n n B. How many subsets of C are there of cardinality n. We are selecting elements for such a

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1 4. [10] Usig a combiatorial argumet, prove that for 1: = 0 = Let A ad B be disjoit sets of cardiality each ad C = A B. How may subsets of C are there of cardiality. We are selectig elemets for such a subset without repletio ot with cocer for order so there are such subsets. Alteratively, let represet the umber of elemets i such a subset that were selected from A. The value of may vary from 0 to. There are such selectios of the elemets from A. Now select which elemets from B will ot be i the subset (the that remai will thus be i the subset). There are of selectig these so ways of selectig the subset ad ways overall. = 0 This must equal.. a. [10] Preset a combiatorial argumet that for all 1: = 3 = 0 Let A = {a, b, c} ad cosider all strigs of legth usig elemets of A. Sice there are three optios for each compoet of the strig, there are 3 such strigs. Alteratively, cosider first cosider the positios of ay c's i the strig. Let represet the umber of o-c's (i.e., a's ad b's) i the strig. Clearly could rage from 0 through. For a fixed value of, there are ways to choose the positios for the o-c's. The for each of the positios, there are two optios (i.e., a or b) for the character i the positio. The remaiig - positios must be occupied by c's. Thus there are ways to assig elemets to the positios with o-c's. The total is ad this must equal 3 = 0 b. [10] Preset a combiatorial argumet that for all oegative itegers p, s, ad satisfyig p + s p p + s = p s p + s p (Hit: Cosider choosig two subsets.)

2 Let a set A have elemets ad cosider how may ways there are to select disjoit subsets B ad C of A so that B has p elemets ad C has s elemets. First we could select the p elemets for B i ways ad the select the s elemets for C from the remaiig p p -p elemets of A~ B i s ways. Together this yields p such p s selectios. Alteratively, we could first select the p+s elemets for B C i p + s p + s ways ad the select the p elemets for B from B C i ways. There are thus p p+ s p + s p such selectios ad this must equal p p s. a. [10] Preset a combiatorial argumet that for all 1: = 1 = 1 (Note: The summatio begis with = 1.) Cosider the cardiality of the set of o-empty subsets of a set A of elemets. For each elemet of A, there are two optios: either be preset i a subset or ot. Thus there are total subsets but oe of these is empty so there are 1 oempty subsets of A. Alteratively, let idicate the cardiality of the subset. Sice we are coutig o-empty subsets, rages from 1 to. For a fixed value of, there are ways of selectig the subset elemets from the total elemets of A. Addig this to iclude all possible cases of, we obtai ad this must = 1 equal 1.

3 b. [10] Preset a combiatorial argumet that for all itegers ad satisfyig 3 = (Hit: Cosider three special elemets.) Cosider the umber of subsets of size of a set B of cardiality. Sice 3, we may select three elemets b, b b of B ad let C = B { b, b, }. Thus C has 1, 3 cardiality -3 ad B = C { b 1, b, b 3 }. We ow there are such subsets. Alteratively, to select elemets of B for a subset there are four optios: all come from C, -1 come from C ad the th is either b,, or, -come from C ad the -1st ad th are exactly two of b 1, b, or, or -3 come from C ad all of b, b, 1 ad b 3 are preset. For the first optio, there are possibilities sice all come from C. For the secod optio, there are 3 possibilities, sice -1 1 elemets are selected from C ad oe from the three of b,, or b. For the third optio, there are 3 possibilities, sice - elemets are selected from C ad oe from the three of b 1, or b is ot selected. Lastly, if -3 come from C ad b, 3 ~ 1 b3 all of b ad b are preset, the there are, b 1, 3 optios. The total is 3 + ad this must equal [10] Preset a combiatorial argumet that for all positive itegers m,, ad r, satisfyig r mi{ m, } : m+ = r m r =0 r. (Hit: Cosider selectig from two sets.) Let A ad B be disjoit sets of cardialities m ad, respectively. Let C = A B ad cosider the umber of subsets of C of cardiality r. Sice m+ C = A + B) = m+, there are r such subsets. Alteratively let be the umber of elemets i a subset that came from A. The value of ca rage from m 0 to r. For a fixed value of, there are ways to select the elemets from b 3 1 b b 3 1 b 3

