In the question below suppose that a word is any string of seven letters of the alphabet, with repeated letters allowed.

Transcription

1 163 Dscrete Mathematcs Revew Use the followg to aswer questo 1: I the questo below suppose that a word s ay strg of seve letters of the alphabet, wth repeated letters allowed 1 How may words beg wth A or B or ed wth A or B? 6 Soluto Number of words begg wth A or B s 6 - two possbltes for the frst letter ad 6 for each of the other 6 letters I the same way umber of words edg wth A or B s 6 6 Number of words both startg wth A or B ad edg wth A or B s 46 - possbltes for the frst letter, two for the last ad 6 for each of letters the mddle Applyg the formula E ( F) = E ( ) + ( f) E ( F) we see that the aswer to the problem s = 4 6 (6 1) = = Use the followg to aswer questo : I the questos below let A be the set of all bt strgs of legth 10 How may bt strgs of legth 10 have exactly sx 0s? Soluto We have to choose 6 places for 0s amog te The umber s C(10, 6) = C(10, 4) = = 10 4! Use the followg to aswer questo 3: I the questos below a club wth 0 wome ad 17 me eeds to form a commttee of sze sx 3 How may commttees are possble f the commttee must cosst of all wome or all me? Soluto Number of commttees of all wome s C(0,6), of all me - C(17, 6) The aswer s C(0, 6) + C(17, 6) = =1136 Page 1

2 4 Fd the umber of subsets of S = {1,,3,,10} that cota exactly fve elemets, the sum of whch s eve Soluto If the sum of the elemets of a subset s eve the the subset cotas 0, or, or 4 odd umbers ad respectvely, or 3, or 1 eve umbers Applyg the multplcato prcple we see that the total umber of such subsets s C(, 0) C(,) + C(, ) C(,3) + C(, 4) C(,1) = = 16 Fd the coeffcet of x y 6 the expaso of (x y) 11 Soluto The correspodg term the expaso sc(11, )( x) ( y) 6 whece the 6 coeffcet of x y s C(11,) = 46 3 = Use the followg to aswer questo 6: I the questos below assume that you have 0 pees ad three jars, labeled A, B, ad C 6 I how may ways ca you put the pees the jars, assumg that the pees are detcal ad each jar must have at least two pees put to t? Soluto If we put two pees each box we wll have 44 detcal (dstgushable) pees to be dstrbuted to 3 labeled (dstgushable) jars The umber of ways to dstrbute dstgushable objects to k dstgushable boxes s (see page 377 ad Theorem o page 373) C ( + k 1, k 1) I our case the umber s C (46,) = 103 Use the followg to aswer questo 7: I the questos below you pck a bt strg from the set of all bt strgs of legth te 7 What s the probablty that the bt strg has more 0s tha 1s? Soluto Let p1, p, p3be the probabltes to have more 0s tha 1s, more 1s tha 0s, ad the same umber of 0s ad 1s, respectvely The p 1 = p, p3 = C(10,)(1/ ) (1/ ), ad p1+ p + p3 = 1 Therefore 10 1 C(10,)/ p1 = = Page

3 Use the followg to aswer questo 8: I the questos below a expermet cossts of pckg at radom a bt strg of legth fve Cosder the followg evets: E 1 : the bt strg chose begs wth 1; E : the bt strg chose eds wth 1; E 3 : the bt strg chose has exactly three 1s 8 Determe whether E ad E 3 are depedet Soluto Let us frst fd the probablty of The total umber of bt strgs of legth 4 s = 3 The umber of strgs edg wth 1 s = 16 Therefore, 16 1 C(,3) pe ( ) = = I a smlar way probablty of E3 s PE ( 3) = = Fally let 3 16 us compute the probablty of E E The umber of strgs edg wth 1 ad havg 3 6 exactly three 1s s C (4,) = 6 whece PE ( E3) = pe ( 1) pe ( 3) = 3 3 are ot depedet E 3 E ad E ad 9 Each of 6 cards has a dfferet letter of the alphabet o t You pck oe card at radom A vowel s worth 3 pots ad a cosoat s worth 0 pots Let X = the value of the card pcked Fd E(X), V(X), ad the stadard devato of X 1 1 Soluto EX ( ) = = V( X) = σ ( X) = V( X) 118 Use the followg to aswer questo 10: I the questos below, descrbe each sequece recursvely Iclude tal codtos ad assume that the sequeces beg wth a 1 Page 3

