Basics of Counting. A note on combinations. Recap. 22C:19, Chapter 6.5, 6.7 Hantao Zhang

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1 Bscs of Countng 22C:9, Chpter 6.5, 6.7 Hnto Zhng A note on comntons An lterntve (nd more common) wy to denote n r-comnton: n n C ( n, r) r I ll use C(n,r) whenever possle, s t s eser to wrte n PowerPont 2 Recp r-permuttons: The numer of wys n whch we cn drw r lls from collecton of n dfferent lls, where the order s mportnt: P(n,r) n! / (n-r)! r-comntons: The numer of wys n whch h we cn drw r lls from collecton of n dfferent lls, where we do not cre out the orderng: C(n,r) n! / r! (n-r)! Tody: we study countng prolems where repettons re llowed,.e. t s possle tht the sme ll s drwn multple tmes.

2 n lls Comntons wthout repetton dfferent ecuse the slots hve lels (dstngushle) r oxes n wys n- wys n-r wys 2 r X X X X 2 r r-permutton wthout repetton (order mportnt) r-comnton wthout repetton (order not mportnt) the lls re not replced when they hve een drwn the sme ecuse the slots hve no lels (ndstngushle) we over-counted y r! Comntons wth repetton r oxes n lls dfferent ecuse the slots hve lels (dstngushle) n wys n wys n wys 2 r 2 r r-permutton wth repetton (order mportnt) r-comnton wth repetton (order not mportnt) the lls re replced when they hve een drwn. Or there s very lrge stc of ndstngushle lls of ech color. the sme ecuse the slots hve no lels (ndstngushle) ths one s trcy! Comntons wth repetton Exmple: We wnt to drw 2 peces of frut from owl tht contns 2 pples, 2 pers, nd 2 ornges. In how mny wys cn we do ths such tht: ) The fruts re numered nd the order mtters. 2) The fruts re numered nd the order does not mtter. 3) The fruts re ndstngushle ut the order mtters. 4) The fruts re ndstngushle nd the order does not mtter. ) 6 * 5 30 P(6,2) 2) 6 * 5 / 2 5 C(6,2) 3) Now there re 3 nds of frut tht we drw wth replcement (snce there re enough of ech nd to e le to pc ny frut t ny drw). Ths s true ecuse drwng pple s no dfferent thn drwng pple 2. It s le there re copes of the sme pple present. Thus: ( ), ( p), ( o), (p ), (p p), (p o), (o ), (o p), (o o). 4) Snce the order doesn t mtter ( p) (p ), ( o)(o ), (p o)(o p). we over-counted 3 peces:

3 Comntons wth repetton How to count the ltter (r-comnton wth repetton)? One strtegy could e to strt from drwng where the order mtters nd try to count the numer of wys we over-counted (lst exmple). However, there s much smrter wy! Three nds of fruts re seprted y two dvders nd two chosen fruts re held y two oxes: We choose two tems to fll the oxes. pples pers ornges ( ) XX, ( p) X X, ( o) X X, (p p) XX, (p o) X X, (o o) XX X X The totl numer of X nd s 4. The numer of wys choosng 2 postons out of 4 s C(4,2) 6. We my lso use 0 for X nd for. Comntons wth repetton How to count the ltter (r-comnton wth repetton)? The prevous exmple cn e generlzed: Choose r lls out of g of lls of n dfferent colors needs (n-) dvders ( ) nd r oxes ( X ). The totl numer of nd X s (nr-). The numer of wys choosng r postons out of (nr-) postons s C(nr-,r) (nr-)! /(r! (n-)!) n 6, r 4: 5 dvders nd 4 oxes (or lls) lls ecome ndstngushle C(9,4) t-strngs! More exmples ) How mny wys re there to select fve lls from csh ox contnng mny \$, \$2, \$5, \$0, \$20, \$50 nd \$00 lls, such tht the lls of the sme vlue re ndstngushle nd the order n whch they re selected s unmportnt. Ths s le drwng colored lls wth replcement. The colors correspond to the vlues. Snce the order doesn t mtter we hve: C(75-,5)462 2) A cooe shop hs 4 nds of cooes nd we wnt to pc 6. We don t cre out the order nd cooes from one nd re ndstngushle. Agn, drwng colored lls wth replcement: colors re nd of cooes. C(64-,6)84. 3

