Discrete Mathematics (Classroom Practice Booklet Solutions)

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2 Discrete Mthemtics (Clssroom Prctice Booklet Solutions). Logic Propositionl Logic. Ans: () Sol: ( (P Q) (P Q) P) ( P Q) ( P Q) P (By Demorgn s lw) (( P Q) ( P Q)) P (By Associtive lw) ( P ( Q Q)) P (By Distributive lw) ( P T) P ( Q Q is tutology) P P ( P T P) T. Ans: (c) Sol: (( P Q) ( P Q)) R By ssocitive lw & distributive lw P.Q Q P P Q P Q R. R. By distributive lw By Demorgn s lw & commuttive lw.r R. Ans: (d) Sol: () L.H.S A (P C) A (P C) 6 ( A P) C Associtive lw (A P) C 6 & Demorgn s lw R.H.S (b) L.H.S (P C) (Q C) ( P C) ( Q C) By 6 ( P Q) C Distributive lw (T ) R R ((P Q) ( PQ)) (T is tutology). Ans: (d) Sol: In Boolen lgebr nottion, the given formul cn be written s P. Q.R Q.R P.R P.Q.R Q P R. (P Q) C By 6 R.H.S (c) A (BC) ~A (~B C) By 6 (~A ~B) C By ssocitive lw (~B ~A) C By commuttive lw ~ B (~A C) By ssocitive lw B (AC) By 6 Boole s contribution in logic firmly estblished the point of view tht logic should use symbols nd the lgebric properties should be studied in logic

3 : : Discrete Mthemtics (d) When A is flse nd B is flse, we hve LHS is true nd RHS is flse.. Ans: (d) LHS RHS Sol: The truth tble of propositionl function in n vribles contin n rows. In ech row the function cn be true or flse. By product rule, number of non equivlent propositionl functions (different truth n tbles) possible 6. Ans: (c) Sol: A set of connectives is sid to be functionlly complete, if equivlent form of every sttement formul cn be written with those connectives. () {, } is functionlly complete becuse the other connectives cn be epressed by these two connectives. (P Q) ( P Q) (P Q) ( P Q) (P Q)( ( P Q) (QP)) (b) {, } is functionlly complete (P Q) ( P Q) (P Q) (P Q) (P Q) (P Q) (Q P) (c) The set {,} is not functionlly complete, becuse, we cnnot epress not opertion using the connectives nd (d) We hve, (p q) (~p q) {, ~} is functionlly complete set. 7. Ans: (b) Sol: Argument I: This rgument is not vlid, becuse it comes under fllcy of ssuming the converse. Argument II: This rgument is vlid, by the rule of modus tollens. Argument III: Let P : It rins nd Q : rik is sick In symbolic form the rgument is P Q P Q This rgument is not vlid, becuse when P is flse nd Q true, the premises re true but conclusion is flse. 8. Ans: (d) Sol: () The given formul is equivlent to the following rgument () ( b) c () ( b) ( c) Proof: () new premise to pply conditionl proof () b (), (), modus ponens

4 : : CSIT Postl Coching Solutions () ( b) (), (), conjunction (6) c (), (), modus ponens The rgument is vlid (c.p) (b) The rgument is () () () b c c d d ~ b () new premise to pply C.P () ~ b (), (), conjunctive syllogism(c.s) (6) c (), (), D.S (7) d (), (6), M.P The given rgument is vlid (conditionl proof) It is not vlid 9. Ans: (b) Sol: From the truth tble ( y) ( ~y) (p q) ~(~p ~q) ~(~p q). Ans: (d) Sol: If {( b) ( c)} hs truth vlue flse, then '' hs truth vlue true, b hs truth vlue true, nd c hs truth vlue flse. For these truth vlues, only the compound proposition given in option (d) hs truth vlue true. (c) The given formul is equivlent to the following rgument () () ( b) (c d) (~b c) Proof:. Ans: (c) Sol: Let us denote the formul by P Q, Where P ( b) c nd Q (b c). () Here Q is flse only when is true, b is flse, nd c is flse. For these truth vlues P hs truth vlue true. () ~ b now premise to pply c.p (P Q) hs truth vlue flse. () ( ~b) (), (), conjunction The given formul is not tutology. () ~( b) (), 7 (b) When is true, b is true nd c is flse; (6) (c d) (), (), D.S (P Q) hs truth vlue true. (7) c (6), simplifiction The given formul is not The rgument is vlid (c.p) contrdiction (c) The given formul true in one cse nd (d) When is flse, b is true nd c is flse, flse in other cses the given formul hs truth vlue flse. The given formul is contingency.

5 : : Discrete Mthemtics. Ans: (b) Sol: We hve (p q) (p q) (q p) (~ p q) (~ q p) Option (b) is true.. Ans: () Sol: Argument I. ~p (q ~w) premise. ~s q premise. ~t premise. (~p t) premise. ~p (),(), disjunctive syllogism 6. q ~w (), (), modus ponens 7. (~s ~w) (), (6), trnsitivity 8. (w s) (7), contrpositive property Argument I is vlid. Argument II We cn not derive the conclusion from the premises, by pplying the rules of inference. Further, when p, q, r, t nd w hs truth vlues true, we hve ll the premises re true but conclusion is flse. Argument II is not vlid. Ans: (d) Sol: () L. H. S ( b c) quivlence ( b) c Associtivity ( b) c quivlence R. H. S. (b) L. H. S ( b) ( c d) quivlence ( b c) d quivlence R. H. S (c) L. H. S. ( ( b)) T R. H. S. ( b) ( b) ( ) b T b T L. H. S R. H. S (d) When is flse nd b is true We hve L. H. S. F nd R. H. S. T Option (d) is not true becuse L. H. S. R. H. S.. Ans: () Sol: The given formul is equivlent to the following rgument. ~ p (q ~w) Premise. ~ s q Premise. ~ (w t) Premise. (~ p t) Premise s

6 : : CSIT Postl Coching Solutions. (w ~ t) ), quivlence 6. w ), Simplifiction 7. ~t ), Simplifiction 8. ~p ), 7) Disjunctive syllogism 9. (q ~w) ), 8) Modus ponens. ~q 9), 6) Modus tollens. s ), ) Modus tollens The rgument is vlid Hence, the given sttement is tutology 6. Ans: (d) Sol: () When p is true, q is flse, nd r is true; the premises re true nd conclusion is flse. Therefore, the rgument is not vlid. (b) When p is flse nd r is true, the premises re true nd conclusion is flse. The rgument is not vlid. (c) When p is true, q is true, r is true, nd s is flse; then the given rgument is not vlid (d) The premises re true only when p is true, q is true, r is true, s is flse nd t is true. Whenever the premises re true, the conclusion is lso true. The rgument is vlid 7. Ans: (c) Sol: If (p q) is contrdiction, then p nd q hve different truth vlues. S is vlid, becuse (p q) hs truth vlue flse. S is vlid, becuse (p q) is true S is not vlid, becuse when p is true nd q is flse,s hs truth vlue flse. S hs truth vlue true, in both the cses, i.e., {p true nd q flse} or {p flse nd q true} S is vlid. 8. Ans: (c) Sol: Argument I is equivlent to the following rgument. {p r, q r, p q} r This rgument is vlid by the rule of dilemm. The given rgument is lso vlid by Conditionl proof (C.P). Argument II is equivlent to the following rgument {p q, p r, p} (q r) ) p q premise ) p r premise ) p new premise to pply (C.P) (q r) ) q (), (), modus ponens

