(ii) Two-permutations of {a, b, c}. Answer. (B) P (3, 3) = 3! (C) 3! = 6, and there are 6 items in (A). ... Answer.

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1 SOLUTIONS Homewor 5 Due /6/19 Exercise. (a Cosider the set {a, b, c}. For each of the followig, (A list the objects described, (B give a formula that tells you how may you should have listed, ad (C verify that the formula ad the list agree. (i Permutatios of {a, b, c}. (A abc, acb, bac, bca, cab, cba (B P (,! (C! 6, ad there are 6 items i (A. (ii Two-permutatios of {a, b, c}. (A ab, ac, ba, bc, ca, cb (B P (, (C 6, ad there are 6 items i (A. (iii Size-two subsets of {a, b, c}. (A {a, b}, {a, c}, {b, c} (B ( /! (C /!, ad there are items i (A. (b For each of the followig, classify the problem as a permutatio or a combiatio problem or either, ad give a aswer usig a usimplified formula. (Aswers should loo, for example, lie 5 or 5!/! istead of P (5, or 0. (i I how may differet orders ca five ruers fiish a race if o ties are allowed? This is a permutatio problem, ad there are 5! 5 1 ways. (ii How may strigs of 1 s ad 0 s of legth seve have exactly three 1 s? This is a combiatio problem: Choosig the places where the 1 s go gives ( 7 7!!! strigs. (iii How may strigs of 1 s ad 0 s of legth seve have three or fewer 1 s? This is a combiatio problem, to be computed i cases depedig o exactly how may 1 s the sequece has: ( ( ( ( ( ! 1 0!! 1 5!! 1 6!1! 1. 7!0! 1

2 (iv How may three-digit umbers are there with o 1 s? (a three-digit umber is somethig lie 1 or 009 or 05 This is either combiatio or permutatio. This is just product rule: there are ie choices for each digit, givig 9 i total. (v How may three-digit umbers are there with o digits repeated? This is a permutatio problem, ad there are P (10, of these. (c For each of the followig, provide your aswers i a usimplified form, ad justify. (i A six-sided dice is rolled 5 times. How may ways could it tur out that a value greater tha (i.e. 5 or 6 is rolled exactly twice? (Hit: first pic which rolls are from {5, 6} (this implies which rolls are from {1,,, } for free, ad the pic the values for the {5, 6}-valued rolls, ad fially pic the values for the other rolls (the {1,,, }-valued rolls. First choose whe the high rolls happe: ( 5 5 possibilities. The choose how the high rolls go: possibilities. The choose how the low rolls go: possibilities. (For example, a roll sequece lie 5,,, 6, 1 is i the category of roll sequece goes HLLHL; high rolls go 5 the 6; low rolls go the the 1. Now use product rule to combie: 5 ( (. (ii If 10 me ad 10 wome show up for oe team of a itramural basetball game, how may ways ca you pic 5 people to play for oe team if there must be at least oe perso of each geder o the team? Brea this ito the cases of how the geder balace wors: me ad 1 wome, me ad wome, me ad wome, or 1 ma ad wome. The use product rule i each of those cases to get ( 10 ( 10 1 ( ( ( 10!!1! ( ( ( 10!!! ( 10!!! ( ( ( 10!!1! (iii How may ways are there for 5 wome ad me to stad i lie? Now how may ways are there for them to stad i lie if the two me do t stad ext to each other? (The me ad the wome are distict idividuals. Ways are there for 5 wome ad me to stad i lie: This is just 7 people stadig i lie, which has 7! possibilities. If the two me do t stad ext to each other: Pic where the me are stadig, of which there are ( 7 6 possibilities. The pic the order of the wome 5! ad the order of the me. Combie usig product rule: ( ( 7 65!.

