Definition: 

Permutation:          An arrangement is called a Permutation. It is the rearrangement of objects or symbols into distinguishable sequences. When we set things in order, we say we have made an arrangement. When we change the order, we say we have changed the arrangement. So each of the arrangement that can be made by taking some or all of a number of things is known as Permutation.

Combination:          A Combination is a selection of some or all of a number of different objects. It is an un-ordered collection of unique sizes.In a permutation the order of occurence of the objects or the arrangement is important but in combination the order of occurence of the objects is not important.

Formula:

Permutation = nPr = n! / (n-r)! Combination = nCr = nPr / r! where,              n, r are non negative integers and r<=n.              r is the size of each permutation.              n is the size of the set from which elements are permuted.              ! is the factorial operator.

Example: take the number of permutations and combinations: n=3; r=1.

  Step 1: Find the factorial of 7.            6! = 6*5*4*3*2*1 = 720

  Step 2: Find the factorial of 6-4.            (6-4)! = 2! = 2

  Step 3: Divide 720 by 2.            Permutation = 720/2 = 360

  Step 4: Find the factorial of 4.            4! = 4*3*2*1 = 24

  Step 5:Divide 360 by 24.            Combination = 360/24 = 15

If  represents the number of combinations of n items taken r at a time,

what is the value of  ?

This principle can be extended to any number of operationsFACTORIAL ‘n’The continuous product of the first ‘n’ natural numbers is called factorial n and is deonoted by n! i.e, n! = 1×2×3x ….. x(n-1)xn.

PERMUTATIONAn arrangement that can be formed by taking some or all of a finite set of things (or objects) is called a Permutation.Order of the things is very important in case of permutation.A permutation is said to be a Linear Permutation if the objects are arranged in a line. A linear permutation is simply called as a permutation.A permutation is said to be a Circular Permutation if the objects are arranged in the form of a circle.The number of (linear) permutations that can be formed by taking r things at a time from a set of n distinct

things   is denoted by  .%

NUMBER OF PERMUTATIONS UNDER CERTAIN CONDITIONS

1. Number of permutations of n different things, taken r at a time, when a particular thng is to be always included in each arrangement , is  .

2. Number of permutations of n different things, taken r at a time, when a particular thing is never taken in each arrangement is  .

3. Number of permutations of n different things, taken all at a time, when m specified things always come together is  .

4. Number of permutations of n different things, taken all at a time, when m specified

never come together is  .

5. The number of permutations of n dissimilar things taken r at a time when k(< r)

particular things always occur is  .

6. The number of permutations of n dissimilar things taken r at a time when k particular things never occur is  .

7. The number of permutations of n dissimilar things taken r at a time when repetition of things is allowed any number of times is 

8. The number of permutations of n different things, taken not more than r at a time,

when each thing may occur any number of times is  .

9. The number of permutations of n different things taken not more than r at a time  .

+ PERMUTATIONS OF SIMILAR THINGS+The number of permutations of n things taken all tat a time when p of them are all

alike and the rest are all different is  .If p things are alike of one type, q things are alike of other type, r things are alike of another type, then the number of

permutations with p+q+r things is  .

CIRCULAR PERMUTATIONS

}1. The number of circular permutations of n dissimilar things taken r at a time is  .

2. The number of circular permutations of n dissimilar things taken all at a time is  .

3. The number of circular permutations of n things taken r at a time in one direction

is  .

4. The number of circular permutations of n dissimilar things in clock-wise direction =

Number of permutations in anticlock-wise direction =  .

COMBINATIONA selection that can be formed by taking some or all of a finite set of things( or objects) is called a CombinationThe number of combinations of n dissimilar things taken r at a time is denoted

by  .

1.

2.

3.

4.

5. The number of combinations of n things taken r at a time in which

a)s particular things will always occur is  .

b)s particular things will never occur is  .

c)s particular things always occurs and p particular things never occur is  .

DISTRIBUTION OF THINGS INTO GROUPS1.Number of ways in which (m+n) items can be divided into two unequal groups

containing m and n items is  .

2.The number of ways in which mn different items can be divided equally into m groups,

each containing n objects and the order of the groups is not important is 3.The number of ways in which mn different items can be divided equally into m groups,

each containing n objects and the order of the groups is important is  .

4.The number of ways in which (m+n+p) things can be divided into three different

groups of m,n, an p things respectively is 

5.The required number of ways of dividing 3n things into three groups of n each =

.When the order of groups has importance then the required number of ways=

DIVISION OF IDENTICAL OBJECTS INTO GROUPSThe total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items   is 

}The number of non-negative integral solutions of the equation  .

The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is 

The number of positive integral solutions of the equation  .