4 10 1,,3 3,4, Soluto We see that, 1,, a = = Therefore a 1 a = ( + 1) + = + 1 ad the sequece ca be descrbed recursvely as a = 1, a = a Use the followg to aswer questo 11: I the questos below solve the recurrece relato ether by usg the characterstc equato or by dscoverg a patter formed by the terms 11 a = 10a 1 1a, a 0 =, a 1 = 1 Soluto The characterstc equato of the above recurrece relato s r + 10r+ 1 = 0 The left part ca be factored as ( r+ 3)( r+ 7) whece the solutos of the characterstc equato are r1 = 3, r = 7, ad the geeral soluto of the recurrece relato s a = α1( 3) + α( 7) From the tal codtos we fd α1+ α = 3α1+ 7α = 1 By multplyg the frst equato by 7 ad subtractg from t the secod equato we get 4α 1 = 1whece α = ad α = α = 1 = 7 Fally a = ( 3) ( 7) 4 4 Page 4

5 1 What form does a partcular soluto of the lear ohomogeeous recurrece relato a = 4a 1 4a + F() have whe F() = ( + 1)? Soluto The assocated homogeeous recurrece relato s a = 4a 1 4a Its characterstc equato r 4r+ 4= 0has soluto of multplcty By theorem 6 o page 469 there s a partcular soluto of our ohomogeeous recurrece relato of the 4 3 form a = ( A + B+ C) = ( A + B + C ) To fd the values of the coeffcets A, B, ad C let us troduce the polyomal G ( ) = A B C a = G( ), a = G( + 1), ad a = 4 G( + ) The Pluggg these expressos to the relato a = 4a 4 a + [( + ) + 1] ad dvdg both parts by 4 we obta the relato G ( + ) = G ( + 1) G ( ) Or G ( + ) G ( + 1) + G ( ) 4 = 0 By expadg the expressos the left part ad combg the lke terms we get ( 1+ 1 A ) + (6B+ 4A 4) + 6B+ C+ 14A= 0, From here we fd A =, B =, ad C = Page

6 13 A market sells te kds of soda You wat to buy 1 bottles How may possbltes are there f you wat at most three bottles of ay kd? Soluto We wll use the formula o page 06 ad follow the patter of Example 1 o the same page Our problem ca be reformulated as follows How may solutos does the equato x 1+ + x10 = 1 have, where x,, 1 x1 0 are oegatve tegers ad x 3, = 1,,10 Let P, = 1,,10, be the property that x > 3 Notce that for ay four dstct dces 1 1 < < 3 < 4 10 we have N( P P ) 0 1 P 3 P 4 = Ideed, otherwse the umber of bottles would be at least 16 Therefore our case the formula o page 06 takes the form N( P P ) = N N( P) + N( PP) N( PPP ) ' ' 1 10 j < j 10 1 < j< k 10 j k Let us ow compute each term the rght part of the formula above We ca fd the total umber N of oegatve teger solutos of the equato x1+ + x10 = 1 by followg the patter Example o page 373 N = C( ,1) = C(1,1) = C(1,9) Next, aga by followg Example o page 373, we see that for ay dex,1 10, whece N P C C ' ( ) = ( ,8) = (17,8) Smlarly for ay ad therefore, N( P ) = 10 C(17,8) 1 10, j,1 < j 10we have N( PP) = C( , 4) = C(13, 4) j 1 < j 10 N( PP) = C(10, ) C(13, 4) j Fally, for ay, j, k,1 < j < k 10 ad NPPP ( j k) = 1 Page 6

7 1 < j< k 10 NPPP ( ) = C(10, 3) j k The aswer s C(1,9) 10 C(17,8) + C(10,) C(13,4) C(10,3) = How may permutatos of all 6 letters of the alphabet are there, that cota at least oe of the words SWORD, PLANT, CARTS? Soluto Let A, B, ad C be the sets of all permutatos whch cota SWORD, PLANT, or CARTS, respectvely Notce that A C = B C = Therefore by cluso excluso prcple for three sets we have N( A B C) = N( A) + N( B) + N( C) N( A B) Let us compute N( A) SWORD ca start from posto 1 up to posto ; each case the remag 1 letters ca be arraged 1! ways Therefore N( A ) = 1! =! Smlarly NB ( ) = N( C= )! Next we compute N( A B) Let us frst look at the case whe SWORD goes before PLANT SWORD ca start from ay posto from 1 to 17, the respectve umbers of ways to place PLANT wll be 17, 16,,1 The remag 16 letters ca be arraged 16! ways There s the same umber of ways to place PLANT frot of SWORD Thus we obta N( A B) = ( )16! = 16! = 18! The aswer s N( A B C) = 3! 18! = Fd the umber of ways to clmb a 1-step starcase, f you go up ether oe or three steps at a tme Soluto Let a be the umber of ways to clmb steps by gog oe or three steps at a tme The we have a recurret relato a = a + a 1 3 wth the tal codtos a1 = a = 1, a3 = From here t s easy to compute drectly that a 1 = 60 Page 7