Download presentation

Presentation is loading. Please wait.

Published byErnest Madren Modified over 3 years ago

1
Permutations and Combinations Rosen 4.3

2
Permutations A permutation of a set of distinct objects is an ordered arrangement these objects. An ordered arrangement of r elements of a set is called an r-permutation. The number of r-permutations of a set with n elements is denoted by P(n,r). A = {1,2,3,4} 2-permutations of A include 1,2; 2,1; 1,3; 2,3; etc…

3
Counting Permutations Using the product rule we can find P(n,r) = n*(n-1)*(n-2)* …*(n-r+1) = n!/(n-r)! How many 2-permutations are there for the set {1,2,3,4}? P(4,2)

4
Combinations An r-combination of elements of a set is an unordered selection of r element from the set. (i.e., an r-combination is simply a subset of the set with r elements). Let A={1,2,3,4} 3-combinations of A are {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}(same as {3,2,4}) The number of r-combinations of a set with n distinct elements is denoted by C(n,r).

5
Example Let A = {1,2,3} 2-permutations of A are: 1,2 2,1 1,3 3,1 2,3 3,2 6 total.Order is important 2-combinations of A are: {1,2}, {1,3}, {2,3} 3 total. Order is not important If we counted the number of permutations of each 2- combination we could figure out P(3,2)!

6
How to compute C(n,r) To find P(n,r), we could first find C(n,r), then order each subset of r elements to count the number of different orderings. P(n,r) = C(n,r)P(r,r). So C(n,r) = P(n,r) / P(r,r)

7
A club has 25 members. How many ways are there to choose four members of the club to serve on an executive committee? –Order not important –C(25,4) = 25!/21!4! = 25*24*23*22/4*3*2*1 =25*23*22 = 12,650 How many ways are there to choose a president, vice president, secretary, and treasurer of the club? –Order is important –P(25,4) = 25!/21! = 303,600

8
The English alphabet contains 21 consonants and 5 vowels. How many strings of six lower case letters of the English alphabet contain: exactly one vowel? exactly 2 vowels at least 1 vowel at least 2 vowels

9
The English alphabet contains 21 consonants and 5 vowels. How many strings of six lower case letters of the English alphabet contain: exactly one vowel? Note that strings can have repeated letters! We need to choose the position for the vowel C(6,1) = 6!/1!5! This can be done 6 ways. Choose which vowel to use. This can be done in 5 ways. Each of the other 5 positions can contain any of the 21 consonants (not distinct). There are 21 5 ways to fill the rest of the string. 6*5*21 5

10
The English alphabet contains 21 consonants and 5 vowels. How many strings of six lower case letters of the English alphabet contain: exactly 2 vowels? Choose position for the vowels. C(6,2) = 6!/2!4! = 15 Choose the two vowels. 5 choices for each of 2 positions = 5 2 Each of the other 4 positions can contain any of 21 consonants. 21 4 15*5 2 *21 4

11
The English alphabet contains 21 consonants and 5 vowels. How many strings of six lower case letters of the English alphabet contain: at least 1 vowel Count the number of strings with no vowels and subtract this from the total number of strings. 26 6 - 21 6

12
The English alphabet contains 21 consonants and 5 vowels. How many strings of six lower case letters of the English alphabet contain: at least 2 vowels Compute total number of strings and subtract number of strings with no vowels and the number of strings with exactly 1 vowel. 26 6 - 21 6 - 6*5*21 5

13
Corollary 1: Let n and r be nonnegative integers with r n. Then C(n,r) = C(n,n-r) Proof: C(n,r) = n!/r!(n-r)! C(n,n-r) = n!/(n-r)!(n-(n-r))! = n!/r!(n-r)!

14
Binomial Coefficient Another notation for C(n,r) is. This number is also called a binomial coefficient. These numbers occur as coefficients in the expansions of powers of binomial expressions such as (a+b) n.

15
Pascal’s Identity Let n and k be positive integers with n k. Then C(n+1,k) = C(n, k-1) + C(n,k). Proof:

16
Let n be a positive integer. Then Proof: We know from set theory that the number of subsets in a set of size n is 2 n. We also know that C(n,k) is the number of subsets of a set of size n that are of size k. counts the number of subsets of every size from 0 (empty set) to n. Therefore the sum must add up to 2 n.

17
Vandermonde’s Identity Proof: Suppose there are n items in one set and m items in a second set. Then the total number of ways to pick r elements from the union of these sets is C(m+n,r). Another way to pick r elements from the union is to pick k elements from the first set and then r-k elements from the second set, where 0 k r. There are C(n,k) ways to pick the k elements from the first set and C(m,r-k) ways to pick the rest of the elements from the second set.

18
Proof: Suppose there are n items in one set and m items in a second set. Then the total number of ways to pick r elements from the union of these sets is C(m+n,r). Another way to pick r elements from the union is to pick k elements from the first set and then r-k elements from the second set, where 0 k r. For any k,there are C(n,k) ways to pick the k elements from the first set and C(m,r-k) ways to pick the rest of the elements from the second set. By the product rule there are C(m,r-k)C(n,k) ways to pick r elements for a particular k. For all possible values of k

19
Pascal’s Triangle 1 1 1 1 12 33 1 1 14641 n’th row, C nk =k = 0, 1, …, n

20
Binomial Theorem Let x and y be variables and let n be a positive integer. Then

Similar presentations

OK

1 CS 140 Discrete Mathematics Combinatorics And Review Notes.

1 CS 140 Discrete Mathematics Combinatorics And Review Notes.

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on human nutrition and digestion article Ppt on causes and effects of revolt of 1857 Ppt on surface water pump Ppt on review of related literature search Ppt on area of equilateral triangle Ppt on personality development and motivation Ppt on natural resources for class 9 Ppt on non ferrous minerals technologies Ppt on selling techniques Ppt on miniature circuit breaker