Worksheet 2.1, Math 455

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Ashlee Bryan
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1 Worksheet, - Math 55 Note that there are any, any ways to arrive to the sae answer for these questions If you got the sae nuber though a different thought process, it is probably right! The vowels are a, e, i, o, u (a How any eleven-letter sequences fro the alphabet contain exactly three vowels? There are ( ways of choosing where the vowels will be in the eleven-letter sequence Then for each of the 8 consonants in the sequence, there will be choices, and for each of the consonants in the sequence, there will be 5 choices So ( (b How any of these have at least one repeated letter? If all letters were different, then I would choose 8 consonants out of, and vowels out of 5, and then I would arrange these letters in whatever order I want So there are ( ( 8 5! different ( sequences with vowels and 8 consonants where no letters are repeated Thus, there are 8 5 ( ( 8 5! letter sequences with exactly three vowels with at least one repeated letter Copute the nuber of ways to deal each of the following 5-card hand in poker Note that the jack has value, queen, king, and an ace can have either value or (a Straight (the values of the cards for a sequence of consecutive integers and do not have all of the sae suit A straight can start with 0 different card values (ace, two,, 0 it cannot start with a jack or higher since then there are not enough bigger cards to have a sequence of five cards Fixing the value of the lowest card fixes the value of the other cards in the straight, so all that is left to choose is the suit for each of the five cards in the hand Each can take any of four suits However, we do not want a straight flush For a straight starting with a certain lowest card, there are four ways that it could be a straight flush: it could be all hearts, all diaonds, all clubs or all spades Thus we get = 0, 00 king the other a single For the single card, we ust also decide which one of the suits it is (For the quadruple, we ust take all four suits out of four So we get ( ( = 6 (b Three of a kind (three of the cards have the sae value and the other two cards have different values In such a hand, there will be a total of three different card values out of We ust choose which one of these three values is a triple (thus autoatically aking the other two singles For the triple, we ust choose suits out of For each of the single cards, we ust choose which one of the suits it is So we get ( ( = 5, 9 (c Full house (a pair and a three of a kind incidentally, the nae of the show that taught e English In such a hand, there will be a total of two different card values out of We ust choose which one of these two values is a triple (thus autoatically aking the other a double For the triple, we ust choose suits out of For the double, we ust choose suits out of So we get ( ( =, 7

2 (d Two distinct atching pairs (but not a full house In such a hand, there will be a total of three different card values out of We ust choose which two of these three values are doubles (thus autoatically aking the other a single For each double, ( we ust choose suits out of For the single, we ust choose suit out of So we get ( ( ( ( =, 55 (e Exactly one atching pair (but no three of a kind In such a hand, there will be a total of four different card values out of We ust choose which one of these four values is a double (thus autoatically aking the others singles For the double, we ust choose suits out of For each single, we ust choose suit out of So we get ( =, 098, 0 (f At least one card fro each suit In such a hand, there will be one suit that there will be two cards fro, and all the other suits will appear once So we ust choose which one of four suits appears twice In that suit, we ust choose two card values out of In each of the other suits, we ust choose card value out of So we ( ( ( ( get ( = 685, 6 (g At least one card fro each suit, but no two values atching In such a hand, there will be one suit that there will be two cards fro, and all the other suits will appear once So we uch choose which one of four suits appears twice In that suit, we uch choose two card values out of In the next suit, we ust choose card out of the values reaining In the next one, out of the 0 values reaining And in the final suit, we ust choose card out of the 9 values reaining So we get ( ( ( ( 0 ( 9 = 08, 880 (h Three cards of one suit, and the other two of another suit, like three hearts and two spades In such a hand, there will be two suits out of four We ust then choose which one of these two suits has three cards (thus autoatically aking the other suit have two cards In the suit that contains three cards, we ust choose card values out of, and in the suit with two cards, we ust choose card values out of We thus get ( ( ( = 67, 696 Suppose in soe lottery gae, one selects six nubers between and n What fraction of all lottery tickets have the property that half the nubers are odd and half are even? We re assuing that the order doesn t atter in this lottery Then one ust choose nubers out of n even nubers and nubers out of n odd nubers In total, one can choose any six nubers out of n So we get ( n ( n ( n 6 Assue that a positive integer cannot have 0 as its leading digit (a How any five-digit positive integers have no repeated digits at all? There are nine choices for the first digit since zero cannot be it Then we ust choose which four of the reaining nine digits (ie, different fro the first one we will use, and then choose in which of! ways to order the So we get 9 (9 =, (b How any have no consecutive repeated digits? Page