The number of ways of choosing r objects from p objects of one kind, q objects of second kind, and so on is the coefficient of  in the expansion

he number of ways of choosing r objects from p objects of one kind, q objects of second kind, and so on, such that one object of each kind may be included is the coefficient of   is the coefficient of  in the expansion

.

%{font-family:verdana}+*TOTAL NUMBER OF COMBINATIONS*+%%{font-family:verdana}1.The total number of combinations

of   things taken any number at a time when  things are alike of one kind,   things are alike of second kind…. things are alike of kind,

is  .%

%{font-family:verdana}2.The total number of combinations

of  things taken one or more at a time when  things are alike of one kind,   things are alike of second kind…. things are alike of kind, is%

.

SUM OF THE NUMBERSSum of the numbers formed by taking all the given n digits (excluding 0) is 

Sum of the numbers formed by taking all the given n digits (including 0) is 

Sum of all the r-digit numbers formed by taking the given n digits(excluding 0) is  %

%{font-family:verdana}Sum of all the r-digit numbers formed by taking the given n digits(including 0) is   

DE-ARRANGEMENT:

The number of ways in which exactly r letters can be placed in wrongly addressed envelopes when n letters are placed in n addressed envelopes

is  .

The number of ways in which n different letters can be placed in their n addressed envelopes so that al the letters are in the wrong envelopes

is  .

IMPORTANT RESULTS TO REMEBERIn a plane if there are n points of which no three are collinear, then

1. The number of straight lines that can be formed by joining them is  .

2. The number of triangles that can be formed by joining them is  .

3. The number of polygons with k sides that can be formed by joining them is  .

In a plane if there are n points out of which m points are collinear, then

1. The number of straight lines that can be formed by joining them is  .

2. The number of triangles that can be formed by joining them is  .

3. The number of polygons with k sides that can be formed by joining them is .

Number of rectangles of any size in a square of n x n is 

In a rectangle of p x q (p < q) number of rectangles of any size is 

In a rectangle of p x q (p < q) number of squares of any size

is 

n straight lines are drawn in the plane such that no two lines are parallel and no three lines three lines are concurrent. Then the number of parts into which these lines divide

the plane is equal to  .

SECTION 3.4 Counting, Permutations and Combinations

Basics of Counting

Permutations

r-permutations

Combinations

Basics of Counting

People may think that counting is easy, and certainly sometimes it is. But some of the aspects of counting are not simple, especially counting a large number of elements. In this case, we need to use some mathematical skills to find out the answer.

Consider counting the number of elements in a list, where each element has an index beginning with some integer m and ending with some integer n, and m < n.  Then there are n - m + 1 elements in the list, from m to ninclusive.  For example, if the first index is 0, and the last indexed element is 99, then you have 99 - 0 + 1 = 100 elements.   This method also works for counting a certain number of elements from only part of the list (for example, the number of elements between index nos. 12 and 43.)

There are two basic counting principles:

Sum Rule Suppose that an operation can be performed by either of two different procedures, with m possible outcomes for the first procedure and n possible outcomes for the

second. If the two sets of possible outcomes are disjoint, then the number of possible outcomes for the operation is m + n.

Product Rule

Suppose that an operation consists of k steps and:

the first step can be performed in n1 ways;

the second step can be performed in n2 ways (ignoring how the first step was performed), . . .; and

the kth step can be performed in nk ways.

Then the whole operation can be performed in n1 * n2 * ........... * nk ways.

Consider the following examples:

Example 1:

A scholarship is available, and the student to receive this scholarship must be chosen from the Mathematics, Computer Science, or the Engineering Department. How many different choices are there for this student if there are 38 qualified students from the Mathematics Department, 45 qualified students from the Computer Science Department and 27 qualified students from the Engineering Department?

Solution: The procedure of choosing a student from the Mathematics Department has 38 possible outcomes, the procedure of choosing a student from the Computer Science Department has 45 possible outcomes, and the procedure of choosing a student from the Engineering Department has 27 possible outcomes. Therefore, there are (38 + 45 + 27 ) 110 possible choices for the student to receive the scholarship.

Example 2:

A man has 10 shirts, 8 pairs of pants and three pairs of shoes. How many different outfits, consisting of one shirt, one pair of pants and one pair of shoes, are possible?

Solution: We have to consider three steps: choose a shirt; choose a pair of pants; and choose a pair of shoes. Choosing a shirt has 10 possible outcomes (as he has ten shirts!), choosing a pair of pants has 8 possible outcomes, and choosing a pair of shoes has 3 possible outcomes. So the number of different outfits is 10 * 8 * 3 = 240.

Question : How many different outfits, consisting of one shirt, one pair of pants, one pair of shoes and one hat, are possible if he has bought two hats? Click here for answer.