7 : 6 : Discrete Mthemtics ) r (), (), modus ponens 6) q r (), (), conjunction Argument II is vlid (Conditionl Proof). 9. Ans: () Sol: The given formul cn be written s ((b) (cd) (b d)) ( c) This formul is vlid, by the rule of destructive dilemm. First order Logic. Ans: () Sol: To negte sttement formul we hve to replce with, with nd negte the scope of the quntifiers. ~{ {P() ~Q()}} {P() Q()} (Use the equivlence (P Q) (P Q)). Ans: (c) Sol: (B () I ()) [ (B() I ())] ( (B() I ())). Ans: () Sol: To negte sttement formul we hve to replce with, with nd negte the scope of the quntifiers. Use the equivlence (P Q) (P Q). Ans: (d) Sol: () vlid (see Reltionship digrm in mteril book) (b) vlid (see Reltionship digrm in mteril book) (c) vlid (see Reltionship digrm in mteril book) (d) not vlid (see Reltionship digrm in mteril book). Ans: (d) Sol: S is true Once we select ny integer n, the integer m n does eist nd n + m n + ( n) S is true, becuse if we choose n the sttement nm m is true for ny integer m. S is flse, for emple, when m the sttement is flse for ll n S is flse, here we cn not choose n m, becuse m is fied.. Ans: (b) & (d) Sol: () L.H.S (A () B()) ( A() B()), 6 A() B(), A() B(), 6 R.H.S

8 : 7 : CSIT Postl Coching Solutions (b) L.H.S { A() B()) ( A() B()) (A() B() R.H.S But converse is not true (b) is flse (c) vlid equivlence (d) not vlid (converse is not true) 6. Ans: (b) Sol: () The given formul is vlid by conditionl proof, if the following rgument is vlid. () { P() Q() } () P() new premise to pply C.P Q() Proof: () P() Q() (), U.S () P() (), U.S () Q() (),(), M.P (6) Q() (), U.S The given formul is vlid (C.P) (b) The sttement need not be true. Let c nd d re two elements in the universe of discourse, such tht P(c) is true nd P(d) is flse nd Q(c) is flse nd Q(d) is flse. Now, the L.H.S of the given sttement is true but R.H.S is flse. The given sttement is not vlid. (c) (P() Q()) ( P() Q()) Indirect proof: ) (P() Q ()) Premise ) ( P() Q ()) New premise to pply Indirect proof ) P() Q() (), Demorgn s lw ) P() (), Simplifiction ) Q() (), Simplifiction 6) P () (),.S 7) Q () (), U.S 8) ( P () Q ()) (6), (7),Conjunction 9) (P() Q()) (8), Demorgn s lw ) (P () Q()) (), U.S ) F (9), (), Conjunction vlid (Indirect proof) S : The rgument is ) y (P(, y) W (, y)) ) W (, b) P(, b) (d) { P() Q() } follows from ( P() Q()) The given sttement is vlid.

9 : 8 : Discrete Mthemtics 7. Ans: (c) Sol: Consider Argument I. {p() q()} premise. [{~p() q()} r()] premise. {p() q()} (), universl specifiction. {~p() q()} r() (), U. S.. ~r() new premise to pply C.P 6. ~{~p() q()} (), (), disjunctive syllogism 7. { p() ~q()} (6), demorgn s lw 8. { p() q()} {p() ~ q()} (), (7) conjunction 9. p() {q() ~q()} (8), distributive lw. p() F from (9). p() from (). {~r() p()} from (),. {~r() p()} from (), U.G The rgument is vlid (C.P) Argument II. [p() {q() r()}] premise. {p() s()} premise. p() s() (),. S. p() (), simplifiction. s() (), simplifiction 6. p() {q() r()} (), U.S 7. q() r() (),(6), modus ponens 8. r() (7), simplifiction 9. r() s() (),(8), conjunction. {r () s()} (9),. G The rgument is vlid (C.P)

10 : 9 : CSIT Postl Coching Solutions 8. Ans: (d) Sol: The given sttement cn be represented by S. Further, S S S Option (d) is correct 9. Ans: (b) Sol: S is equivlent to the following rgument ) P() premise ) {P() Q() premise Q() Here, we cnnot combine () nd (), to get the conclusion, becuse in both the formule eistentil quntifiers re used. ) P() from () nd (), by modus ponens, ) P() from () S is vlid S : ) P() premise ) Q() premise {(P() Q() () Q() (),.S. () P() () U.S. () P() Q() (), (), conjunction (6) {P() Q()} (), U.G. S is vlid S is equivlent to the following rgument ) P() premise ) {P() Q() premise Q(). Ans: (d) Sol: I) Let D () : is doctor C () : is college grdute G () : is golfer The given rgument cn be written s ) {D () C ()} ) {D() G ()} {G () C()} ) {D() G ()} ), istentil Specifiction ) {D () C()} ), Universl Specifiction ) D () ), Simplifiction

11 : : Discrete Mthemtics 6) G () ), Simplifiction 7) C () ), ), Modus ponens Conjunction 8) C () G ()} 7),6), Conjunction 9) {G () C()} 8), istentil Generliztion The rgument is not vlid II) Let M () is mother N () is mle P () : is politicin The given rgument is ) {M() N ()} ) {N () P ()} {P() M ()} ) N () P () ), istentil Specifiction ) M () N () ), Universl Specifiction ) N () ), Simplifiction 6) P () ), Simplifiction 7) M () ), ), Modus tollens 8) {P() M ()} 6), 7), Conjunction 9) {P () M ()} 8), istentil Generliztion The rgument is vlid.. Ans: (b) Sol: S is flse. For. There is no integer y such tht is divisor of y S is true. If we choose,, then the sttement is true for ny integer y S is true. If we choose,, then the sttement is true for ny integer y S is flse, becuse there is no integer y which is divisible by ll integers.. Ans: (b) Sol: P(): 7 +, Q():, S ) For, Q () true nd R() is flse. S is not true. S ) For, Q() is true nd R() is true. S is true

12 : : CSIT Postl Coching Solutions S ) For the integer, P() is flse P() R() is true for integer S is true.. Ans: (c) Sol: () When y, the given sttement is flse (b) When the given sttement is flse (c) Solving the equtions + y nd y 8 we get nd y The given sttement is true (d) Solving the equtions y 7 nd + y we get nd y The given sttement is flse. Ans: (b) Sol: Argument I:. A (),. { A () Q () Q(). A() from (), by eistentil specifiction. ~ {A () Q ()} from (), by universl specifiction. (~ A () ~ Q ()) (), demorgn s lw 6. ~ Q() from () nd () by disjunctive syllogism. given rgument is not vlid. Argument II:. { {(P() Q ()) R ()},. Q ()} R(). {(P() Q ()) R ()}, from () by eistentil specifiction.. Q() from (), by universl specifiction. P() Q () from () by ddition 6. R() from () nd () by modus ponens 7. R() from (6) by eistentil genrliztion Argument II is vlid.