3 There are nine choices for the first digit since zero cannot be it Then for the next digit, it can be any digit but the one that we chose for the first digit, so there are nine choices Siilarly for the third, fourth and fifth digit So we get 9 5 = 59, 09 (c How any have at least one run of consecutive repeated digits (for exaple, 555, or 555, but not? There are 90,000 five-digit positive integers (fro 00,000 to 999,999, or if you prefer you have nine choices for the first digit and 0 for each subsequent digit We know that 59,09 have no repeated digits whatsoever, so there are 90, , 09 = 0, 95 5 (a You decide to go riding a century with your bike for the first tie Afraid that you will starve, you decide to bring not one, not two, but THREE Clif bars, one in each of your jersey back pockets You like variety, so you want to bring along three different flavors of Clif bars aong the 9 equally delicious flavors that exist What is the probability that you bring a brownie bar, a acadaia bar and a peanut butter bar if the likelihood of picking any flavor is equal? Explain carefully every part of the forula that you coe up with First calculate how any different assortents you could bring There are 9 choices for the first bar, 8 for the second (since you do not want to eat the sae thing again AND 7 for the third However, you are overcounting each assortent 6 ties because this counts each assortent in each possible order (brownie, acadaia, peanut butter and brownie, peanut butter, acadaia and, and there are! ways of ordering the So in total, there are = 9! 6!! = 9! (9!! = ( 9 different assortents Since any assortent is as likely as the next, the probability that you bring a brownie bar, a acadaia bar and a peanut butter bar is ( 9 (b You will first eat the Clif bar in your left pocket, then the one in your iddle pocket and finish with the one in your right pocket Being a Clif bar connoisseur, you actually feel that eating first a brownie bar, then a acadaia bar and finally a peanut butter bar has nothing to do with first eating a brownie bar, then a peanut butter bar and finally a acadaia bar As a Clif bar snob, how any different Clif bar tasting enus can you create for your ride? Again, explain in inute detail every part of the forula that you coe up with Since order here atters, there truly is 9 choices for the first bar, 8 for the second AND 7 for the third So there are different tasting enus = 9! 6! = 9! (9! =! ( 9 (c As uch as you love Clif bars, you ust adit that you also like Kind bars and that they soehow feel a bit healthier For your upcoing usual 50 ile ride, you know that a Clif bar and a Kind bar will suffice Given that there are 9 different varieties of Kind bars, how any Clif bar/kind bar duos exist? Again, explain your answer fully You want to buy any one of 9 Clif bars AND one of 9 Kind bars So there are 9 9 duos (d An even hungrier friend decides to go on the century with you and wants to bring bars, Kind or Clif This person being uch ore noral than you, they don t care whether soe of the bars are the sae (or even all of the sae nor how any are Clif and how any are Kind How any different assortents of bars are there for this philistine? Your answer should be as whole as a whole wheat bar which neither Clif or Kind offers, that would be gross Page

