Main Menu Main Menu (Click on the topics below) Permutation Example 1 Example 2 Circular Permutation Permuting r of n objects Example 1 Example 2 Example 3 Example 4 Addition Rule Example 1 Example 2 Difference Rule Example 1 Inclusion Exclusion Rule Example 1 Symmetric Relations Simple graphs Click on the picture

Permutations Sanjay Jain, Lecturer, School of Computing

Permutations Permutation of a set of objects is an ordering of the objects in a row. Example: {A, B, C} ABC, ACB, BAC, BCA, CAB, CBA

Permutations Theorem: Suppose a set A has n elements (where n  1). Then the number of permutations of A is n!= n*(n-1)*(n-2)*…*1. Proof: Job: select a permuation T 1 : Select the 1st element in the row ---> n ways T 2 : Select the 2nd element in the row ---> n-1 ways …….. T n : Select the nth element in the row ---> 1 way Total number of permutations: n*(n-1)*…..*1

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Examples Number of anagrams of SINGAPORE: This is same as premuting 9 distinct elements. 9!

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Examples Number of anagrams of SINGAPORE which have “SING” as a substring: We can think of “SING” as one element. Thus there are a total of 6 elements to be permuted (“SING”,A,P,O,R,E). 6!

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Examples Letters of “SING” appear together, but not necessarily in that order. T1: First permute “SING”, A, P, O, R, E T2: Permute letters of “SING”. T1 can be done in 6! ways. T2 can be done in 4! ways. Total number of anagrams with the constraint: 6!*4!

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Circular Permutations A CB BA C B C A A BC

How many circular permutations are there? Note that each circular permutation has n different row permutations (by starting at different objects in the circle) Convention 0!=1

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Permuting r of n objects Suppose 1  r  n. An r-permutation of a set of n elements is an ordered selection of r elements from the set. The number of r-permutations of a set of n elements is denoted by P(n,r) nPrnPr

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Permuting r of n objects Theorem: Suppose n, r are integers with 1  r  n. P(n,r) = n*(n-1)*…..(n-r+1) = n!/(n-r)! For r=0, we take P(n,0)=1.

Permuting r of n objects Proof: T 1 : Select the 1st element in the row T 2 : Select the 2nd element in the row …….. T r : Select the rth element in the row T 1 can be done in n ways. T 2 can be done in n-1 ways. ….. T r can be done in n - (r - 1) = n - r + 1 ways. Total number of r-permutations are: n * (n - 1) * ….. * (n - r + 1)

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Example Suppose there are 350 students. In how many ways can one select president, secretary and treasurer if no person can hold two posts? Permuting 3 of 350 objects. P(350,3)

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Example Suppose A and B are finite sets. How many different functions from A to B are 1--1? A = {a 1, a 2,…, a n }. B = {b 1, b 2,..., b m } if n > m: No 1--1 functions from A to B if n  m: Want to select f(a 1 ), f(a 2 ),…, f(a n ) from the set B All distinct. Thus, we are finding a n-permutation from a set of m objects. P(m,n) ways.

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Example How many bijective functions are there from A to B? A = {a 1, a 2,…, a n }. B = {b 1, b 2,..., b m } If m  n: zero bijective functions. If m=n: Want to select f(a 1 ), f(a 2 ),…, f(a n ) from the set B All distinct. We are selecting n out of n objects. P(n,n)=n!

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Example How many Hamiltonian circuits are there in K 5 ? Assume that we start at a fixed vertex. T 1 : Pick first vertex in HC (fixed to be v 1 ) T 2 : Pick second vertex in HC. T 3 : Pick third vertex in HC. T 4 : Pick fourth vertex in HC. T 5 : Pick fifth vertex in HC. Total number of HC starting at a fixed vertex: 1*4*3*2*1=4! T 1 : 1 T 2 : 4 T 3 : 3 T 4 : 2 T 5 : 1

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The Addition Rule Theorem: Suppose a finite set A equals the union of k distinct mutually disjoint sets A 1, A 2,…, A k. That is A= A 1  A 2  ….  A k, and, for i  j, A i  A j = . Then #(A) = #(A 1 ) + #(A 2 ) + …. + #(A k ).

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Example Suppose I can go from SIN to KL by bus, train or plane. There are 8 flights daily 2 morning and 2 evening trains, daily 1 bus daily In how many ways can one go from SIN to KL on a particular day Answer: 8+(2+2)+1 ways

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Example In Fortran identifiers consist of 1 to 6 characters where the first character must be English letter and others either English letter or a digit. How many different identifiers are possible.

First step: we calculate how many identifiers of length k are there (where 1  k  6) T 1 : Pick the first character T 2 : Pick the second character ….. T k : Pick the kth character. T 1 : 26 ways T 2 : 36 ways ….. T k : 36 ways 26*36 k-1 identifiers of length k Second step: Total number of identifiers= 26* * * …..+ 26*36 5

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The Difference Rule Theorem: If A is a finite set and B  A, then #(A - B) = #(A) - #(B).

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Example How many three digit numbers have at least one digit repeated? A---Set of three digit numbers B--- Set of three digit numbers which have no digit repeated. #(A) = 9*10*10 #(B) = 9*9*8 #(A - B) = 9*10*10 - 9*9*8

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Inclusion Exclusion Rule Theorem: If A, B and C are finite sets, then #(A  B) = #(A) + #(B) - #(A  B) #(A  B  C) = #(A) + #(B) + #(C) - #(A  B) - #(A  C) - #(B  C) + #(A  B  C)

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Example Class of 50 students. 30 know Pascal 18 know Fortran 26 know Java 9 know both Pascal and Fortran 16 know both Pascal and Java 8 know both Fortran and Java 47 know at least one of the three languages. Question: How many know all three languages? P: set of students who know Pascal F: set of students who know Fortran J: set of students who know Java

Example #(P  F  J) = #(P) + #(F) + #(J) - #(P  F) - #(P  J) - #(F  J) + #(P  F  J) 47= #(P  F  J) #(P  F  J) =

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