13 : : Discrete Mthemtics. Ans: (b) Sol: (). {P() Q()} Premise. ~Q() Premise. ~Q() () istentil Specifiction (.S). {P() Q()} (), Universl Specifiction (U.S). P() (), (), Disjunctive Syllogism (D.S) 6. P() (), istentil Generliztion (.G) The rgument is not vlid (b). ~P() Premise. {P() Q()} Premise. ~P() (),.S. P() Q() (), U.S From () nd () we cnnot derive the conclusion ~Q() The rgument is not vlid (c). {P() Q()} Premise. Q() Premise. P() Q() (), U.S. Q() (), U.S From () nd (), we cnnot derive the conclusion. The rgument is not vlid (d). P() Premise. ~{P() Q()} Premise. ~{P() Q()} (),.S. P() (), U.S. ~Q() (), (), Conjunctive Syllogism 6. ~Q() (),.G The rgument is vlid

14 : : CSIT Postl Coching Solutions 6. Ans: (c) Sol: () If is not free then {W A()} W A() is vlid Refer pge of mteril book. (b) L.H.S {A() W} { A() } W A() W (c) { A() W} { A () W} 6 { A() } W using () { A() } W 6 R.H.S (d) vlid refer derivtion of option (c).. Combintorics. Ans: Sol: {n(a B C)} sum of the elements in ll the regions of the digrm. n(a B C) n( A B C) B A C. Ans: () Sol: n(bpc) n(b) + n(p) + n(c) n(bp) n(pc) n(bc) + n(bpc) + + n(bp) n(pc) n(bc) + n(b P) + n(pc) +n(bc) 7 Number of students received wrds in ectly two subjects 7 8. Ans: () Sol: Number of students received wrds in only one subject 8 Option () is correct.

15 : : Discrete Mthemtics. Ans: (d) Sol: Given: M e c b g C + b + c + d d b + c +d p. m. c f p + m + c (+b+c+d+e+f+g)+(b+c+d)+ 7 7 P n(f G) n(g S) n(f S) + n(f G) + n(g S) + n(f S) 9 ( + ) + (y + ) + (z + ) 9 + y + z 7 Number of students who cn spek tlest two lnguges + y + z + III. Number of students who cn spek ectly two lnguges + y + z 7 IV. Number of students who cn spek only one of the lnguges + b + c 8 ( + y + z + ) 8 6. Ans: (d) Sol: I. Number of students who cn spek tlest one lnguge (number of students who cn spek none of the lnguges) 8 II. G F b y z c S n (F G S) n(f) + n(g) + n(s) n (F G) n(g S) n(f S) + n (F G S) 7. Ans: 8 Sol: n( ) Number of integers divisible by nd Number of integers divisible by L.C.M of nd n( 7) Number of integers in the set divisible by L.C.M of {7,, } Number of integers in the set divisible by Required number of integers 8

16 : : CSIT Postl Coching Solutions 8. Ans: 96 Sol: In binry mtri of order we hve elements. ch element we cn choose in wys. By product rule, Required number of mtrices Ans: 88 Sol: An nglish movie nd telugu movie cn be selected in (6)(8) 8 wys A telugu movie nd hindi movie cn be selected in (8).() 8 wys A hindi movie nd n nglish movie cn be selected in ()(6) 6 movies Required number of wys Ans: 78 Sol: Number of -digit integers 9 Number of -digit integers with distinct digits (9) (9) 8 Number of -digit integers with distinct digits (9).(9).(8) 68 Required number of integers Ans: 67 Sol: Cse(i): If the first digit is 6, then ech of the remining digits we cn choose in 9 wys. Cse(ii): If the first digit is not 6, then first digit we cn choose in 8 wys, digit 6 cn pper in wys nd ech of the remining digits we cn choose in 9 wys. Required number of integers (9)(9)(9) + (8)()(9)(9) 67. Ans: 9 Sol: Consider n integer with digits. Digit cn pper in wys Digit cn pper in wys Digit cn pper in wys ch of the remining digits we cn choose in 7 wys. By product rule, Required number of integers ()()()(7)(7) 9. Ans: 89 Sol: ch Tennis mtch elimintes one plyer nd we hve to eliminte 89 plyers. We hve to conduct 89 mtches.. Ans: Sol: ch element of A cn pper in the subsets in wys.

17 : 6 : Discrete Mthemtics Cse : The element ppers in P but does not pper in Q. Cse : The element ppers in Q nd does not pper in P. Cse : The element does not pper in P nd does not pper in Q. By product rule, Required number of wys n. Ans: Sol: Required number of wys Number of onto functions possible from persons to rooms C(, ) + C(, ). () + 6. Ans: P(,6) Sol: Required number of wys Number of wys we cn mp the 6 persons to 6 of the books P(,6) 7. Ans: () Sol: The women cn spek s group in wys. The women group cn spek with other men in wys. Required number of wys Ans: 88 Sol: First girls cn sit round circle in wys. Now there re distinct plces mong the girls, for the boys to sit. Therefore, the boys cn sit in P(, ) wys. By product rule, Required number of wys.p(, ) Ans: Sol: Consider 8 positions in row mrked,,,.., 8. Cse : Boys cn sit in odd numbered positions in wys nd girls cn sit in even numbered positions in wys. Cse : Boys cn sit in even numbered positions in wys nd girls cn sit in odd numbered positions in wys. Required number of wys. +. Ans: Sol: Number of signls we cn generte using flg Number of signls we cn generte using two flgs P(,). nd so on. Required number of signls + P(,) + P(,) + P(,) + P(,)

18 : 7 : CSIT Postl Coching Solutions. Ans: 6 Sol: ch book we cn give in wys. By product rule, Required number of wys 6. Ans: Sol: ch digit of the integer we cn choose in wys. By product rule, Required number of integers. Ans: 6 Sol: Required number of permuttions, 6... Ans: Sol: Required number of binry sequences 6.. Ans: Sol: Required number of outcomes. 6. Ans: Sol: Required number of wys.. 7. Ans: Sol: Required number of wys number of ordered prtitions 8. Ans: 9.. Sol: Required number of wys Number of unordered prtitions of set in to subjects of sme size (.... ) Ans: 6 Sol: We cn select men in C(, ) wys. Those men cn be pired with women in P(, ) wys. Number of possible selections C(, ). P(, ). () 6. Ans: Sol: The zeros cn pper in the sequence in C(,) wys. The remining 7 positions of the sequence cn be filled with ones in only one wy. Required number of binry sequences C(,).. Ans: Sol: Consider string of 6 ones in row. There re 7 positions mong the 6 ones for plcing