3 (d For each of the followig idetities, (A explai i words why it maes sese give what it represets, ad the (B verify it algebraically usig the formulas for permutatio or combiatio. (For example, a aswer for (A might start out looig lie P (, 1 meas..., ad a aswer for part (B should loo lie a calculatio that starts with P (, (i P (, 1 (A This is the umber of ways to pic oe thig out of, of which there are possibilities. Order does t come i to play, sice there s oly oe thig. (B P (, 1! ( 1! (ii P (, 0 1 (A This is the umber of ways to pic othig out of, of which there is oly oe possibility. (B P (, 0!! 1 (iii P (, 1 P (, ( (A P (, 1 is the umber of ways to choose 1 thigs from i order. Sice it s i order, you ca choose the first thigs first, ad the choose the last thig separately. After thigs, there are to choose from.! (B P (, 1 ( (1!! (!/( P (, ( (iv ( 1 (A This is the umber of ways to pic oe thig out of, of which there are possibilities. (B ( 1! ( 1!1! (v ( 1 (A This is the umber of ways to choose everythig all at oce out of a group of. There is oe way to do this pic everythig. (B (! 1 (sice 0! 1 (!! (vi ( 0 1

4 (A This is the umber of ways to choose othig from thigs. There is oe way to do that - do t tae aythig. (B ( 0! 1 (agai, sice 0! 1 ( 0!0! (vii ( ( (A This is because choosig thigs from is the same as picig thigs from to exclude. (B (! (!!!!(! ( Exercise. For each of the followig, be sure to iclude how the pigeohole priciple or its geeralized versio apply i your justificatios, or why either of them do. (1 The lights have goe out ad you re diggig through a uorgaized soc drawer filled with umatched blac socs ad brow socs (otherwise roughly idetical. (a If you re pullig them out at radom, how may socs do you eed to tae out to esure you have a matchig pair if there are 10 of each id of soc? How about if there are 0 of each? 100 of each? It does t matter how may socs you have to choose from as log as you have eough to apply geeralized pigeohole priciple. I ay case, the colors are the boxes ad the socs you ve piced are the items, ad so you wat to solve / for the miimal :.. (b Agai pullig at radom, how may socs do you eed to tae out to esure you have a matchig brow pair if there are 10 of each id of soc? How about if there are 0 of each? 100 of each? This is ot pigeohole priciple sice you eed brow i particular. This is asig to guaratee that a specific box has at least object. So you eed to tae ito accout that you might pull out all of the blac socs before you get a sigle brow soc. Therefore you eed to pic 1 if there are 10 of each id of soc, if there are 0 of each, ad 10 if there are 100 of each.. ( Explai why, out of ay set of four itegers, at least two have the same remaider whe divided by. Here the elemets are the itegers ad the remaiders are the boxes. There are oly three possible remaiders whe dividig by : 0, 1, ad. Sice you ve piced objects, the pigeohole priciple tells us at least of them go i the same box, i.e. that of them have the same remaider ( A recet estimate showed that the US ad Caada together (which share the coutry code 1 have approximately 1,000,000 phoe lies i use. What is the miimum umber of area codes eeded to mae that possible?

5 The boxes are the area codes ad the 7-digit phoe umbers are the objects. You have to esure that there are eough of them so that o more tha the umber of possible 7-digit phoe umbers are forced ito the same area code. Sice there are possible 7-digit phoe umbers, you wat to solve for the miimal such that ( / This is ( Let f : A B be a fuctio betwee fiite sets such that A > B. Explai why f caot possibly be ijective. (Cosider the sizes of the preimages {f 1 (b b B}. Here, the elemets of B are the boxes ad the elemets of A are the objects. A fuctio f is a assigmet of what objects go ito what boxes (a object a is i box b if a is i the preimage of b, i.e. f(a b. Sice there are more objects tha boxes, pigeohole priciple tells us that some box has more tha oe object i it, i.e. some elemet of b has more tha oe elemet of a i its preimage. Therefore the fuctio is ot ijective (5 Explai why, i ay sequece of cosecutive itegers, at least oe of them must be divisible by. (Start with, say, as a example. Lie i part (c, the remaiders are the boxes ad the cosecutive itegers are the boxes. Let a, a 1, a,..., a 1 be a collectio of cosecutive igeters. Ad suppose the least of these itegers has remaider r whe divided by (so that a r, with 0 r <. Now suppose for the sae of cotradictio that oe of them have a remaider of 0. Note that whe you icrease a iteger z by 1, you icrease its remaider by 1; uless z 1 is divisible by, i which case you drop it to 0. Sice oe of the cosecutive itegers is divisible by, ot oly do oe of them have a remaider of 0, but movig up the chai, you ever drop the remaider to 0. So the itegers have distict remaiders. However, if oe of the itegers are divisible by, the we re oly usig 1 out of the boxes. So pigeohole priciple tells us that some box has at least elemets, i.e. there are two of these itegers that have the same remaider whe divided by. This is a cotradictio. Thus oe of ay cosecutive itegers must be divisible by Exercise. (a Expad (x y 5 ad (x y 8 usig the biomial theorem. By the biomial theorem, we have ( ( ( (x y 5 x 5 x y x y 0 1 ( 5 x y ( 5 xy x 5 5x y 10x y 10x y 5xy y 5 ; ad ( ( ( ( ( (x y 8 x 8 x 7 y x 6 y x 5 y x y 0 1 ( ( ( ( x y 5 x y 6 xy 7 y ( 5 y 5 5 x 8 8x 7 y 8x 6 y 56x 5 y 70x y 56x y 5 8x y 6 8xy 7 y