4 Comntons wth repetton n s numer of dstnct clsses of ojects n the orgnl g! - r-permutton wthout repetton - order mtters (r dstngushle slots) - wthout replcement (n dstngushle ojects) n! / (n-r)! - r-comnton wthout repetton - order does not mtter (r ndstngushle slots) - wthout replcement (n dstngushle ojects) n! / r! (n-r)! - r-permutton wth repetton - order mtters (r dstngushle slots) - wth replcement (n dstnct clsses of ndstngushle ojects) n r - r-comnton wth repetton - order does not mtter (r ndstngushle slots) - wth replcement (n dstnct clsses of ndstngushle ojects) (nr-)! / r! (n-)! Another exmple How mny dfferent non-negtve nteger solutons for the vrles x, x2, x3, x4 wth x x2 x3 x4 0? A soluton le x, x2 0, x3 4, x4 5 dvdes 0 nto four prts: (, 0, 4, 5) or X XXXX XXXXX. We need three dvders ( ) to dvde 0 oxes ( X ) nto four prts. The numer of wys of choosng three slots out of 03 slots s C(04-, 3) C(3, 3) 286. Another exmple How mny dfferent nteger trples (, j, ) where < < j < < n? Let n 5. A trple le (3, 3, 5) my e vewed s XX X whch hve three oxes ( X ) holdng ojects from fve clsses ( thru 5) nd we need four dvders to seprte these fve clsses. In generl, we need n- dvders ( ) to dvde three oxes ( X ) nto n prts. The numer of wys of choosng three slots out of (n-)3 slots s C(n2, 3) (n2)(n)n/3. 2 4

5 Another exmple How mny tmes wll the code B n the followng progrm e executed? For : to n For j : to For : to j B(,j,) end for end for end for Answer: C(n2, 3) (n2)(n)n/3. 3 Boo-Shelve Prolems In how mny wys cn you put n dfferent oos on dfferent shelves? (shelves cn hold ll oos). Soluton: Plce oos one y one. Frst oo: on shelves. Second oo: to the left or rght of exstng oo, or on empty shelve (2- ). Thrd oo: Two cses: () The frst two oos on one shelve: 3 wys on the sme shelf nd - wys on n empty shelve. 3(-) 2. () The frst two oos not one shelve: 2 wys wth oo nd two wys wth oo2, (-2) wys on n empty shelve. 22(-2) 2. Fourth oo: 3 wys. Totl: ()(2)... (n-) (n-)! / (-)! Boo-Shelve Prolems In how mny wys cn you put n dfferent oos on dfferent shelves? (shelves cn hold ll oos). Soluton 2: There re n! wys to put them nto sequence. For ech sequence, we need to cut the sequence nto susequences usng - dvders. how mny t-strngs re there wth - (dvders) nd n X (oos): C(n-,-). Totl: C(n-,-) n! (n-)!/(-)! 5

6 Susets If ll the elements of set S re lso elements of set T, then S s suset of T For exmple, f S {2, 4, 6} nd T {, 2, 3, 4, 5, 6, 7}, then S s suset of T Ths s specfed y S T Or y {2, 4, 6} {,2,3,4,5,6,7} If S s not suset of T, t s wrtten s such: S T For exmple, {, 2, 8} {,2,3,4,5,6,7} 6 Susets Any set s suset of tself! Gven set S {2, 4, 6}, snce ll the elements of S re elements of S, S s suset of tself Ths s nd of le syng 5 s less thn or equl to 5 Thus, for ny set S, S S The empty set s suset of ll sets (ncludng tself!) Recll tht ll sets re susets of themselves All sets re susets of the unversl set. 7 Proper Susets If S s suset of T, nd S s not equl to T, then S s proper suset of T LetT{0,,2,3,4,5} If S {, 2, 3}, S s not equl to T, nd S s suset of T A proper suset s wrtten s S T Let R {0,, 2, 3, 4, 5}. R s equl to T, nd thus s suset (ut not proper suset) or T Cn e wrtten s: R T nd R T(orjustRT) Let Q {4, 5, 6}. Q s nether suset or T nor proper suset of T 8 6

7 Set crdnlty The crdnlty of set s the numer of elements n set Wrtten s A Exmples Let R {, 2, 3, 4, 5}. Then R 5 0 Let S {, {}, {}, {, }}. Then S 4 Ths s the sme notton used for vector length n geometry A set wth one element s sometmes clled sngleton set 9 Power sets Gven the set S {0, }. Wht re ll the possle susets of S? Theyre: (s t s suset of ll sets), {0}, {}, nd {0, } The power set of S (wrtten s P(S)) s the set of ll the susets of S P(S){, {0}, {}, {0,} } Theorem: If S n, then P(S) 2 n 20 Tuples In 2-dmensonl spce, t s (x, y) pr of numers to specfy locton In 3-dmensonl (,2,3) s not the sme s (3,2,) spce, t s (x, y, z) trple of numers In n-dmensonl spce, t s y n-tuple of numers (2,3) Two-dmensonl spce uses prs, or 2-tuples Three-dmensonl spce uses trples, or 3-tuples Note tht these tuples re ordered, unle sets thexvlue hs to come frst 2 x 7