Permutations

D E F I N I T I O N S

Permutation Given n different elements in a set, any ordered arrangement of these elements is called a permutation.

r-permutation An r-permutation of a set of n elements is an ordered selection of r elements taken from the set of n elements. It is denoted by P(n, r).

Permutations are arrangements of the objects within a set. For example, the set of elements a, b and c has six permutations.

abc,   bac,    bca,   acb,   cab,   cba

Imagine that we have a set of n elements, how can we find the number permutations of the set?

Try to view creating a permutation as an n-step operation:

1st step: choose the first element. If there are n elements, then there are n possible choices for the first element.

2nd step: choose the second element. Since one element has already been chosen and placed, there are n-1 possible choices remaining.

3rd step: choose the third element. Since two elements have already been chosen and placed, there are n-2 possible choices remaining.

. . . . .

nth step: choose the nth element. Since n-1 elements have already been chosen and placed for the preceding steps, there is only one choice left in the set.

Hence by the product rule, there are n * (n - 1) * (n - 2) * ...... * 2 * 1 ways to perform the entire operation. In other words, there are n! permutations of a set of n elements.

r-permutations

Now, consider the following example:

Given a set of 3 elements a, b, and c. there are six ways to select two elements from the set and write them in order.

ab,    ba,    bc,    cb,    ca,     ac

This is called a 2-permutation of the set {a, b, c}.

If we are given a set of n elements, and we have to select r elements from the set where r < n, what is the r-permutation of the set?

1st step: choose the first element. If there are n elements, then there are n possible choices for the first element.

2nd step: choose the second element. Since one element has already been chosen and placed, there are n-1 possible choices remaining.

3rd step: choose the third element. Since two elements have already been chosen and placed, there are n-2 possible choices remaining.

. . . . .

rth step: choose the rth element. This is the last step, and since there are n - r + 1 elements remaining, there are clearly n - r + 1 ways left to select an element from the set.

Therefore by the product rule, there are n * (n - 1) * (n - 2) * ....... * (n - r + 1) ways to perform the entire operation.

P(n, r) = n * (n - 1) * (n - 2) * ....... * (n - r + 1)

or equivalently,

P(n, r) = 

Try to figure out how this equation equals n * (n - 1) * (n - 2) * ....... * (n - r + 1).  Click here for explanation.

Combinations

D E F I N I T I O N

r-combinations

An r-combination from a set of n elements is a unordered selection of r elements from the set, where n and r are nonnegative integers with r < n. r-combination is

denoted by the symbol  , read "nchoose r,"  and denotes the number of subsets of size r  (r-combinations) that can be chosen from a set of n elements.

In some books and calculators, the symbols C(n, r), nCr, Cn, r, or nCr are used instead

of  .

Consider the following example:

Let L = {Discrete Structures, Assembly Language, C Programming, Computer organization, File Processing}. A student can take three of the five classes in L in one quarter.

A.  List all 3-combinations of L.

B.  Find  .

Answers:

A.  {Discrete Structures, Assembly Language, C Programming}{Discrete Structures, Assembly Language, Computer Organization}{Discrete Structures, Assembly Language, File Processing}{Discrete Structures, C Programming, Computer Organization}{Discrete Structures, C Programming, File Processing}{Discrete Structures, Computer Organization, File Processing}{Assembly Language, C Programming, Computer Organization}{Assembly Language, C Programming, File Processing}{Assembly Language, Computer Organization, File Processing}{C Programming, Computer Organization, File Processing}

There are ten 3-combinations in L.

B.    is the number of 3-combinations of a set with five elements, by part A,   = 10.

If we are given a set with n elements, what is the number of r-combinations from that set? Consider obtaining an r-permutation:    one can first select r elements, which can

be done in   ways, and then arrange them, which can be done in r! ways.

Hence, P(n, r) =   r!,      = P(n, r)/r!

Since P(n, r) =  ,   

 = 

Consider the following example:

Suppose you have a group of 10 children consisting of 4 girls and 6 boys.

A.  How many four-person teams can be chosen that consist of two girls and two boys? B.  How many four-person teams contain at least one girl?

Answers:

A.  To solve this problem, we have to do two things:

First choose two girls; then choose two boys.

There are   ways to choose two girls out of the four and   ways to choose two boys out of the six. Hence by the product rule,

Number of teams that contain two girls and two boys

=   *   

=   * 

=    *    

= 

=  90

B.  The number of teams containing at least one girl =

total number of teams of four - number of teams of four that do not contain any girls.

Top of Form

Definition: 

Permutation:          An arrangement is called a Permutation. It is the rearrangement of objects or symbols into distinguishable sequences. When we set things in order, we say we have made an arrangement. When we change the order, we say we have changed the arrangement. So each of the arrangement that can be made by taking some or all of a number of things is known as Permutation.