19 : 8 : Discrete Mthemtics the zeros. The zeros cn be plced in C(7,) wys. Required number of binry sequences C(7,) C(7,). Ans: Sol: To meet the given condition, we hve to choose distinct deciml digits nd then rrnge them in descending order. We cn choose distinct deciml digits in C(,) wys nd we cn rrnge them in descending order in only one wy. Required number of wys C(,).. Ans: n(n ) Sol: We hve n persons. Number of hndshkes possible with n persons C(n,) If ech person shkes hnds with only his/her spouse, then number of hndshkes possible n Required number of hndshkes C(n, ) n n(n ). Ans: 9 Sol: In chess bord, we hve 9 horizontl lines nd 9 verticl lines. A rectngle cn be formed with ny two horizontl lines nd ny two verticl lines. Number of rectngles possibles C(9,). C(9,) (6)(6) 96 Number of squres in chess bord very squre is lso rectngle. Required number of rectngles which re not squres Ans: 8 Sol: Between H nd R, we hve 9 letters. We cn choose letters in C(9,) wys nd then rrnge them between H nd R in lphbeticl order in only one wy. Required number of letter strings C(9,) Ans: Sol: We cn choose 6 persons in C(, 6) wys We cn distinct 6 similr books mong the 6 persons in only one wys Required number of wys C(, 6). C(, ) 7. Ans: Sol: Required number of wys V(,) V(n,k) C(n +k, k) V(,) C(,) C(,)

20 : 9 : CSIT Postl Coching Solutions 8. Ans: Sol: To meet the given condition, let us put bll in ech bo, The remining blls we cn distribute in V(,) wys. Required number of wys V(,). C(,) C(,) 9. Ans: Sol: The number of solutions to the inequlity is sme s the number or solutions to the eqution Where 6 The required number of solutions V(6,) C(,) C(,). Ans: Sol: Let y +, y, y + The given eqution becomes y + y + y Number of solutions to this eqution V(, ) C(, ) Required number of solutions. Ans: 6 Sol: Let X units digit, X tens digit nd X hundreds digit Number of non negtive integer solutions to the eqution X + X + X is V (, ) C (, ) C (, ) 66 We hve to eclude the cses where X i (i,, ) Required number of integers Ans: Sol: Let y +, y +, y +, y + nd y + 6 Required number of solutions Number of non negtive integer solutions to the eqution y + y + y + y + y. V(, ) C( +, ) C(, ) C(, ). Ans: Sol: To meet the given conditions, let us put books on ech of the shelves. Now we re left with books to distribute mong the shelves. Which ever wy we distribute the remining books, the number of books on ny shelf cnnot eceed. Required number of wys V (, ) C( +, ) C (, ). Ans: () Sol: This is similr to distributing n similr blls in k numbered boes, so tht ech bo contins tlest one bll.

21 : : Discrete Mthemtics If we put bll in ech of the k boes, then we re left with (n k) blls to distribute in k boes. Required number of wys V(k, n k) C(k + n k, n k) C(n, k ). Ans: 8 Sol: Required number of wys Number of non negtive integer solutions to the eqution, where i (i,,, ) This is equivlent to number of non negtive integer solutions to the eqution Required number of wys V(, 6) Where V(n, k) C(n + k, k) V(, 6) C(9, 6) C(9, ) 8 6. Ans: S, S, S, S 6 Sol: Averge number of letters received by n prtment A 8. Here, A 9 nd A 8 By pigeonhole principle, S nd S re necessrily true. S follows from S nd S 6 follows from S. S nd S need not be true. 7. Ans: 97 Sol: If we hve n pigeon holes, then minimum number of pigeons required to ensure tht tlest (k+) pigeons belong to sme pigeonhole kn + For the present emple, n nd k+ 9 Required number of persons kn + 8() Ans: 6 Sol: By Pigeonhole principle, Required number of blls kn + () Ans: 9 Sol: The fvourble colors to drw 9 blls of sme color re green, white nd yellow. We hve to include ll red blls nd ll green blls in the selection of minimum number of blls. For the fvourble colors we cn pply pigeonhole principle. Required number of blls (kn + ) Where k + 9 nd n (8 + ) 9. Ans: Sol: Suppose 6, Minimum number of blls required kn +6 where k + 6 nd n. () + 6

22 : : CSIT Postl Coching Solutions Which is impossible < 6 Now, minimum number of blls required + (kn + ) where k + 6 nd n + () +. Ans: 7 Sol: For sum to be 9, the possible -element subsets re{,9}, {, 8},{, 7},{, 6},{, } If we tret these subsets s pigeon holes, then ny subset of S with 6 elements cn hve t lest one of these subsets. Since we need two such subsets, the required vlue of k 7.. Ans: 7 Sol: If we divide number by the possible reminders re,,,, 9. Here, we cn pply pigeonhole principle. The 6 pigeonholes re {}, {}, {,9}, {,8}, {,7}, {,6} In the first two sets both + y nd y re divisible by. In the remining sets either + y or y us divisible by. The minimum number of integers we hve to choose rndomly is 7.. Ans: Sol: very positive integer n cn be written s, n k m where m is odd nd k. Let us cll m the odd prt. If we tret the odd numbers,,,, s pigeonholes then we hve pigeonholes. very element in S hs on odd prt nd ssocited with one of the pigeonholes. The minimum vlue of k. Ans: 6 Sol: For the difference to be, the possible combintions re {, 6}, {, 7}, {, 8}, {, 9}, {, }. If we tret them s pigeonholes, then we hve pigeonholes. By pigeonhole principle, if we choose ny 6 integers in S, then the difference of the two integers is.. Ans: Sol: The distinct prime fctors of re, nd. Required number of +ve integers ()....

23 : : Discrete Mthemtics 6. Ans: 8 Sol: The distinct prime fctors of 8 re, nd. Required number of +ve integers (8) Ans: 88 Sol: The distinct prime fctors of re 7 nd 9. Required number of +ve integers () Ans: 6 Sol: Required number of functions number of derngements possible with 6 elements D Ans: (i) (ii) 76 (iii) (iv) 89 (v) 9 (vi) Sol: (i) Number of wys we cn put letters, so tht no letter is correctly plced D (ii) Number of wys in which we cn put letters in envelopes Number of wys we cn put the letters so tht no letter is correctly plced D Required number of wys D 76 (iii)number of wys we cn put the letters correctly C(,) The remining letters cn be wrongly plced in D wys. Required number of wys C(,) D () (iv) Number of wys in which no letter is correctly plced D Number of wys in which ectly one letter is correctly plced C(,) D Required number of wys D + C (,)D + (9) 89 (v) There is only one wy in which we cn put ll letters in correct envelopes. Required number of wys 9 (vi) It is not possible to put only one letter in wrong envelope. Required number of wys

24 : : CSIT Postl Coching Solutions 6. Ans: (i) 96 (ii) Sol: (i) The derngements of first letters in first plces D Similrly, the lst letters cn be dernged in lst plces in D wys. The required number of derngements D D () () 96 (ii) Any permuttion of the sequence in which the first letters re not in first plces is derngement. The first letters cn be rrnged in lst plces in wys. Similrly, the lst letters of the given sequence cn be rrnged in first plces in wys. Required number of derngements. 6. Ans:!.D 6 Sol: First time, The books cn be distributed in wys. Second time, we cn distribute the books in D wys. Required number of wys D 6 6. Ans: Sol: Let bn n The given recurrence reltion becomes b n+ b n The solution is b n ( n ) n n, 6. Ans: () Sol: The recurrence reltion is n n n () The chrcteristic eqution is t Complementry function C. n Here, is chrcteristic root with multiplicity. Let prticulr solution (c n + d n) Substituting in (), (cn + d n) {c (n ) + d(n )} n n c + d n c + d c nd d P. S n n The solution is n C + n n () Using the initil condition, we get C Substituting C vlue in eqution (), we get n n n + 6. Ans: (b) Sol: Cse : If the first digit is, then the remining digits we cn choose in n wys Cse : If the first digit is nd second digit is, then the remining digits we cn choose in n wys.

