MATH10101 2018/19: poblems fo supevisio i week 08 Q1. Let A be a set. SOLUTIONS (i Pove that the fuctio c: P(A P(A, defied by c(x A \ X, is bijective. (ii Let ow A be fiite, A. Use (i to show that fo each {0, 1,..., }, the estictio of c oto ( the( subset P (A gives a bijectio betwee P (A ad P (A. Hece coclude that without usig the factoial fomula fo the biomial coefficiet. (iii Now check that, usig the factoial fomula. Q1 - solutio. (i Obseve that c(c(x A \ (A \ X X fo all subsets X of A. Thus, c c Id P(A. This meas that the fuctio c is ivetible, ad its ivese is c itself. Recall that a fuctio is ivetible, if ad oly if it is a bijectio. Hece c is a bijectio. (ii Let A. Note that fo a subset X of A, X if ad oly if c(x. Hece the estictio of c oto P (A maps P (A to P (A, ad its ivese, which is the estictio of c oto P (A, maps P (A to P (A. We theefoe have mutually ivese fuctios betwee P (A ad ( P (A. ( Hece these sets ae i bijectio, implyig P (A P (A, which ewites as. This agumet uses oly the defiitio of a biomial coefficiet, ot its factoial fom. (! (iii The factoial fomula gives! (!, which is clealy equal to! (!!, the factoial expessio fo. Q2. (i Pove, usig the Biomial Theoem, that fo all Z, 2. (ii Now pove the same statemet without usig the Biomial Theoem by cosideig a set A with A ad calculatig the cadiality of P (A. (iii Check that the statemet i (i is tue fo by diect evaluatio of both sides. Q2 - solutio. (i Note that Theoem. This is the same as 2. 1 1 equals (1+1 by the Biomial (ii P (A is the collectio of all subsets of A, i.e. P (A. It is a disjoit uio. Recall fom the lectues that the cadiality of a disjoit uio is the sum of the cadialities.
Aothe esult fom the lectues is hece 2 P (A P (A 2, P (A by defiitio of the biomial coefficiets. ( (iii, 1 + + 10 + 10 + + 1 32 2. Q3.. Usig the Biomial Theoem, pove that with Q2 to evaluate the sums eve ad 1 odd, ( 1 0. Use this esult alog. Check that both of you aswes ae coect fo 4 by diect calculatio (this must be pat of you witte aswe! Q3 - solutio. The Biomial Theoem states that fo all x, y we have (x + y x y. Choose x 1 ad y 1 to get as equied. Add Subtact to get 0 ( 1 + 1 ( 1 2, fom Q2, to ( ( 1 0 (just poved to get 2 2 (1 + ( 1 (1 ( 1 eve 2 2 odd Hece the biomial umbes fo eve add up to 2 1, ad the biomial umbes fo odd also add up to 2 1. ( 4 4 4 4 4 Example: 4, + + 1 + + 1 8, + 4 + 4 8. 0 2 4 1 3..
100 ( 100 Q4. Use the Biomial Theoem to calculate 3 2 100 2. [Aswe:.8 100 ] Q4 - solutio. 3 2 2 1 3 2 2 2 1 9 2, which by the Biomial Theoem equals 1 34 (9 + 2. It emais to put 100. Q. Let A be a fiite set ad let Q(A {(C, D P(A P(A : C D}. Pove that Q(A 3 A. Q - solutio. Solutio 1: Let A ad let us cout pais (C, D of subsets ( of A whee C D ad additioally D. A set D with D ca be chose i ways, ad to each such choice of ( D thee coespods 2 choices of C, sice C is a abitay subset of D. Hece thee ae 2 pais (C, D P(A P(A such that C D ad D. The total umbe of elemets of Q(A is obtaied by summig ove all possible : that is, Q(A 2 (1 + 2 3. Solutio 2: Give a pai (C, D of subsets of A such that C D, defie the fuctio f : A {0, 1, 2} by 0, a / D, f(a 1, a D \ C, 2, a C. Ifomally, fo a A, f(a is the umbe of eties out of two eties of the pai (C, D which cotai a. It tus out that each fuctio f Fu(A, {0, 1, 2} coespods to exactly oe pai (C, D Q(A, amely C {a A : f(a 2} ad D {a A : f(a 1}. We have thus costucted a bijectio betwee Q(A ad the set Fu(A, {0, 1, 2}. The latte set has cadiality 3 A (a esult fom the couse, hece so does Q(A. Q.. Fid the quotiet q ad emaide o dividig the followig umbes by 17: (i 1; (ii 1; (iii 100; (iv 100.