Combination:          A Combination is a selection of some or all of a number of different objects. It is an un-ordered collection of unique sizes.In a permutation the order of occurence of the objects or the arrangement is important but in combination the order of occurence of the objects is not important.

Formula:

Permutation = nPr = n! / (n-r)! Combination = nCr = nPr / r! where,              n, r are non negative integers and r<=n.              r is the size of each permutation.              n is the size of the set from which elements are permuted.              ! is the factorial operator.

Example:Find the number of permutations and combinations: n=6; r=4.

  Step 1: Find the factorial of 6.            6! = 6*5*4*3*2*1 = 720

  Step 2: Find the factorial of 6-4.            (6-4)! = 2! = 2

  Step 3: Divide 720 by 2.            Permutation = 720/2 = 360

  Step 4: Find the factorial of 4.            4! = 4*3*2*1 = 24

  Step 5:Divide 360 by 24.            Combination = 360/24 = 15

The above example will help you to find the Permutation and Combination manually.

Basics Of Permutation And Combination — Presentation Transcript

1. BASICS OF PERMUTATION AND COMBINATION by : DR. T.K. JAIN

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2. What is factorial ? 5! = 5*4*3*2*1 thus factorial 5 means we will multiply

5 to all the numbers lower to it.

3. What is permutation? Permutation means number of arrangements

(keeping the order into mind). Thus A,B,C can give you following permutations :

ABC,ACB,BAC,BCA,BAC,CAB,CBA

4. What is combination? In combination, the order it not considered, thus

A,B,C will have only one combination ABC or BAC or BCA (whatever you call, it is

one combination) combination means together, thus when some digits are together,

it is a combination, it is not importnat who comes first or second or third.

5. How many numbers can you make out of 4 different digits? Answer : 4!

remember, we have to solve it as N!, thus if we have four digits 1,2,3,4, we can

make 24 numbers out of it. Verify it :

1234,1243,1324,1342,1423,1432,2134,2143, ...so on

6. How many permutations are possible from 4 digits taking 3 at a time?

Total digits = N = 4 digits taken = r = 3 N! / (N-r)! =4! / (4-3)! = 24 answer

7. You have beads of 5 colours, how many necklace can you form? For

such questions, use formula of circular permutation = ½ * (n-1)! =1/2 * (5-1)! =12

answer

8. How many combinations can be formed from 5 digits taking 3 at a time?

Formula of combination : n! / (r! (n-r)!) n = 5 r=2 5!/(2! *3!) =120/12 =10 answer

9. How many words can you form from C O L L E C T I O N THERE ARE

10 DIGITS. Some are similar (L, C, O, are coming twice). For such the formula is :

N! / (X!Y!Z!) N = 10, X = 2, Y = 2, Z = 2 (X, Y and Z are for number of repeatitions)

= 10!/ (2! * 2!* 2!)

10. How many combinations are possible out of 6 boys, taking one or all

or some of them together? Formula for taking one or all or some of them is : 2^n –

1 here we can take 1, 2, 3, 4,5, or 6 boys at a time. We cannot take 0 boys. So the

formula is 2^n – 1 = 2^6 – 1 = 63 answer

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PermutationFrom Wikipedia, the free encyclopedia

For other uses, see Permutation (disambiguation).

The 6 permutations of 3 balls

In mathematics, the notion of permutation is used with several slightly different meanings, all related to the act

of permuting (rearranging) objects or values. Informally, a permutation of a set of objects is an arrangement of

those objects into a particular order. For example, there are six permutations of the set {1,2,3}, namely (1,2,3),

(1,3,2), (2,1,3), (2,3,1), (3,1,2), and (3,2,1). One might define an anagram of a word as a permutation of its

letters. The study of permutations in this sense generally belongs to the field of combinatorics.

The number of permutations of n distinct objects is n×(n − 1)×(n − 2)×...×2×1, which number is called

"n factorial" and written "n!".

Permutations occur, in more or less prominent ways, in almost every domain of mathematics. They often arise

when different orderings on certain finite sets are considered, possibly only because one wants to ignore such

orderings and needs to know how many configurations are thus identified. For similar reasons permutations

arise in the study of sorting algorithms in computer science.

In algebra and particularly in group theory, a permutation of a set S is defined as a bijection from S to itself (i.e.,

a map S → S for which every element of S occurs exactly once as image value). This is related to the

rearrangement of S in which each element s takes the place of the corresponding f(s). The collection of such

permutations form a symmetric group. The key to its structure is the possibility to compose permutations:

performing two given rearrangements in succession defines a third rearrangement, the composition.

Permutations may act on composite objects by rearranging their components, or by certain replacements

(substitutions) of symbols.