8 Crtesn products A Crtesn product s set of ll ordered 2- tuples where ech prt s from gven set Denoted y A x B, nd uses prenthess (not curly rcets) For exmple, 2-D Crtesn coordntes re the set of ll ordered prs Z x Z Recll Z s the set of ll ntegers Ths s ll the possle coordntes n 2-D spce Exmple: Gven A {, } nd B { 0, }, wht s ther Crtesn product? C A x B { (,0), (,), (,0), (,) } 22 Crtesn products Note tht Crtesn products hve only 2 prts n these exmples (lter exmples hve more prts) Forml defnton of Crtesn product: AxB{(,) A nd B} Theorem: A x B A B. 23 Crtesn products All the possle grdes n ths clss wll e Crtesn product of the set S of ll the students n ths clss nd the set G of ll possle grdes Let S { Alce, Bo, Chrs } nd G { A, B, C } D { (Alce, A), (Alce, B), (Alce, C), (Bo, A), (Bo, B), (Bo, C), (Chrs, A), (Chrs, B), (Chrs, C) } The fnl grdes wll e suset of ths: { (Alce, C), (Bo, B), (Chrs, A) } Such suset of Crtesn product s clled relton (more on ths lter n the course) 24 8

9 Crtesn products There cn e Crtesn products on more thn two sets A 3-D coordnte s n element from the Crtesn product of Z x Z x Z 25 The Bnoml Theorem Theorem: Gven ny numers nd nd ny nonnegtve nteger n, n n ( ) 0 n. n The Bnoml Theorem Proof: Use nducton on n. Bse cse: Let n 0. Then ( ) 0 nd Therefore, the sttement s true when n 0. 9

10 0 Proof, contnued Inductve step Suppose the sttement s true when n for some 0. Then ( ) ( )( ) ( ) Proof, contnued 0 Proof, contnued Therefore, the sttement s true when n. Thus, the sttement s true for ll n 0.. 0

11 Exmple: Bnoml Theorem Expnd ( ) 8. C(8, 0) C(8, 8). C(8, ) C(8, 7) 8. C(8, 2) C(8, 6) 28. C(8, 3) C(8, 5) 56. C(8, 4) 70. Exmple: Bnoml Theorem Therefore, ( ) Exmple: Clcultng.0 8 Compute.0 8 on clcultor. Wht do you see?

12 Exmple: Clcultng.0 8 Compute.0 8 on clcultor. Wht do you see? Exmple: Clcultng ( 0.0) 8 8(0.0) 28(0.0) 2 56(0.0) 3 70(0.0) 4 56(0.0) 5 28(0.0) 0) 6 8(0.0) 0) 7 (0.0) 0) Exmple: Approxmtng (x) n Theorem: For smll vlues of x, ( x) n nx. ( x) n n 2 3 ( x) nx x x. nd so on. n( n ) 2 nx x. 2 n( n ) 2 n( n )( n 2) 6 2

13 Exmple For exmple, ( x) 8 8x 28x 2 when x s smll. Compute the vlue of ( x) 8 pproxmton when x.03. Do t gn for x nd the Expndng Trnomls Expnd ( c) 3. Expndng Trnomls Expnd ( c) 3. ( c) 3 (( ) c) 3 ( ) 3 3( ) 2 c 3( )c 2 c 3, ( ) 3( )c 3( )c 2 c 3. 3

14 Expndng Trnomls c 6c 3 2 c 3c 2 3c 2 c 3. Wht s the pttern? Expndng Trnomls ( c) 3 ( 3 3 c 3 ) 3( 2 2 c 2 2 c c 2 c 2 ) 6c. The Multnoml Theorem Theorem: In the expnson of ( ) n, the coeffcent of n 2 n 2 n s n! n! n! n! 2 4

15 Exmple: The Multnoml Theorem Expnd ( cd) 3. The terms re 3, 3, c 3, d 3, wth coeffcent 3!/3! d d d d 2 2, 2 c, 2 d, 2, 2 c, 2 d, c 2, c 2, c 2 d, d 2, d 2, cd 2, wth coeffcent 3!/(!2!) 3. c, d, cd, cd, wth coeffcent 3!/(!!!) 6. Exmple: The Multnoml Theorem Therefore, ( c d) c 3 d c 3 2 d c 3 2 d 3c 2 3c 2 3c 2 d 3d 2 3d 2 3cd 2 6c 6d 6cd 6cd. Exmple: The Multnoml Theorem Fnd( 2 ) 4. 5

16 Another Prolem If we expnd the expresson ( 2 3c) 4, wht wll e the sum of the coeffcents? 6

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