25 : : Discrete Mthemtics Cse : If the first two digits re zeros, then ech of the remining digits should be. By sum rule, The recurrence reltion for n is n n + n + 6. Ans: (c) Sol: Let n number of n-digit quternry sequences with even number of zeros Cse : If the first digit is not, then we cn choose first digit in wys nd the remining digits we cn choose in n wys. By product rule, number of quternry sequences in this cse is n. Cse : If the first digit is, then the remining digits should contin odd number of zeros. Number quternry sequences in this cse is ( n n ) By sum rule, the recurrence reltion is n n + ( n n ) n n + n 66. Ans: () Sol: The given recurrence reltion is ( + 6) n. The chrcteristic eqution is t t + 6 t, Complementry function C. n + C. n Prticulr solution 6 n n 6 () 6 The solution is n C n + C n + Using initil conditions, we get C nd C n ( n ) + Now, ( ) Ans: (d) Sol: Cse: (i) If the first squre is not red then it cn be colored in wys nd the remining squres cn be colored in n wys. Cse (ii) If the first squre is colored in red, then second squre cn be colored in two wys nd remining squres cn be colored in n wys. By sum rule, the recurrence reltion is n n + n n ( n + n ) 68. Ans: 867 Sol: n n + (n ) n + ( ) n + ( ) + ( + ) n + ( ) + ( + + ) n + ( n )

26 : : CSIT Postl Coching Solutions 7 + n (n+) (n +) 7 + () () () Ans: (b) Sol: The chrcteristic eqution is t t t The solution is n C n C Using the initil conditions, we get C nd C 7. Ans: (d) Sol: The recurrence reltion cn be written s ( + ) n n+ The uiliry eqution is t t + t, C.F. (C + C n) n n P.S. n n+ The solution is n C + C n + n+ n 7. Ans: () Sol: Cse : If the first digit is, then number of bit strings possible with consecutive zeros, is n. Cse : If the first bit is nd second bit is, then the number of bit strings possible with consecutive zeros is n. Cse : If the first two bits re zeros nd third bit is, then number of bit strings with consecutive zeros is n Cse : If the first bits re zeros, then ech of the remining n bits we cn choose in wys. The number of bit strings with consecutive zeros in this cse is n. The recurrence reltion for n is n n + n + n + n. 7. Ans: () Sol: Replcing n by n+, the given reltion cn be written s n+ n + (n + ) n+ ( ) n 6 (n +) n.() The chrcteristic eqution is t t complementry function C n Let prticulr solution is n n (cn +d) where c nd d re undetermined coefficients. Substituting in the given recurrence reltion, we hve

27 : 6 : Discrete Mthemtics n (c n + d) n {c(n )+d} n n (c n +d) {c(n )+d} n quting coefficients of n nd constnts on both sides, we get c nd d 6 Prticulr solution n ( n 6) Hence the solution is n C n (n + 6) n () C 6 C n ( n ) (n + 6) n 7. Ans: () Sol: Cse(i): If the first bit is, then the required number of bit strings is n Cse(ii): If the first bit is, then ll the remining bits should be zero The recurrence reltion for n is n n + 7. Ans: (b) Sol: n n + (n ) n + n + + n n n n n 7. Ans: Sol: Let +, b + y, c + z, nd d + w The trnsformed system is + y + z + w 7 where, y, z, w The generting function f() ( + + ) ( ) ( ) ( ) n C (n, n) n Required number of solutions Coefficient of 7 in f() C (, ) C (7, ) + 6 C (, ) Ans: () Sol: Required generting function f() ( ) n n. n 77. Ans: 86 Sol: The generting function for the given eqution is F (X) (X + X + X + X + X + X 6 ) The coefficient of X in F(X) is the nswer in our problem

28 : 7 : CSIT Postl Coching Solutions F (X) X ( + X + + X ) 6 X X X X ( X 6 ) ( X) (X X + 6X 6 X + X 8 ). n C (n,n)x Coefficient of X C (, ) C (8, ) 78. Ans: (d) Sol: Required generting function f() Ans: () ( ) ( + ) (Binomil theorem) Sol: ( ) ( ). [( ) ] [ ] n C n 9,n Coefficient of 7 C(6, 7) C(6, 9) n n 8. Ans: 8 Sol: The generting functions for the problem is f() ( ) ( ) ( ) ( ) ( ) ( ) ( ) n C n,n The required number of solutions coefficient of in f() C(, ) C(7, ) Ans: (c) Sol: The generting function is f() ( ) + ( ) +. ( ) n

29 : 8 : Discrete Mthemtics. Ans: () Sol: For ny simple grph, ( G) V (G) V (G) () (6) (G). (G). Ans: 9 Sol: By sum of degrees of regions theorem, if degree of ech verte is k, then k V V (8) V 9. Ans: (c) Sol: If degree of ech verte is k, k V k V () V ( k,,,) k V or or 8 or 6 only option (c) is possible.. Ans: (e). Grph Theory Sol: () {,,,,, } Here, sum of degrees, n odd number. The given sequence cnnot represent simple non directed grph (b) {,,,, } In simple grph with vertices, degree of every verte should be. The given sequence cnnot represent simple non directed grph. (c) {,,,,, 6, 6} Here we hve two vertices with degree 6. These two vertices re djcent to ll the other vertices. Therefore, verte with degree is not possible. Hence, the given sequence cnnot represent simple non directed grph. (d) {,,,.., n } Here, we hve n vertices, with one verte hving degree n-. This verte is djcent to ll the other vertices. Therefore, verte with degree is not possible. Hence, the given sequence, cnnot represent simple non directed grph. (e) A grph with the degree sequence {,,,, } is shown below. c d b e

30 : 9 : CSIT Postl Coching Solutions 6. Ans: 8 Sol: Here, degree of ech verte is By sum of degrees theorem, V V () V 8 The minimum number of vertices G cn hve 8 7. Ans: Sol: G is tree By sum of degrees theorem, n. + () +. () +. () n + 8 ( v ) (n ) n + 8 n + 6 n 8. Ans: 8 Sol: G hs 8 vertices with odd degree. For ny verte vg, Degree of v in G + degree of v in G 8 If degree of v in G is odd, then degree of v in G is lso odd. If degree of v in G is even, then degree of v in G is lso even. Number of vertices with odd degree in G 8 9. Ans: 7 Sol: By sum of degrees theorem, if degree of ech verte is tmost K, then K V () Ans: (d) Sol: () Sum of the degree of the vertices () 7 An odd number. The grph is not simple grph (b) Mimum number of edges possible in simple grph with vertices C(, ) (c) Sum of the degrees n An odd number The grph is not simple grph (d) A connected grph with n vertices nd n edges is tree. A tree is simple grph.. Ans: Sol: In the grph, ll the cycles re of even length. G is biprtite grph. Chromtic number of ny biprtite grph is.