In elementary combinatorics, the name "permutations and combinations" refers to two related problems, both

counting possibilities to select k distinct elements from a set of n elements, where for k-permutations the order

of selection is taken into account, but for k-combinations it is ignored. However k-permutations do not

correspond to permutations as discussed in this article (unless k = n).

Contents

  [hide] 

1     History   

2     Generalities   

o 2.1      In group theory   

o 2.2      In combinatorics   

3     Permutations in group theory   

o 3.1      Notation   

o 3.2      Product and inverse   

4     Permutations in combinatorics   

o 4.1      Ascents, descents and runs   

o 4.2      Inversions   

o 4.3      Counting sequences without repetition   

5     Permutations in computing   

o 5.1      Numbering permutations   

o 5.2      Algorithms to generate permutations   

5.2.1      Random generation of    

permutations

5.2.2      Generation in lexicographic    

order

5.2.3      Generation with minimal    

changes

o 5.3      Software implementations   

5.3.1      Calculator functions   

5.3.2      Spreadsheet functions   

o 5.4      Applications   

6     See also   

7     Notes   

8     References   

[edit]History

The rule to determine the number of permutations of n objects was known in Hindu culture at least as early as

around 1150: the Lilavati by the Indian mathematician Bhaskara II contains a passage that translates to

The product of multiplication of the arithmetical series beginning and increasing by unity and continued to the

number of places, will be the variations of number with specific figures.[1]

A first case in which seemingly unrelated mathematical questions were studied with the help of permutations

occurred around 1770, when Joseph Louis Lagrange, in the study of polynomial equations, observed that

properties of the permutations of the roots of an equation are related to the possibilities to solve it. This line of

work ultimately resulted, through the work of Évariste Galois, in Galois theory, which gives a complete

description of what is possible and impossible with respect to solving polynomial equations (in one unknown)

by radicals. In modern mathematics there are many similar situations in which understanding a problem

requires studying certain permutations related to it.

[edit]Generalities

The notion of permutation is used in the following contexts.

[edit]In group theory

In group theory and related areas, one considers permutations of arbitrary sets, even infinite ones. A

permutation of a set S is a bijection from S to itself. This allows for permutations to be composed, which allows

the definition of groups of permutations. If S is a finite set of n elements, then there are n !  permutations of S.

[edit]In combinatorics

Permutations of multisets

In combinatorics, a permutation is usually understood to be a sequence containing each element from a finite

set once, and only once. The concept ofsequence is distinct from that of a set, in that the elements of a

sequence appear in some order: the sequence has a first element (unless it is empty), a second element

(unless its length is less than 2), and so on. In contrast, the elements in a set have no order; {1, 2, 3} and {3, 2,

1} are different ways to denote the same set. In this sense a permutation of a finite set S of n elements is

equivalent to a bijection from {1, 2, ... , n} to S (in which any i is mapped to the i-th element of the sequence), or

to a choice of a total ordering on S (for which x < y if x comes before y in the sequence). In this sense there are

also n !  permutations of S.

There is also a weaker meaning of the term "permutation" that is sometimes used in elementary combinatorics

texts, designating those sequences in which no element occurs more than once, but without the requirement to

use all elements from a given set. Indeed this use often involves considering sequences of a fixed length k of

elements taken from a given set of size n. These objects are also known as sequences without repetition, a

term that avoids confusion with the other, more common, meanings of "permutation". The number of such k-

permutations of n is denoted variously by such symbols as n Pk, nPk, Pn,k, or P(n,k), and its value is given by

the product[2]

which is 0 when k > n, and otherwise is equal to

The product is well defined without the assumption that n is a non-negative integer, and is of

importance outside combinatorics as well; it is known as the Pochhammer symbol (n)k or as the k-th

falling factorial power nk of n.

If M is a finite multiset, then a multiset permutation is a sequence of elements of M in which each

element appears exactly as often as is its multiplicity in M. If the multiplicities of the elements

of M (taken in some order) are  ,  , ...,   and their sum (i.e., the size of M) is n, then the

number of multiset permutations of M is given by the multinomial coefficient

[edit]Permutations in group theory

Main article: Symmetric group

In group theory, the term permutation of a set means a bijective map, or bijection, from that set

onto itself. The set of all permutations of any given set S forms a group, with composition of maps

as product and the identity as neutral element. This is the symmetric group of S. Up to

isomorphism, this symmetric group only depends on the cardinality of the set, so the nature of

elements of S is irrelevant for the structure of the group. Symmetric groups have been studied

most in the case of a finite sets, in which case one can assume without loss of generality

that S={1,2,...,n} for some natural number n, which defines the symmetric group of degree n,

written Sn.