4 A ad ways to select the remaiig r r elemets from B, thus r m+ total ways. This must equal =0 r r. 3. [10] Preset a combiatorial argumet that for all positive itegers : 3 = = 0. Cosider as a model strigs of legth usig the characters from the set { abc,,}. For each positios there are 3 optios so there are 3 such strigs. Alteratively, let represet the umber of positios i the strig ot occupied by a (i.e., thus, occupied by either b or c ). The value of ca vary betwee 0 ad. For a fixed umber of b s ad c s, there are ways to determie the positios to be occupied by the b s ad c s ad the choices (either b or c ) for each of these positios, for a total of possibilities. The remaiig positios must be occupied by a s. Summig over all possible values of. We have such = 0 strigs ad this must equal 3. Examiatio 1 Solutios CS [5] For 3, how may diagoals does a covex polygo with extreme poits have? (Cosider a covex polygo give by extreme poits < P, P, 1..., P > i couterclocwise order A diagoal is a lie segmet coectig two o-adjacet extreme poits.) P 3... P P1 P For each of the extreme poits there are -3 distict extreme poits that o-adjacet. This would yield ( 3) edpoits of the diagoals. Sice each diagoal has two edpoits, there are ( 3) diagoals of a covex polygo with extreme poits.. a. [10] Preset a combiatorial argumet that for all 1: ()! ( 1) ( 3) 3 1=!

5 Cosider the set of all partitios of a set of cardiality ito pairs. For the left side, begi with ay permutatio of the elemets. The first elemet o the permutatio is i some pair ad there are 1 choices for its pair-mate. Removig these two from the permutatio, the ext elemet permutatio is also i some pair ad there are 3 choices for its pair-mate. The process cotiues util there are just two elemets left i the permutatio, ad they form the last pair. This yields ( 1) ( 3) 3 1 differet such partitios. Now cosider the right had side. There are ()! differet permutatios of the of the elemets. Pair the first elemet with the secod, the third with the fourth, etc. This yields a partitio ito pairs. However, the order amog the pairs is irrelevat to the partitio ad thus for every array of pairs there are differet permutatios. Lastly, the order amog the pairs, is also irrelevat, so a set of pairs could be arraged i! differet orders. Thus the umber of partitios ito pairs that igores order withi ad amog pairs is ( )! ad this must equal ( 1) ( 3) 3 1.! b. [10] Preset a combiatorial argumet that for all oegative itegers ad satisfyig + = Let set A have cardiality ad b ad c be distict elemets ot cotaied i A. Cosider the subsets of A {} b { c} of cardiality. For the left had side, we recogize that A {} b {} c has cardiality +, so there are + such subsets. Alteratively, cosider that a subset wither has all elemets comig from A, exactly 1 elemets comig from A, or A, exactly elemets comig from A. If all elemets come from A, there are. If exactly 1 elemets come from A, there are ways to select those elemets ad the two choices, b or, 1 to complete the subset. If exactly elemets come from A, there are ways to select those elemets ad the both b ad must be selected to complete the subset. The total is + + ad this must 1 + equal. 3. Preset a combiatorial argumet that for all positive values of m,, ad r:

6 m+ = r r = 0 m r m+ Cosider distict sets A ad B of cardialities m ad, respectively. There are r subset of A B of size r. Alteratively, for ay such subset, there must be some r- elemets of A ad elemets of B for a value of betwee 0 ad r. For a fixed there m r are m such subsets ad thus overall. r r = 0 3. a. [10] Usig a combiatorial argumet, prove that for 1: 1 = 1 = 1 (Hit: Let A ad B be disjoit sets of cardiality. Cosider pairs <C, a> where C A B, C has cardiality, ad a C A.) Usig the otatio of the hit, first choose a ad the choose C ~ { a}. There are choices for a (sice # A = ) ad there remai -1 elemets i A B ~ { a}. Thus, 1 there are total choices. Alteratively, let = # ( A C). The value of ca 1 rage from 1 (sice a A C ) to. For a fixed, there are choices for A C, choices from that for a, ad choices for C B. The total is = 1 b. [10] Usig a combiatorial argumet, prove that for 1: = 0 = Usig the same otatio as above, cosider choosig just a set C A B of cardiality, There are such choices. Alteratively, let be the umber of elemets i A C : ca rage from 0 to. For a fixed, there are ways of choosig A C, ad sice there are - elemets i B C there are elemets i B ~ C, ad hece ways of choosig them. Choosig B ~ C however is equivalet to choosig