31 : : Discrete Mthemtics. Ans: Sol: In the given grph, ll the cycles re of even length. G is biprtite grph nd every biprtite grph is -colorble Chromtic number of G.. Ans: Sol: G is disconnected grph with two components, one component is the complete grph K nd the other component is the trivil grph with only n isolted verte Chromtic number of G. Ans: (b) Sol: n n/ + n n/ + + n { n/ + n/ } 6 n n Ans: (c) Sol: Chromtic number of K n n If we delete n edge in K, then for the two vertices connecting tht edge we cn ssign sme color. Chromtic number 9 The grph hs 9 vertices. The mimum number of vertices we cn mtch is 8. A mtching in which we cn mtch 8 vertices is { b, c d, e f, g h} Mtching number of the grph 7. Ans: Sol: The given grph is K, Mtching number 8. Ans: Sol: The given grph is b c d e f h g If we delete the edge {,b} then the grph is str grph. If we mtch with b, then in the remining vertices we cn mtch only two vertices. Mtching number 9. Ans: Sol: Let us lbel the vertices of the grph s shown below d c 6. Ans: c b Sol: b h g i d f e There re miml mtchings s given below { d, b c}, { c, b d} nd { c d}

32 : : CSIT Postl Coching Solutions. Ans: Sol: The given grph is b d c The miml mtchings re { b, c d}, { c, b d}, { d, b c}. Ans: Sol: The grph hs miml mtchings, 6 mtchings with one edge, nd mtching with no edges. Number of mtchings. Ans: Sol: G is complete grph on 7 vertices. n Mtching number of K n Mtching number of K 7. Ans: () Sol: If n is even, then biprtite grph with mimum number of edges is k n/,n/ n Mtching number of G If n is odd, then biprtite grph with mimum number of edges k m,n n n Where m nd n Mtching number of G n, if n is even n, if n is odd Mtching number of G. Ans:, n Sol: The grph cn be lbeled s e c b d c is cut verte of the grph G. verte connectivity of G K(G) G hs no cut edge. dge connectivity (G).. () We hve, (G) (G) () From () nd (), we hve (G). Ans:, Sol: The grph G cn be lbeled s g d e c f b h G hs no cut edge nd no cut verte. By deleting the edges {d, e} nd {b, h} we cn disconnect G.

33 : : Discrete Mthemtics (G) By deleting the vertices b nd d, we cn disconnect G. K (G) 6. Ans: & Sol: The grph G cn be lbeled s b c d The verte d is cut verte of G. K (G) e f We hve (G) (G)... () G hs no cut edge nd by deleting ny two edges of G we cnnot disconnect G. (G) 7. Ans: S, S & S Sol: S : This sttement is true. g h Proof: Suppose G is not connected G hs tlest connected components. Let G nd G re two components of G. Let u nd v re ny two vertices in G We cn prove tht there eists pth between u nd v in G. Cse: u nd v re in different component of G. Now u nd v re not djcent in G. u nd v re djcent in G Cse: u nd v re in sme component G of G. Tke ny verte wg. Now u nd v re djcent to w in G. There eists pth between u nd v in G. Hence, G is connected. S : The sttement is flse. we cn give counter emple. d b d b G G Here, G is connected nd G is lso connected. S : Suppose G is not connected Let G nd G re two connected components of G. Let vg deg(v) n n (G) n Now V(G ) Similrly, n V(G ) Now, V(G) V(G ) + V(G ) V(G) n + Which is contrdiction c G is connected. c

34 : : CSIT Postl Coching Solutions S : If G is connected, then the sttement is true. If G is not connected, then the two vertices of odd degree should lie in the sme component, By sum of degrees of vertices theorem. There eists pth between the vertices. 8. Ans: S, S & S Sol: The grph G cn be lbeled s d c b e The number of vertices with odd degree is. S nd S re true C is cut verte of G. Hmiltonin cycle does not eists. By deleting the edges {, c} nd {c, e}, there eists Hmiltonin pth b c d e 9. Ans: S, S & S Sol: The grph G cn be lbeled s e d c b The number of vertices with odd degree uler pth eists but uler circuit does not eist. There eists cycle pssing through ll the vertices of G. b c d e is the Hmiltonin cycle of G. The Hmiltonin pth is b c d e. Ans: S & S Sol: The number of vertices with odd degree S nd S re true. To construct Hmiltonin cycle, we hve to delete two edges t ech of the vertices nd f. Then, we re left with edges nd 6 vertices. G hs neither Hmiltonin cycle nor Hmiltonin pth.. Ans: (B) Sol: S is flse. We cn prove it by giving counter emple. Consider the grph G shown below d c b e e is cut verte of G. But, G hs no cut edge S is flse. We cn prove it by giving counter emple. For the grph K shown below, b The edge {, b} is cut edge. But K hs no cut verte.

35 : : Discrete Mthemtics. Ans: Sol: If G hs K components, then V K 6 V 7 V. Ans: (b) Sol: A -regulr grph G hs perfect mtching iff every component of G is n even cycle. S nd S re true. S need not be true. For emple the complete grph K hs perfect mtching but K hs no cycle. S need not be true. For emple G cn hve two components where ech component is K.. Ans: Sol: In simple grph with n vertices nd K components, > n k n k Required minimum number of edges n k n k Where n nd k. Ans: (d) Sol: G hs ectly two vertices of odd degree. Therefore, uler pth eists in G but uler circuit does not eist. In Hmiltonin cycle, degree of ech verte is. So, we hve to delete edges t verte d nd one edge t ech of the vertices nd g. Then we re left with 8 vertices nd 6 edges. Therefore, neither Hmilton cycle eists nor Hmiltonin pth eists. 6. Ans: (b) Sol: G hs cycles of odd length Chromtic number of G (G) () For the vertices c nd h we cn use sme color C The remining vertices from cycle of length 6. A cycle of even length require only two colors for its verte coloring. For vertices, d nd f we cn pply sme color C For the vertices {b, e, g} we cn use sme color C (G) A perfect mtching of the grph is b, c d, e f, g h Mtching number Hence, chromtic number of G + Mtching number of G + 7

36 : : CSIT Postl Coching Solutions 7. Ans: (c) Sol: S need not be true. Consider the grph Here, we hve 6 vertices with degree, but the grph is not connected. S need not be true. For the grph given bove, uler circuit does not eist, becuse it is not connected grph. A simple grph G with n vertices is necessrily connected if (G) S is true. 8. Ans: () n. Sol: Verte connectivity of G k(g) (G) (G) By sum of degrees theorem V Minimum number of edges necessry 9. Ans: (b) Sol: Here, G is complete grph with k+ vertices. Number of edges C(k+, ). Ans: Sol: If verte connectivity of G is, then degree of ech verte in G is. By sum of degrees theorem, V.. Set Theory. Ans: (c) Sol: () A (A B) A (Absorption lw) Option () is flse (b) A (A B) A (Absorption lw) Option (b) is flse (c) (A B) (A B) A (B B ) Distribution lw A Option (c) is true. Ans: (b) Sol: If S {} then () P(S) S {} option () is flse (b) P(S) P(S) {, {}} Option (b) is true (c) If S {, b} then P(S) S option (c) is flse (d) flse Refer option (b)