Any subgroup of a symmetric group is called a permutation group. In fact by Cayley's

theorem any group is isomorphic to some permutation group, and every finite group to a

subgroup of some finite symmetric group. However, permutation groups have more structure than

abstract groups, allowing for instance to define the cycle type of an element of a permutation

group; different realizations of a group as a permutation group need not be equivalent for this

additional structure. For instance S3 is naturally a permutation group, in which any transposition

has cycle type (2,1), but the proof of Cayley's theorem realizes S3 as a subgroup of S6 (namely

the permutations of the 6 elements of S3 itself), in which permutation group transpositions get

cycle type (2,2,2). So in spite of Cayley's theorem, the study of permutation groups differs from

the study of abstract groups.

[edit]Notation

There are three main notations for permutations of a finite set S. In Cauchy's two-line notation,

one lists the elements of S in the first row, and for each one its image under the permutation

below it in the second row. For instance, a particular permutation of the set {1,2,3,4,5} can be

written as:

this means that σ satisfies σ(1)=2, σ(2)=5, σ(3)=4, σ(4)=3, and σ(5)=1.

In one-line notation, one gives only the second row of this array, so the one-line notation for

the permutation above is 25431. (It is typical to use commas to separate these entries only if

some have two or more digits.)

Cycle notation, the third method of notation, focuses on the effect of successively applying

the permutation. It expresses the permutation as a product of cycles corresponding to

the orbits (with at least two elements) of the permutation; since distinct orbits are disjoint, this

is loosely referred to as "the decomposition into disjoint cycles" of the permutation. It works

as follows: starting from some element x of S with σ(x) ≠ x, one writes the sequence

(x σ(x) σ(σ(x)) ...) of successive images under σ, until the image would be x, at which point

one instead closes the parenthesis. The set of values written down forms the orbit (under σ)

of x, and the parenthesized expression gives the corresponding cycle of σ. One then

continues choosing an element y of S that is not in the orbit already written down, and such

that σ(y) ≠ y, and writes down the corresponding cycle, and so on until all elements

of S either belong to a cycle written down or are fixed points of σ. Since for every new cycle

the starting point can be chosen in different ways, there are in general many different cycle

notations for the same permutation; for the example above one has for instance

Each cycle (x1 x2 ... xl) of σ denotes a permutation in its own right, namely the one that

takes the same values as σ on this orbit (so it maps xi to xi+1 for i < l, and xl to x1), while

mapping all other elements of S to themselves. The size l of the orbit is called the length

of the cycle. Distinct orbits of σ are by definition disjoint, so the corresponding cycles are

easily seen to commute, and σ is the product of its cycles (taken in any order).

Therefore the concatenation of cycles in the cycle notation can be interpreted as

denoting composition of permutations, whence the name "decomposition" of the

permutation. This decomposition is essentially unique: apart from the reordering the

cycles in the product, there are no other ways to write σ as a product of cycles (possibly

unrelated to the cycles of σ) that have disjoint orbits. The cycle notation is less unique,

since each individual cycle can be written in different ways, as in the example above

where (5 1 2) denotes the same cycle as (1 2 5) (but (5 2 1) would denote a different

permutation).

An orbit of size 1 (a fixed point x in S) has no corresponding cycle, since that

permutation would fix x as well as every other element of S, in other words it would be

the identity, independently ofx. It is possible to include (x) in the cycle notation for σ to

stress that σ fixes x (and this is even standard in combinatorics, as described in cycles

and fixed points), but this does not correspond to a factor in the (group theoretic)

decomposition of σ. If the notion of "cycle" were taken to include the identity

permutation, then this would spoil the uniqueness (up to order) of the decomposition of a

permutation into disjoint cycles. The decomposition into disjoint cycles of the identity

permutation is an empty product; its cycle notation would be empty, so some other

notation like e is usually used instead.

Cycles of length two are called transpositions; such permutations merely exchange the

place of two elements.

[edit]Product and inverse

Main article: Symmetric group

The product of two permutations is defined as their composition as functions, in other

words σ·π is the function that maps any element x of the set to σ(π(x)). Note that the

rightmost permutation is applied to the argument first, because of the way function

application is written. Some authors prefer the leftmost factor acting first, but to that end

permutations must be written to the right of their argument, for instance as an exponent,

where σ acting on x is written xσ; then the product is defined by xσ·π=(xσ)π. However this

gives a different rule for multiplying permutations; this article uses the definition where

the rightmost permutation is applied first.

Since the composition of two bijections always gives another bijection, the product of

two permutations is again a permutation. Since function composition is associative, so is

the product operation on permutations: (σ·π)·ρ=σ·(π·ρ). Therefore, products of more

than two permutations are usually written without adding parentheses to express

grouping; they are also usually written without a dot or other sign to indicate

multiplication.