6 6 (b Substitute x z ad y to calculate (z. By the biomial theorem, we have (z (z (z ( 6(z ( (z( z 5 z z z (c What idetity do you get if you substitute x 1 ad y 1 i the biomial theorem? Chec your idetity (lie i the previous problem for. Evaluatig the biomial theorem at x 1 ad y 1 says 0 (1 1 (x y x1 y ( x y x1 y 1 ( 1 ( 1 ( ( 1 0 ( ( 1 0 For example, whe, ( ( 1 0 ( 0 ( Sice ( 1 ( ( ( ( ( (d Usig the biomial theorem to prove combiatorial idetities. (i Use the biomial theorem to explai why 0 ( The chec ad examples of this idetity by calculatig both sides for. (Hit: substitute x y 1.

7 7 Evaluatig the biomial theorem at x y 1 says I particular, for, ( 0 (1 1 (x y x1 y ( 0 ( x y x1 y1 ( (. 1 1 ( 1 ( ( ( (ii Use the biomial theorem to explai why ( 1 0 ( (. The chec ad examples of this idetity by calculatig both sides for. (Hit: what other examples ca you thi of of itegers that sum to?. Evaluatig the biomial theorem at x ad y 1 says ( 1 (x y x y ( x y x y 1 ( ( ( 1 ( ( ( 1, ( ( (. I particular, for, ( ( ( 1 ( 0 sice ( 1 ( 1 ( 1 ( ( 1 ( 1 ( 1 ( ( 1, ( ( ( ( ( (

8 8 (e Give a formula for the coefficiet of x i the expasio of (x 1/x 100, where is a iteger. Usig the biomial theorem, evaluate at y 1/x x 1. So the coefficiet of x correspods to x a (1/x 100 a where a (100 a a 100. So a ( 100/. So x has coefficiet 0 whe is odd or ( (100/ whe is eve ad betwee 100 ad 100. For a smaller example, cosider (x 1/x x x (1/x 6x (1/x x(1/x (1/x x x 6x 0 x x. Chec: ( ( ( ( ( 1,, 6,,. ( / ( / ( 0/ ( / ( / Alteratively, ote that (x 1/x 100 ( (1/x(x ( 1 x 100 (x x 100 (x By the biomial theorem, 100 ( ( 100 (x (x x. So ( ( 100 (x 1/x 100 x 100 x x 100. So the coefficiet o x l is ( whe l 100 for ad is 0 otherwise (l100 (which agrees with our solutio above (f As a useful coutig tool, we have (so far oly defied ( for o-egative itegers ad. But i 8. (p. 59, Def., the boo defies exteded biomial coefficiets as follows: Let u R ad Z 0. The exteded biomial coefficiet ( u is defied by ( { u u(u 1(u (u 1/! if > 0 1 if 0. For example, ( 0. (i Compute ( ( π, 1/ (, ad 7/ ( 0.8( We have ( π π(π 1(π (π /!.05; ( 1/ (1/(1/ 1/! 1/8; ad ( 7/ 1, by defiitio, sice 0 here. 0