37 : 6 : Discrete Mthemtics. Ans: (c). Ans: (c) Sol: If A then Sol: Let X A A 9 Cse : If is even number then it cn P(A A) 9 pper in two wys i.e., either A B or B A. Ans: (, c & d) Sol: A I II V IV VII VI III B Cse : If is odd number then it cn pper in two wys i.e., A B or A B By product rule, required number of subsets C I, II,.., VII re regions () (A B) C {I, IV} {IV, V,VI, VIII} {I} A (C B) {I, II, IV, V} {IV, VII} {I, II, V} Option () is flse (b) A (B C) (A B) (A C) is true by Demorgn s lw (c) A (B C) {I, II, IV, V} {II, III} {I, IV, V} A (C B) {I, II, IV, V} {IV, VII} {I, II, V} Option (c) need not be true (d) Similrly show tht option (d) is not true 6. Ans: (c) Sol: If A B (A B) C Then A B But (A B) A B U (A B) U Where U is universl set 7. Ans: (d) Sol: () Let AB A AB A B (b) (AB) B A (B B) A A (c) A C B C A B (cncelltion lw)

38 : 7 : CSIT Postl Coching Solutions (d) LHS A B (A B) (A B) RHS (A B) (A B) (A B) L.H.S R.H.S 8. Ans: (c) Sol: S ) Let A {}, B {}, C {} Now (AB) (BA) But A B S is not true S ) Let A {}, B {}, C {, } Now AC BC C But A B S is not true S ) Let A. Consider the two cses Cse: C (A C) ( (A C)) (B C) (A C B C) B.() Cse: C (A C) (B C) B ( C) () A B (Form () nd ()) Similrly we cn show tht B A. A B Hence, S is true 9. Ans: 686 Sol: A symmetric reltion on A with ect ordered pirs cn be in one of the following wys. i) The reltion my contin digonl pirs. We cn choose digonl pirs in C(7, ) wys. ii) The reltion my contin digonl pirs nd non-digonl pirs. Number of symmetric reltions in this cse is C(7, ). C(, ) iii) The reltion my contin nondigonl pirs. Number of symmetric reltions in this cse is C(, ) By sum rule, Required number of symmetric reltions C(7, ) + C(7, ). C(, ) + C(, ) Ans: (d) Sol: Cse : If the reltion R contins only digonl pirs, then R is symmetric nd trnsitive but not irrefleive. Cse : If the reltion R contins some non digonl pirs Let (, b) R

39 : 8 : Discrete Mthemtics (b, ) R ( R is symmetric) Now (, b) R nd (b, ) R (, ) R ( R is trnsitive) R is not irrefleive. Options () & (b) need not be true, for emple. If A {,, } then the reltion R {(, ), {, ), (, ), (, )} is symmetric nd trnsitive but neither refleive nor irrefleive. Option (b) need not be true for emple. If A {,, }, then the reltion R {(,), (, ), (, )} is symmetric, trnsitive nd not irrefleive.. Ans: (c) Sol: We hve, divides A R A R is refleive Let R b is divisor of b or b is divisor of b is divisor of or is divisor of b b R R is symmetric R is not trnsitive, for emples R 6 nd 6 R, but is not relted to. Hence, R is comptibility reltion on A.. Ans: (c) Sol: We hve, g.c.d of (, ) R is not refleive Let R y g. c. d of (, y) g. c. d. of (y, ) y R R is symmetric R is not trnsitive, For emple g. c. d. of (, ) nd g. c. d. of (, ) but g. c. d. of (, ). Ans: (c) Sol: Let A {,, } nd R {(, ), (, )} S {(, ), (, )} R S {(, ), (, ), (, ) (, )} Here, R nd S re nti-symmetric reltions on A, But (R S) is not nti-symmetric. (R S) need not be nti-symmetric. Any subset of nti-symmetric reltion is lso nti-symmetric, nd (R S) R. (R S) is nti-symmetric Hence, only option (c) is true.. Ans: (D) Sol: () R is not refleive. For emple, is not relted to. i.e., (, ) R (b) R is not irrefleive. For emple, is relted to. i.e., (, ) R

40 : 9 : CSIT Postl Coching Solutions (c) R is not symmetric. For emple, is relted to but is not relted to. (d) R is nti-symmetric. i.e., If ( R b nd b R ) then ( b), b A i.e., For ny two integers. If { b nd b } then b. Ans: (d) Sol: Two positive integers nd b re reltively prime, if g. c. d of nd b is. i) R is not refleive. For emple g. c. d. of {, } is not relted ii) Let R b g. c. d. of {, b} g. c. d. of {b, } b R R is symmetric iii) R is not trnsitive. For emple, R nd R but is not relted. iv) R is not nti-symmetric. For emple ( R ) nd ( R ) v) R is not irrefleive. For emple (, ) R R 6. Ans: () Sol: Symmetric closure of R {(, ), (, ), (, ), (, )} Trnsitive Symmetric closure of R A A 7. Ans: (d) Sol: [] { ( ) is divisible by } {k + k is integer} {, 6,,, 6,,,.} [] [ 6] 8. Ans: (d) Sol: Z is not relted to Z R is irrefleive Let R b b b b R R is symmetric 9. Ans: (b) Sol: S {,,,, 6, 9,, 8, 6} () R is prtil order on ny set of positive integers. Option () is true (b) In the set S, nd re not comprble. Therefore R is not totl order on S. (c) R is not symmetric. For emple is divisor of, then is not divisor of. R is not n equivlence reltion

41 : : Discrete Mthemtics (d) R is prtil order. Therefore R is refleive nd trnsitive.. Ans: (c) Sol: If S is ny set of positive integers, then R is R prtil order on S. [A; R] is poset The gretest lower bound of nd does not eist. [A ; R] is not lttice. In the poset ll the prime numbers re miniml elements of A with respect to R.. Ans: Sol: The Hsse digrm is 8 6 Number of prtil orders of type A 6 Number of prtil orders of type B Number of prtil orders of type C Number of prtil orders of type D 6 Number of prtil orders of type Required number of prtil orders Ans: Sol: The prtil orders re R {(, ), (, ), (, )} R {(, ), (, ), (, )} R {(, ), (, )}. Ans: () Sol: D {,,,, 6, 8,, } The Hsse digrm of the poset is given below Number of edges 8 6. Ans: 9 Sol: The possible Hsse digrms re B C A D S : Number of edges in the Hsse digrm is. S is true S : If nd b re complement of ech other, then Join of nd b I nd Meet of nd b O

42 : : CSIT Postl Coching Solutions Here Join of nd I Complement of S : Join of nd 8 I Meet of nd 8 O Complement of 8 S : Join of nd 6 I Meet of nd 6 O Complement of 6. Ans: (b) Sol: S {,, 9, 7, 8, } (S, *) is totlly ordered set. A totlly ordered set is lwys distributive lttice but not Boolen lgebr 6. Ans: (c) Sol: The Hsse digrm of the poset is 8 6 The poset is lttice, becuse for every pir of elements lub nd glb eist. 7. Ans: () Sol: We hve 8.7. A product of distinct prime numbers [D 8 ; ] is Boolen lgebr 8. Ans: (c) Sol: The join of d nd g h Upper bound. The meet of d nd g c Lower bound Complement of d g 9. Ans: (d) Sol: The Hsse digrm is 8 6 If we delete the element, the resulting sublttice is not distributive. The given lttice is not distributive. For the element, there is no complement. The given lttice is not complemented lttice.. Ans: 7 Sol: The Hsse digrm is shown below. D 6 {,,,,, 6,,,,, 6} 6 6 Number of edges in the digrm 7