The identity permutation, which maps every element of the set to itself, is the neutral

element for this product. In two-line notation, the identity is

Since bijections have inverses, so do permutations, and the inverse σ−1 of σ is again

a permutation. Explicitly, whenever σ(x)=y one also has σ−1(y)=x. In two-line

notation the inverse can be obtained by interchanging the two lines (and sorting the

columns if one wishes the first line to be in a given order). For instance

In cycle notation one can reverse the order of the elements in each cycle to

obtain a cycle notation for its inverse.

Having an associative product, a neutral element, and inverses for all its

elements, makes the set of all permutations of S into a group, called the

symmetric group of S.

Every permutation of a finite set can be expressed as the product of

transpositions. Moreover, although many such expressions for a given

permutation may exist, there can never be among them both expressions with

an even number and expressions with an odd number of transpositions. All

permutations are then classified as even or odd, according to the parity of the

transpositions in any such expression.

Composition of permutations corresponds to multiplication of permutation matrices.

Multiplying permutations written in cycle notation follows no easily described

pattern, and the cycles of the product can be entirely different from those of the

permutations being composed. However the cycle structure is preserved in the

special case of conjugating a permutation σ by another permutationπ, which

means forming the product π·σ·π−1. Here the cycle notation of the result can be

obtained by taking the cycle notation for σ and applying π to all the entries in it.

[3]

One can represent a permutation of {1, 2, ..., n} as an n×n matrix. There are

two natural ways to do so, but only one for which multiplications of matrices

corresponds to multiplication of permutations in the same order: this is the one

that associates to σ the matrix M whose entry Mi,j is 1 if i = σ(j), and 0

otherwise. The resulting matrix has exactly one entry 1 in each column and in

each row, and is called a permutation matrix.

Here (file) is a list of these matrices for permutations of 4 elements. The Cayley

table on the right shows these matrices for permutations of 3 elements.

[edit]Permutations in combinatorics

In combinatorics a permutation of a set S with n elements is a listing of the

elements of S in some order (each element occurring exactly once). This can

be defined formally as a bijection from the set { 1, 2, ..., n } to S. Note that

if S equals { 1, 2, ..., n }, then this definition coincides with the definition in

group theory. More generally one could use instead of { 1, 2, ..., n } any set

equipped with a total ordering of its elements.

One combinatorial property that is related to the group theoretic interpretation

of permutations, and can be defined without using a total ordering of S, is

the cycle structure of a permutation σ. It is the partition of n describing the

lengths of the cycles of σ. Here there is a part "1" in the partition for every fixed

point of σ. A permutation that has no fixed point is called a derangement.

Other combinatorial properties however are directly related to the ordering of S,

and to the way the permutation relates to it. Here are a number of such

properties.

[edit]Ascents, descents and runs

An ascent of a permutation σ of n is any position i < n where the following value

is bigger than the current one. That is, if σ = σ1σ2...σn, then i is an ascent

if σi < σi+1.

For example, the permutation 3452167 has ascents (at positions) 1,2,5,6.

Similarly, a descent is a position i < n with σi > σi+1, so

every i with   either is an ascent or is a descent of σ.

The number of permutations of n with k ascents is the Eulerian number  ;

this is also the number of permutations of n with k descents.[4]

An ascending run of a permutation is a nonempty increasing contiguous

subsequence of the permutation that cannot be extended at either end; it

corresponds to a maximal sequence of successive ascents (the latter may be

empty: between two successive descents there is still an ascending run of

length 1). By contrast an increasing subsequence of a permutation is not

necessarily contiguous: it is an increasing sequence of elements obtained from

the permutation by omitting the values at some positions. For example, the

permutation 2453167 has the ascending runs 245, 3, and 167, while it has an

increasing subsequence 2367.

If a permutation has k − 1 descents, then it must be the union of k ascending

runs. Hence, the number of permutations of n with k ascending runs is the

same as the number   of permutations with k − 1 descents.[5]

[edit]Inversions

Main article: Inversion (discrete mathematics)

An inversion of a permutation σ is a pair (i,j) of positions where the entries of a

permutation are in the opposite order:   and  .[6] So a descent

is just an inversion at two adjacent positions. For example, the

permutation σ = 23154 has three inversions: (1,3), (2,3), (4,5), for the pairs of

entries (2,1), (3,1), (5,4).

