphabetical order How may are there? abcd abcd abdc acbd acdb adbc adcb bacd badc bcad bcda bdac bdca cabd cadb cbad cbda cdab cdba dabc cadb dbac dbca dcab dcba Problem 11. Here s a method for 5 letters. (a) First, pick oe of the letters. How may ways ca you do this? 5 differet letters to pick (b) The arrage the remaiig letters. How may ways ca you do this? (Use the previous problem) There are differet ways to rearrage letters (c) Now stick the letter you picked at the start of the arragemet. This gives you a 5 letter arragemet. How may ways ca you rearrage 5 letters? There are therefore 5 = 10 ways to rearrage 5 letters based o this method. 7

8 LAMC hadout 1/0/13 8 Problem 1. How may ways ca you arrage 6 letters? Explai how you got your aswer i a full setece. (You may leave your solutio as a formula) There are 6! ways. We kow that there are 5! ways to arrage 5 letters. First pick a letter (6 ways to do this) ad the arrage the other 5 letters as you wish. The attach the removed letter to the frot of whatever sequece you started with. This gives you a total of ways of rearragig 5 letters. 6 5! = 6! Problem 13. Usig the previous problems as ispiratio, how may ways ca you arrage letters? Explai your process i full seteces. There are! ways of arragig letters. First pick oe letter. There are ways to do this. The rearrage the remaiig 1 letters. There are ( 1)! ways to rearrage these letters. This gives a total of ways to rearrage letters. ( 1)! =! For shorthad, we will write ad call this value factorial.! = Problem 1 (Choices ad Rearagemets). Suppose Morga wats to select 5 people from the classroom of 5 to help him with a problem. Morga has the most peculiar method for selectig studets (a) First, Morga puts all 5 people i a lie. How may ways ca Morga do this? (You may leave your solutio By the previous problem, there are 0! ways to rearrage 0 people (b) Morga the asks himself, How may differet lies would have had the same 5 people i the frot 8

9 LAMC hadout 1/0/13 9 He cocludes that either the first 5 people would have bee i a differet order, or the last 0 people would have bee i a differet order. How may differet lies have the same first 5 people i the frot? By alterig the first 5 people, we get 5! combiatios, ad by alterig the last 0 people, we get 0! combiatios, for a total of 5! 0! combiatios. (c) Morga the otes the followig # ways to pick 5 from 5 = # ways to arrage 5 people # arragemets that have 5 chose studets at frot How may ways ca Morga choose 5 studets from 5? (You may leave this aswer as a formula) We take 5! ad the divide it by 5! 0!. The 5! 5! 0! Problem 15. Usig Morga s Method, compute these followig values (1) ( )! = 6!! () ( ) 5 5! = 10 3!! (3) ( ) 6 6! = 15!! () ( ) 6 3 6! 3! 3! = 0 9

10 LAMC hadout 1/0/13 10 Problem 16. Usig full seteces, explai why ( )! = k (! (! Usig Morga s method, we see that the umber of arragemets is exactly the desired amout. Problem 17 (0!). What is ( 5 5)? If we use the above formula to compute ( 5 5), what should 0! be? 1 Problem 18 (Summed Choice, Secod time Aroud). (a) What is the umber of ways of pickig ay umber of items from items. (Hit- At every item, decided whether or ot you wat to pick it.) There are ways of pickig ay umber of items from items. (b) What is the umber of ways of pickig 0 items, or 1 item, or items, or 3 items... or items? (Hit- Sum up the possibilities! ) There is ( ( 0) ( 1) ( )... ) ways to do so. (c) Why should these two quatities be equal. Set them equal to each other to get a familiar lookig idetity. This idetity makes sese, because the umber of ways of pickig out ay umber of items is the same as the umber of ways of pickig up oe, or two, or... items. So we set them equal ad get = ( 0 ) ( 1 ) ( )... ( ) 10