43 . Ans: Sol: Number of functions possible on A Number of functions which re either - or on-to + Required number of functions. Ans: (c) Sol: Let f (,b) f (c,d) (+b, b) (c+d, c d) +b c+d () b c d..() Adding () nd (), we get c Subtrcting () from (), we get b d (,b) (c,d) f is one-to-one Let f (,b) (c, d) (+b, b) (c,d) +b c nd b d c d nd b For ech (c,d) B c d We hve, c d c d (,b), B such tht f(,b) (c,d) f is on-to. Hence, f is bijection. : : Discrete Mthemtics. Ans: Sol: The only equivlence reltion on A which is lso surjection is the digonl reltion on A.. Ans: (b) Sol: Given tht f f we hve (f o f ) (f o f) ( f f ) f(f()) f(+k) + k(+k) + k + k k. Ans: () Sol: (f o g) () f{g()} f And (g o f) g{f()} g But (f o g) I B nd (g o f) I A (f o g) (g o f) 6. Ans: (d) Sol: () Let (g o f) (g o f) b g{f()} g(f(b)} f() f(b) ( g is ) b ( f is ) (g o f) is

44 : : CSIT Postl Coching Solutions (b) Let c C Since g is on-to, then eists n element bb, such tht g(b) c (i) Since f is on-to, there eists n element A, such tht f() b..(ii) From (i) nd (ii), we hve To ech element cc, there eists n element A, such tht (g o f) c (g o f) is on-to (c) If (g o f) is, then f is lso. Otherwise g is not function. (d) If (gof) is on-to then f need not be on-to. A f B g C (d) The set {, 8} is closed with respect to 8. Therefore, it is sub group. 8. Ans: Sol: Let e be the identity element. * e A e e Let b inverse of b * e b b Inverse of b c b c b Here (gof) is on-to, but f is not on-to. 7. Ans: (c) Sol: () We hve, 9 7 Identity element Inverse of 7 (b) The set {,, 7} is closed with respect to 9. Therefore it is sub group of G. (c) We hve, Identity element Inverse of 8 9. Ans: (c) Sol: Order of the cyclic group Number of genertors in G () Number of positive integers which re less thn nd coprime to. Ans: (c) Sol: () The set {, } is closed w.r.t. the given binry opertion. It is subgroup of G (b) The set {, } is closed w.r.t. the given binry opertion. It is subgroup of G.

45 : : Discrete Mthemtics (c) The set {, } is not closed w.r.t. the given binry opertion. The set is not subgroup of G. (d) The set {, } is closed w.r.t. the given binry opertion. It is sub group of G.. Ans: (c) Sol: We hve (A*B) P(S) A,B P(S) * is closed opertion. The symmetric difference opertion * is ssocitive. The empty set P(S) nd is identity element. We hve A - A A P(S) (P(S),*) is group.. Ans: (c) Sol: From the tble b * b b b is identity element The second row is b c nd The second column is b c Now c cnnot pper in the first row nd rd column. * c nd * c b The first row is c b Third row is b c. Ans: (c) Sol: {,} is closed with respect to * {,} is group.. Ans: (c) Sol: (*b) (*b)*(*b) *(b*b)* (*)*(b*b) ( G is belin) * b. Ans: () Sol: Let e be the identity element * e + e e e( ) e 6. Ans: (b) Sol: * e + 7. Ans: () Sol: () 7 8. Ans: (b) Sol: We hve inverse of 7 Q {} (e is identity element) * b g.c.d of {, b} D, bd * is closed opertion on D * is ssocitive on D

46 : : CSIT Postl Coching Solutions We hve *. Ans: (d) The identity element The inverse of ny element in D ecept does not eist. (D ;*) is monoid but not group. 9. Ans: (c) Sol: Let e be the identity element. * e R + e + e We hve, * e + + ( ) Inverse of ( ) 6 Sol: () sum nd product of ny two even numbers re lso even. Further, ddition nd multipliction opertions re ssocitive. (A,.) nd (A, +) re semi-groups. Sum of odd numbers is lwys even. + is not closed opertion on B. (B, +) is not semi-group.. Ans: () Sol: f( + y) +y. y f(). f(y) f is homomorphism. Further f() is bijection. f is n isomorphism.. Ans: (d) Sol: We hve * y y N * is closed opertion (*b)*c b b c c bc *(b*c) * b c c b * is not ssocitive on N. Hence, (N, *) is not semi-group.. Ans: (d) Sol: The identity element 6 We hve, 8 6 is genertor. Ans: (c) Sol: Four numbers cn be selected out of in C wys. : vent tht the four numbers re consecutive. Fvourble cses to : (,,), (,,,), (,,,)...(7,8,9,) whose number is 7 P(). Probbility 7 C 7 Required probbility P P()

47 : 6 : Discrete Mthemtics Ans: (c) Sol: The smple spce is squre whose sides re unit segments of the coordinte es. The figure whose set of points correspond to the outcomes fvourble to the event y is bounded by the grphs of the function nd y, y nd is shown below. Required Probbility re of the shded region d. Ans: (c) C(, ) O(, ) Sol: Let S. Then either P, Q, or P, Q, or P, Q or P, Q. Out of the bove four cses, three cses re fvourble to the event P Q. B(, ) A(, ) The required probbility. Ans: (b) Sol: Let A The event tht ppers in first throw B The event tht sum is 6 The cses fvourble to B re {(,, 6), (, 6, ), (6,, ), (, 6, 6), (6,, 6), (6, 6, )} A B {(,, 6), (, 6, )} Required probbility P(A B) n A B n B 6. Ans: () Sol: The totl number of five digit numbers formed by,,, nd (without repetition) A number is divisible by if the lst two digit number (i.e., tens nd unit plce) is divisible. The lst two digit number must be:,, nd ( cses). With lst two digits fied, the other three plces cn be rrnged in (6) wys. The number of fvourble cses Probbility

48 : 7 : CSIT Postl Coching Solutions 6. Ans: (c) b Sol: Let A where, b, c nd d c d cn tke vlues or. Totl number of such mtrices 6 Let be the event tht A is non singulr. det A. i.e., tlest one of the two numbers & d is zero or tlest one of the two numbers b & c is zero. The mtrices whose determinnts re non zero re:,, 8. Ans:. Sol: If A nd B re independent then P(A B) P(A). P(B).6... () By Addition theorem of probbility P(A B) P(A) + P(B) P(A B).6 P(A) + P(B).6 P(A) + P(B).8... () From () & (), we get P(A) P(B). 9. Ans: () Sol: We hve 8 P(A) P(B) P(C) 6,, P(A B) P(A C) P(B C) 9 6 P() Ans: (d) Sol: Totl number of tringles tht cn be formed by using the vertices of regulr hegon 6 C. Among these, there re only two equilterl tringles. Required probbility Thus P(A B) P(A) P(B) P(A C) P(A)P(C) P(B C) P(B)P(C) which indictes tht A, B, nd C re pir wise independent. However, since the sum of two numbers is even, {A B C) nd P(A) P(B) P(C) 8