7 B C ad thus there are ways to choose B C. The total is ad this = 0 must equal. 3. a. [10] Usig a combiatorial argumet, prove that for ad m : + m = m + + Let A ad B be disjoit sets of cardialities ad m, respectively. We see to determie how may subsets of two elemets there are i A B. Sice the + m cardiality of A B is + m, there are such subsets. Alteratively, we could obtai such a subset by selectig oe elemet from each of A ad B, by selectig both elemets from A, or by selectig both elemets from B. There are m + + m + ways of doig this ad, therefore = m + +. b. [10] Usig a combiatorial argumet, prove that for itegers m,, p 1: p p p p m ( + m) = = 0 Let A ad B be disjoit sets of cardialities ad m, respectively. We see to determie how may strigs of legth p there are cosistig of elemets of A B. Sice the cardiality of A B is + m, there are + m optios for each of p positios i the sequece, so there are ( + m) p such sequeces. Alteratively, let deote the umber of positios i the sequece occupied by elemets of A. The p value of varies from 0 to p. For a fixed value of, there are ways to select these positios ad the optios for each of the positios. For each of the p positios occupied by elemets of B, there are m optios, thus p p p p m for the fixed value of ad m p overall. This must equal = 0 ( + m) p. 3. a. [10] Usig a combiatorial argumet, prove that for ad m : + m = m + +

8 Cosider subsets of two elemets from the uio of disjoit subsets A ad B with + m cardialities ad m, respectively. Sice #( A B) = + m, there are subsets of size two. Alteratively, cosider that either oe elemet comes from each of A ad B, both from A, or both from B. These ca be doe i m,, ad ways, respectively, ad the total is m + +. We coclude that + m = m + + b. [10] Usig a combiatorial argumet, prove that for 1: 1 = = 1 (Hit: Let A be a set of cardiality. Cosider pairs <B, a> where B A ad a A ~ B.) Employig the otatio from the hit, ad cosiderig the left side of the equatio first, there are choices for a ad the 1 subsets from the remaiig 1 elemets. Alteratively, let be the umber of elemets i { a} B. The value of could rage from 1 through. For a fixed value of, there are ways to choose{ a} B, ad the choices from this for a (with the remaiig chose elemets formig B ). There are total ways of doig this ad this must = 1 equal a. [10] Usig a combiatorial argumet, prove that for 1 ad m : ( m 1) = m = 0 Cosider strigs of legth selected from the itegers { 1,,..., m} with repetitio allowed. For each of positios there are m choices, so there are m such strigs. Alteratively, let idicate the umber of copies of m i the strig. The value of varies from 0 to. For a fixed value of there are selectios for the placemet of the m s ad the ( m 1) choices for the itegers { 1,,..., m 1} i

9 each of the remaiig positios. Thus there are ( m 1) such strigs with copies of m, ad ( m 1) = 0 overall. This must equal m. b. [10] Usig a combiatorial argumet, prove that for 0 :!( )! =! Cosider permutatios of legth selected from the itegers { 1,,..., }. There are! such permutatios. Alteratively, let satisfy 0 ad for ay permutatio first select the positios to be occupied by { 1,,..., }. There are such selectios. Now permute the values { 1,,...,} - there are! such permutatios. Fially, permute the values { + 1, +,..., }, which ca be doe i ( )! ways, ad place them ito the positios of the permutatio otoccupied by the values from { 1,,..., }. Thus, there are!( )! such permutatios ad this must equal!.