11 LAMC hadout 1/0/ SHOPPING FOR CANDY The grocery store holds types of cady, which are coveietly listed i alphabetical order, ad each cady bar costs 1. (a) Almod Joys (b) Butterfigers (c) Cady Caes (d) Dumdums Problem 19. Suppose Isaac goes to the store to buy some cady. He has oe dollar. How may ways could he purchase cady? He ca purchase cady i differet ways. Problem 0. Suppose Derek goes i with dollars. How may ways could he purchase cady? (Remember that Derek ca purchase of the same kid of cady if he wats?) He ca either purchase differet kids of cady, or of the same kid of cady. If he purchases differet kids of cady, there are ( ) ways to do this (6). If he purchases 1 kid of cady, there is ( 1) ways to do this, or. This gives a total of 10 ways. Problem 1. Morga, the big speder, eters the store with 3 dollars. How may ways could he purchase cady? He ca either choose 3 of the same cady ( 1) or of 1 cady, ad 1 of the other ( ( ( ) ) or 3 differet kids of cady 3) for a total of 0 possibilities. Problem. Joatha decides that he wats to build a cady shoppig robot. He kows that the cady i the store is lied up i rows alog the aisle, ad so programs a robot that ca do oe of two thigs: Pick up a cady Bar (P) Move alog the aisle to the ext cady bar row (M) Every program fiishes with the robot ext to the Dumdum row of cadies. The cady is arraged i alphabetical order. 11

12 LAMC hadout 1/0/13 1 Suppose Joatha programed the robot to pick up cady bars. Such a program may look like this P MMP M Which would pick up a Almod Joy, advace rows, ad the pick up a Cady Cae, the advace 1 row to the Dumdum aisle ad stop. (a) Describe what the program MP P MM picks up It would pick up Butterfigers (b) Write a program that picks up Dumdums The program picks up Dumdums MMMP P (c) If the Robot is to pick up items, how log does the program have to be? program has to have a legth of 5 The (d) How may programs ca be writte that pick up items? 5 choose programs Problem 3. Usig Joatha s Robot, figure out the umber of ways to pick out items from the grocery store There would be ( ) 7 ways, because the program would have to have legth of 7, ad picks would eed to be placed. 1

13 LAMC hadout 1/0/13 13 Problem (Choose versus Choice with repetitio). (a) Suppose Joatha visits a grocery store with differet types of items, ad wats to purchase k items, ad oly wats to buy at most 1 of each type of item. The, usig the choose otatio, how may differet kids of purchases ca he make? There are simply ( ways. (b) Suppose Johatha is willig to buy more tha oe of each type of item. The how may purchases ca he make? (Hit: Cout the umber of programs that there are for a robot to purchase k items at a grocery store with differet types of cady. How log does the program have to be?) Cosider a robot program. It must have k 1 commads, ad k of those commads must be purchases, for a total of ( ) k 1 k programs Problem 5 (Summed Choice, Third time Aroud). Suppose that I have a classroom with studets, ad I eed to select k of them. Oe way to do this selectio is to decide whether or ot to pick the tallest studet i the class, ad the pick the other studets. (a) If I select the tallest studet i the class, the I oly eed to select k 1 more studets, ad the I ca pick the remaiig k 1 studets. How may ways are there to do this (You may use choice otatio) There are ( 1 k 1) ways. (b) If I do ot select the tallest studet i the class, the I eed to select k studets. How may ways are there to do this? There are ( ) 1 k ways. (c) With the above problems i mid, fill i the blaks with the appropriate values: ( ) ( ) ( ) 1 1 = k ( ) ( ) ( ) 1 1 = k k k 1 13

14 LAMC hadout 1/0/13 1 (d) Usig the above idetity, show ( ) ( ) ( ) ( ) ( ) ( ) = Rearrage the sum ad you ll see every term shows up twice (e) How does this show us that ( ) ( )... 1 ( ) = 0 1