CS 70 July 15 2020 Counting 2 1 Combinatorial Theorem and Exclusion Simple Inclusion 2 and Exclusion Inclusion 3 4 Derangement's Sampling 5 and 6 Star Bars e nu 2 nm no t T XZ 2mg The number of subsets LHS 2 X 2 as of Is g n RHS or n 0 2 or or I subsets iz y F b 2 nu e Nnn 2 no Y e
1 Combinatorial Theorem xn y 7 Mlb Inn yn Hey a distribute my ball among I red to How and y blue bins bins XLxty LHS acxey X City w make subsequent choices Hey'T possibilities f h I 20 RHS inred.ae rni u i i i fin r.edeuef Enix Hy Indy xn X h yn e 7 x x y no e i.si xn Eno City E 2n i II i g
2 Simple Inclusion 1 Enclusion for disjoint sets A and B Rule sum 1AAB led to count the number of elements of A V B IA UBI Alt 1B I we have when A and common element B have Enclusion Rude e Inclusion k IA ABI 1AvB A AB A um i'It'm II in I IAnBl are counted twice subtract IA ABI
Example How many lb digit Phone numbers have first or second 5 as their digit stood first digit 5 A numbers with as as 9 second digit 1131 10 with 5 numbers B as g lo 5 A 5 B with 5 first and second digit AAB numbers as 108 IAAB 1 Is 18 18 108 IAU Bls IAI 11131 IA Three way inclusion Exclusion Rule It Set Beg LAU BU GI IAI 1431 14 IAABI land Brel Ian 1 Brett HAD
Exclusion principle 3 Inclusion Ah Ai sets off Air Ajit I V i Ait flail IA ah Ain l L yn E 2 Hiv to n 4 Derangement see Permutations of l n n no item in its proper How many permutations where fixed Points place or Derangement Example number of derangement 123 If Tooo Derangement Yes 231
count permutations we can count the complement.ro at least fined Points with one I is fined Points Permutation where 2 slg 2,3 least with Complement Permutations at one fined Points 1A VA zu Ast IA UAzUA3l IA in Az I IA il 11A de IA 31 f IAN Aste IA A Azn Ag 1AM Azl 2 2 I I I 2 I t 4 this from the total Permutations Subtract 4 2 Derangement 3 n items For h permutations Permutations with at least one fined point nan I him YELII Efi of Aintidt th cn u Gynt Inn o 7 Ch y
t 4 In 21 of derangement h 7 Cn D e C un nu m I h un 1 e ye eosin n xd Y LT n xte x Eo t.fi or37n h W n a Roughly 0 37 of the Permutations are derangements 5 Sampling out of Assume sample K items n as Without replacement o order matters Kel nx Ch y x n 2 Ah ht a Ch K order does not matter divid by number of orders Second rule k nd K h K
with replacement K order matters n h X h X x n second rule o orderamatters can use we doesn't a of on how many depends Problem each item choose we orderings 4 chosen string ABCD For IAC D ordering 4z n n a ordered elements Different number of each unordered to map Another example can Alice and Bob 5 split How many ways of 5 dollars Pick Alice or Bob For each divid out order 25 and
1 CA A A A A A 5 B D i I E 1B Aga A A B A 4 5 B I gLAgAgAyBgA li is i.i A I 5 13 45 Ag BoB B B g B S A O t B B B B B second of counting is rule no good here Another example can Alice Bob and Eve split How ma 5 ways Alice's dollars from Bob's Idea separat and then Bob's from Eve's Assume dollars 5 are stars.o MAYBE see IfI YI II's.IR manddbajars Bob 3 Eve 3 split Alice D ftp.ff
Zeroth Rule Counting there If is a set they have between one to two one mapping the Same size How many different ask So can we and of 5 stars bars sequence are 2 00000 there µ A A a a a x a e Ig Positio in which 7 positions to Place bars 2 7 to do this 7 choose 2 e way z I to split 5 among 3 People ways 6 Star and Bars k dollars among n people split to ways replacement from where K with n order doesn't matter split to the K correspondence bars 1 n I I I I stars
from which to Positions choose 1 nyk bar positions n 1 ni D
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