Q - solutio. (i 1 17 0 + 1 ad 0 1 < 17, so q 0, 1; (ii 1 17 ( 1 + 1 ad 0 1 < 17, so q 1, 1; (iii 100 17 + 1 ad 0 1 < 17, so q, 1; (iv 100 17 ( + 2 ad 0 2 < 17, so q, 2; Recall that by the Divisio Theoem, the aswe i each case is uique. Impotatly, the emaides ae always o-egative. Q7. (i Usig oly the defiitio of divides,, caefully pove the statemet a, b, q, d Z, ( d a d b ( d b d (a bq. (ii Dispove the followig false statemet: a, b, q, d Z, ( d a d b ( d a d (a bq. Q7 - solutio. (i Let a, b, q, d be abitay iteges. To pove the implicatio, we assume that d a ad d b. By defiitio of, this meas that a md ad b d fo some iteges m,. The a bq (m qd whee m q is a itege, hece by defiitio of agai, d (a bq. Also, d b by assumptio. We have poved that d b ad d (a bq, as equied. We ow pove the implicatio. Assume that d b ad d (a bq, meaig b d ad a bq sd fo some iteges, s. The a a bq + bq (s + qd, ad sice s + q is a itege, this meas that d a. Also, d b by assumptio. We have poved that d a ad d b as equied. (ii To dispove the uivesal statemet, we use a couteexample. Let a 0, b 1234, q 0 ad d 0. The the statemet d a ad d b is false as 0 does ot divide 1234, but the statemet d a ad d (a bq is tue, sice 0 0. Hece the equivalece,, does ot hold fo these a, b, q, d. Examples whee oe of the iteges is 0 ae possible (but ot ecessay.
MATH10101 2018/19: optioal execises o biomial coefficiets SOLUTIONS Opt1. I a cad game you ae dealt a had of cads fom a omal playig deck of 2 cads. i How may diffeet hads ae possible? ii How may hads will cotai all fou aces? iii How may hads will cotaio heats? iv How may hads will cotai at least oe spade? ( 2 Opt1 - solutio. (i ; ( 48 (ii (a had cotais 4 aces ad a futhe 9 cads fom a set of 2 4 48 cads; 9 ( 39 (iii (-elemet subsets of the set of 39 cads which ae ot heats; ( 2 39 39 (iv (sice thee ae hads which cotaio spades. NB: ( thee is ot much poit i tyig to wite these umbes as decimals (but fo efeece, 2 30900 > 10 11. Opt2. Expad (4x 3y. Opt2 - solutio. (4x 3y ( ( (4x + (4x 4 ( 3y + (4x 3 ( 3y 2 1 2 ( ( + (4x 2 ( 3y 3 + (4x ( 3y 4 + ( 3y 3 4 1024x 3840x 4 y + 70x 3 y 2 4320x 2 y 3 + 120xy 4 243y Opt3. Use the Biomial Theoem to calculate 3! (!.
Opt3 - solutio. 1! (3 + 8!. 3!(! 1! Opt4. Calculate: (i 2 0 + 2 1 + + 0 1 (ii ( 2 0 + ( 2 1 + + 0 1 ( 2 ; ( ( 2. 3!!(! 1! 3 Opt4 - solutio. (i The expessio is a paticula case of the expessio a b 0 + ( ( 0 a 1 b 1 + + a 0 b which by the Biomial Theoem is equal to (a + b. Namely, 1 to obtai the sum i (i, we put, a 1 ad b 2 1. By the Biomial Theoem, the sum equals (1 + 2 1 (3/2 729/4. (ii Similaly we put, a 1 ad b 2 i the Biomial Theoem. The sum is equal to (1 + ( 2 ( 1 1. Opt. Fid x > 0 that satisfy 4 ( 4 (i x 2 4 ; (ii x 2 3 ( 3 3. Opt - solutio. i ii 4 ( 4 4 (1 + 4 4 2 2, so x 2. 3 ( 3 3 (1 + 3 3 4 3 8 2, so x 8. Opt. Use the factoial fomula fo the biomial coefficiet to pove that 1 1 fo all 1.
Opt - solutio.!! (!! ( 1! (!! ( 1! (! ( ( 1! 1 ( 1! (( 1 ( 1!. 1 Opt7. Fid the umbe of subsets of {1, 2,..., 10} which cotai 1 ad do ot cotai 10. Opt7 - solutio. Fist of all, subsets which do ot cotai 10 ae all subsets of {1, 2,..., 9}. Thee ae 2 9 12 of these. But we eed to cout oly subsets of {1, 2,..., 9} that cotai 1. If X is a subset of {1, 2,..., 9} such that X cotais 1, the X \ {1} is a subset of {2,..., 9}. Vice vesa, if Y is a abitay subset of {2,..., 9}, the X {1} Y is a subset of {1, 2,..., 9} which cotais 1. Hece we established a bijective coespodece betwee subsets of {1, 2,..., 9} which cotai 1 ad all subsets of {2,..., 9}. Hece the umbe of subsets of {1, 2,..., 9} which cotai 1 is equal to the umbe of all subsets of {2,..., 9}. This umbe is P({2,..., 9} 2 {2,...,9} 2 8 2.