Permutations and Combinations Rosen 4.3. Permutations A permutation of a set of distinct objects is an ordered arrangement these objects. An ordered arrangement.

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Permutations and Combinations Rosen 4.3

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…

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)

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).

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)!

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)

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

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

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

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

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

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

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)!

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.

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:

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.

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.

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

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

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

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