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Permutations of Selected Elements

Counting Techniques: Permutations of Selected Elements Addition Rule, Difference Rule, Inclusion/Exclusion Rule. Permutations of Selected Elements. Typical situation: A chairman, a secretary and a treasurer are to be chosen in a committee of 7 people.

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Permutations of Selected Elements

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  1. Counting Techniques: Permutations of Selected Elements Addition Rule,Difference Rule,Inclusion/Exclusion Rule

  2. Permutations of Selected Elements • Typical situation: A chairman, a secretary and a treasurer are to be chosen in a committee of 7 people. • Question: In how many different ways can it be done? • Definition: An r-permutation of a set S of n elements is an ordered selection of r elements taken from S. • The number of all r-permutations of a set of n elements is denoted P(n,r) . • In the example above we want to find P(7,3).

  3. How to compute P(n,r) • Theorem:P(n,r) = n(n-1)(n-2)…(n-r+1) or, equivalently, • Proof: Forming an r-permutation of a set of n-elements is an r-step operation:  Step 1: Choose the 1st element ( n different ways).  Step 2: Choose the 2nd element ( n-1 different ways). …  Step r: Choose the rth element (n-r+1 different ways). Based on the multiplication rule, the number of r-permutations is n∙(n-1)∙…∙(n-r+1) .

  4. Examples of r-permutations 1. Choosing a chairman, a secretary and a treasurer among 7 people: P(7,3) = 7∙6∙5 = 210 . 2. Suppose Jim is already chosen to be the secretary. Q: How many ways a chairman and a treasurer can be chosen? A: P(6,2) = 6∙5 = 30 . 3. In an instance of the Traveling Salesman Problem, the total number of cities = 10; this time the salesman is supposed to visit only 4 cities (including the home city). Q: How many different tours are possible? A: P(9,3) = 9∙8∙7 = 504 .

  5. The Addition Rule • Suppose a finite set A equals the union of k distinct mutually disjoint subsets A1, A2, …, Ak . Then n(A) = n(A1) + n(A2) + … + n(Ak) • Example: How many integers from 1 through 999 do not have any repeated digits? Solution: Let A = integers from 1 to 999 not having repeated digits. Partition A into 3 sets: A1=one-digit integers not having repeated digits; A2=two-digit integers not having repeated digits; A3=three-digit integers not having repeated digits. Then n(A)=n(A1)+n(A2)+n(A3) (by the addition rule) = 9 + 9∙9 +9∙9∙8 = 738. (by the multipl. rule)

  6. The Difference Rule • If A is a finite set and B is a subset of A, thenn(A-B) = n(A) – n(B) . • Example: Assume that any seven digits can be used to form a telephone number. Q: How many seven-digit phone numbers have at least one repeated digit? Let A = the set of all possible 7-digit phone numbers; B = the set of 7-digit numbers without repetition. Note that BA . Then A-B is the set of 7-digit numbers with repetition. n(A-B) = n(A) – n(B) (by the difference rule) = 107 – P(10,7) (by the multiplication rule) = 107 – 10! / 3! = 10,000,000 – 3,628,800/6 = = 10,000,000 – 604,800 = 9,395,200

  7. C A A B B The Inclusion/Exclusion Rule for Two or Three Sets • If A, B and C are finite sets then  n(A  B) = n(A) + n(B) – n(A  B)  n(A  B  C) = n(A) + n(B) + n(C) - n(A  B) – n(A  C) – n(B  C) + n(A  B  C)

  8. Example on Inclusion/Exclusion Rule (2 sets) • Question: How many integers from 1 through 100 are multiples of 4 or multiples of 6 ? • Solution: Let A=the set of integers from 1 through 100 which are multiples of 4; B = the set of integers from 1 through 100 which are multiples of 6. Then we want to find n(A  B). First note that A  B is the set of integers from 1 through 100 which are multiples of 12 . n(A  B) = n(A) + n(B) - n(A  B) (by incl./excl. rule) = 25 + 16 – 8 = 33 (by counting the elements of the three lists)

  9. Example on Inclusion/Exclusion Rule (3 sets) • 3 headache drugs – A,B, and C – were tested on 40 subjects. The results of tests: 23 reported relief from drug A; 18 reported relief from drug B; 31 reported relief from drug C; 11 reported relief from both drugs A and B; 19 reported relief from both drugs A and C; 14 reported relief from both drugs B and C; 37 reported relief from at least one of the drugs. Questions: 1) How many people got relief from none of the drugs? 2) How many people got relief from all 3 drugs? 3) How many people got relief from A only?

  10. C A B Example on Inclusion/Exclusion Rule (3 sets) S We are given: n(A)=23, n(B)=18, n(C)=31, n(A  B)=11, n(A  C)=19, n(B  C)=14 , n(S)=40, n(A  B  C)=37 Q1) How many people got relief from none of the drugs? By difference rule, n((A  B  C)c ) = n(S) – n(A  B  C) = 40 - 37 = 3

  11. Example on Inclusion/Exclusion Rule (3 sets) Q2)How many people got relief from all 3 drugs? By inclusion/exclusion rule: n(A  B  C) = n(A  B  C) - n(A) - n(B) - n(C) + n(A  B) + n(A  C) + n(B  C) = 37 – 23 – 18 – 31 + 11 + 19 + 14 = 9 Q3)How many people got relief from A only? n(A – (B  C)) (byinclusion/exclusion rule) = n(A) – n(A  B) - n(A  C) + n(A  B  C) = 23 – 11 – 19 + 9 = 2

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