4 This friend is willing to eat any of the 99=8 bars offered by both copanies So for each bar they eat, they have 8 choices: 8 for the first, 8 for the second, 8 for the third AND 8 for the fourth So one ight think that there are = 8 = 5086 choices for the Unfortunately, by doing this, order atters and we are overcounting different assortent different nubers of ties (for exaple, brownie, brownie, brownie, peanut butter is counted ties, whereas brownie, peanut butter, acadaia, int chocolate is counted!= ties Instead, let s think of how any duplicated bars they can bring: they can bring either different bars, of the sae bars and two other different bars, of the sae bars and two other bars that are also the sae, three of the sae bars and different bar OR of the sae bar If they bring different bars, there are ( 8 different assortents that exist If they bring bars that are the sae and then two extra bars that are different, there are 8 different assortents since you can choose what three flavors to bring AND then decide which of the three flavors you duplicate Alternatively, you can say you have 8 choices for the flavor that you have two of and then you ust pick two extra flavors of the 7 flavors reaining, so 8 (7 If they bring of the sae bars and two other bars that are the sae (but different fro the first two, there are 8 different assortents since you choose two flavors out of the 8 to bring (and then you double each, so no choice has to be ade further If they bring of the sae and one different bar, then there are 8 assortents since you choose two flavors out of the 8 AND then decide which of the two you bring three of Alternatively, you can say there are 8 flavors to choose fro for the one you bring three of, AND then 7 flavors left for the extra bar, so 8 7 Finally, if they bring of the sae bar, they ust siply decide which of the 8 flavors to bring, so there are 8 choice So in total, there are assortents ( 8 ( 8 ( 8 ( 8 8 = 9, 900 (e Despite being abashed by the lack of discernent of your friend when it coes to snack bars, you appreciate the fact that they are willing to go ride 00k with you and ostly that they put up with you, period You offer to bring three Clif bars for the, and strongly advise the to buy a Kind bar as their final bar to have a balanced bar offering In this case, how any different assortents of bars exist for your friend? Be at least as clear as the translucent packaging of Kind bars For the three Clif bars you bring for the, the situation is as in (d Either you bring the three of the sae, two of the sae and one different OR three different Clif bars If you bring the three of the sae, there are 9 different assortents If you bring the two of the sae and one different, there are ( ( 9 different assortents Finally, if you bring the three different bars, there are 9 different assortents So in total, there are 9 ( ( 9 9 = 0 different assortents you bring But we are not done: you bring the one of 0 Clif bar assortents AND they bring any of 9 different Kind bars So in total, there are different assortents 0 9 = Use a cobinatorial arguent to prove that there are exactly n different subsets of a set of n eleents (Do not use the binoial theore For each eleent, you decide whether it is in the subset or not So you have two choices for the first eleent AND two choices for the second eleent AND two choices for the third eleent AND AND two choices for the nth eleeent So we have n different subsets of a set of n eleents Page

5 7 Use algebraic ethods to prove the cancellation identity: if n and k are non-negative integers and is an integer with n, then Expanding the left-hand side, we get Expanding the right-hand side, we get ( ( n k k ( ( n n = k n! k! (n k! k!!(k! = n! (n k!!(k! n!!(n! (n! (k!(n (k! = n!!(n! (n! (k!(n k! = n! (n k!!(k! 8 You go bikecaping Out of the n different Clif bars you have at hoe, you bring k for your trip While on the trip, you select to bring on a hike Show how you can count the nuber of possible cobinations in two ways so that the cancellation identity of the previous proble appears You choose k out of the n bars you have at hoe to bring and then you choose out of the k bars you brought to go hiking So we get ( n k k( Another way of getting the sae is to choose which bars you will bring on the hike, and then choose which k out of the n bars reaining to bring for the rest of your trip (so that you have k bars in total So we get ( n ( n k 9 Use induction to show that n ( k ( = n if and n are non-negative integers We proceed by induction on n If n = 0, then we get ( ( 0 = If = 0, then both sides are equal to If is anything else, then we get zero on both sides Suppose it holds for n We ll show it for n First separate your su by pulling out the ter for n n ( n = By the induction hypothesis, this is equal to ( n n ( k which we have seen in class is equal to ( n as desired ( n, n ( k Page 5

Here are two situations involving chance:

Obstacle Courses 1. Introduction. Here are two situations involving chance: (i) Someone rolls a die three times. (People usually roll dice in pairs, so dice is more common than die, the singular form.)