9 9 (ii Verify algebraically that if is a positive iteger ad 0, the ( ( 1 ( 1. Chec this idetity for 5 ad (you should probably do this first. Proof. For 0, the both sides are equal to 1. Otherwise, we have ( ( ( 1( ( 1! ( 1( ( 1 ( 1! ( 1 X(X 1(X (X 1,! where X 1, flippig the order of multiplicatio, as desired. i particular, sice X 1 ( 1 1, ( 1 ( 1, (iii BONUS: The exteded biomial theorem states that for ay real umber u, we have ( u (1 x u x. i0 Now recall from calculus the Taylor series expasio (1 x 1 x i ( 1 i. i0 Chec that the first terms (i 0, 1, of our ow Taylor series expasio match the first terms of the exteded biomial theorem expasio (whe u 1. Fially, verify that this example matches correctly for all terms by showig that ( 1 ( 1 for ay Z 0. Proof. Note that ( 1 1( 1 1( 1 ( 1 1! 1( ( (! ( 1! ( 1.! So the exteded biomial theorem says that ( 1 (1 x 1 x as desired. i0 ( 1 x, i0 (iv BONUS: Show i that the Taylor series expasio for (1 x matches the exteded biomial theorem for.

10 10 Proof. We have ( ( 1( ( 1! ( ( 5 ( ( ( 1( (! ( 1( ( ( ( 5 ( ( ( 1( ( ( 1(! ( 1!( 1( (! ( 1( ( 1 ( 1( ( 1. So the exteded biomial theorem says that (1 x i0 ( x ( 1( ( 1 x i0 i0 ( 1( ( x. O the other had, we have Thus d dx (1 x 1 (1 x ; so that d dx (1 x 1 d ( (1 x (1 x. dx d (1 x 1 (1 x 1 dx ( 1 d dx ( 1 x 0 1 ( d ( 1 dx x 0 1 ( ( 1 ( 1x 1 1 ( 1 ( ( 1x 0 0 ( 1( ( x, which matches what we got from the exteded biomial theorem.

11 Exercise 5. (a Explai the example provided for the proof of Vadermode s i the otes usig words. I provig Vadermode s idetity, we showed that you ca cout the size r subsets of A B i two ways, where A, B m, ad r, m. The first way was to cout them all at oce. I this example, whe r, these are the subsets listed i the table. The secod way to cout them is tae size subsets of A ad size r subsets of B ad uio them. These correspod to the differet colums of the table (the first colum is taig o elemets from B, the secod colum is taig oe elemet from B, ad so o (b Substitute m r ito Vadermode s idetity to show that ( (, ad chec this idetity for. We have For, we get ( m r mr j0 0 ( r ( ( m j r j j0 ( ( j j j0 ( ( 6 ( ( j 0 ( 1 mr j0 ( (. j (c Cosider the idetity ( for itegers 1. (i Verify this idetity for 5 ad. We have ( ( ( 1 1 (

12 1 (ii Explai why this idetity is true usig a combiatorial argumet. [Hit: Cout, i two differet ways, the umber of ways to pic a subset with elemets from a set with elemets, alog with a distiguished elemet of that -elemet subset. For example, out of people, pic a committee of people ad choose someoe o that committee to orgaize their meetigs. ] We ca cout the umber of ways to choose a committee of people from ad a perso to ru that committee i two ways. First, choose the committee ad the someoe to ru it from amogst those people. There are ( ways to do this. O the other had, you could fist choose the perso to ru the committee, ad the choose the other 1 members from the remaiig 1 people. There are ( 1 ways to do this. 1 Sice ( ( ad 1 1 are two differet ways of expressig the size of the same set, they must be equal. (iii Illustrate your combiatorial proof usig the set A {a, b, c} (so that ad. I the first way of coutig, we have committee {a, b} with leader a, committee {a, b} with leader b, committee {a, c} with leader a, committee {a, c} with leader c, committee {b, c} with leader b, ad committee {b, c} with leader c. I the secod way of coutig, we have leader a with other member b, leader b with other member a, leader a with other member c, leader c with other member a, leader b with other member c, ad leader c with other member b. Either way, there are 6 choices. (iv Verify the idetity algebraically usig the formula (!/((!!. Algebraically, we have (! ((!!! ((!( 1! ( 1! (( 1 ( 1!( 1! ( 1. 1