Sometimes an inversion is defined as the pair of values (σi,σj) itself whose

order is reversed; this makes no difference for the number of inversions, and

this pair (reversed) is also an inversion in the above sense for the inverse

permutation σ−1. The number of inversions is an important measure for the

degree to which the entries of a permutation are out of order; it is the same

for σ and forσ−1. To bring a permutation with k inversions into order (i.e.,

transform it into the identity permutation), by successively applying (right-

multiplication by) adjacent transpositions, is always possible and requires a

sequence of k such operations. Moreover any reasonable choice for the

adjacent transpositions will work: it suffices to choose at each step a

transposition of i and i + 1 where i is a descent of the permutation as modified

so far (so that the transposition will remove this particular descent, although it

might create other descents). This is so because applying such a transposition

reduces the number of inversions by 1; also note that as long as this number is

not zero, the permutation is not the identity, so it has at least one

descent. Bubble sort and insertion sort can be interpreted as particular

instances of this procedure to put a sequence into order. Incidentally this

procedure proves that any permutation σ can be written as a product of

adjacent transpositions; for this one may simply reverse any sequence of such

transpositions that transforms σ into the identity. In fact, by enumerating all

sequences of adjacent transpositions that would transform σ into the identity,

one obtains (after reversal) a complete list of all expressions of minimal length

writing σ as a product of adjacent transpositions.

The number of permutations of n with k inversions is expressed by a Mahonian

number,[7] it is the coefficient of Xk in the expansion of the product

which is also known (with q substituted for X) as the q-factorial [n]q! .

[edit]Counting sequences without repetition

In this section, a k-permutation of a set S is an ordered sequence

of k distinct elements of S. For example, given the set of letters

{C, E, G, I, N, R}, the sequence ICE is a 3-

permutation, RINGand RICE are 4-permutations, NICER and REIGN are

5-permutations, and CRINGE is a 6-permutation; since the latter uses all

letters, it is a permutation of the given set in the ordinary combinatorial

sense. ENGINE on the other hand is not a permutation, because of the

repetitions: it uses the elements E and N twice.

Let n be the size of S, the number of elements available for selection. In

constructing a k-permutation, there are n  possible choices for the first

element of the sequence, and this is then number of 1-permutations. Once

it has been chosen, there are n − 1 elements of S left to choose from, so a

second element can be chosen in n − 1 ways, giving a total n × (n − 1)

possible 2-permutations. For each successive element of the sequence,

the number of possibilities decreases by 1  which leads to the number of

n × (n − 1) × (n − 2) ... × (n − k + 1) possible k-permutations.

This gives in particular the number of n-permutations (which contain

all elements of S once, and are therefore simply permutations of S):

n × (n − 1) × (n − 2) × ... × 2 × 1,

a number that occurs so frequently in mathematics that it is given

a compact notation "n!", and is called "n factorial". These n-

permutations are the longest sequences without repetition of

elements of S, which is reflected by the fact that the above

formula for the number of k-permutations gives zero

whenever k > n.

The number of k-permutations of a set of n elements is

sometimes denoted by P(n,k) or a similar notation (usually

accompanied by a notation for the number of k-combinations of a

set of nelements in which the "P" is replaced by "C"). That

notation is rarely used in other contexts than that of counting k-

permutations, but the expression for the number does arise in

many other situations. Being a product of k factors starting

at n and decreasing by unit steps, it is called the k-th falling

factorial power of n:

though many other names and notations are in use, as

detailed at Pochhammer symbol. When k ≤ n the factorial

power can be completed by additional

factors: nk × (n − k)! = n!, which allows writing

The right hand side is often given as expression for the

number of k-permutations, but its main merit is using the

compact factorial notation. Expressing a product

of k factors as a quotient of potentially much larger

products, where all factors in the denominator are also

explicitly present in the numerator, is not particularly

efficient; as a method of computation there is the

additional danger of overflow or rounding errors. It

should also be noted that the expression is undefined

when k > n, whereas in those cases the number nk of k-

permutations is just 0.

[edit]Permutations in computing

[edit]Numbering permutations

One way to represent permutations of n is by an

integer N with 0 ≤ N < n!, provided convenient methods

are given to convert between the number and the usual

representation of a permutation as a sequence. This

gives the most compact representation of arbitrary

permutations, and in computing is particularly attractive

when n is small enough that N can be held in a machine

word; for 32-bit words this means n ≤ 12, and for 64-bit

words this means n ≤ 20. The conversion can be done

via the intermediate form of a sequence of

numbers dn, dn−1, ..., d2, d1, where di is a non-negative

integer less than i (one may omit d1, as it is always 0,

but its presence makes the subsequent conversion to a

permutation easier to describe). The first step then is

simply expression of N in thefactorial number system,

which is just a particular mixed radix representation,

where for numbers up to n! the bases for successive

digits are n, n − 1, ..., 2, 1. The second step interprets

this sequence as a Lehmer code or (almost

equivalently) as an inversion table.

Rothe diagram

for

i ＼ σi 1 2 3 4 5 6 7 8 9

Lehmer code

1 × × × × × • d9 = 5

2 × × • d8 = 2

3 × × × × × • d7 = 5

4 • d6 = 0

5 × • d5 = 1

6 × × × • d4 = 3

7 × × • d3 = 2

8 • d2 = 0

9 • d1 = 0

inversion table

3 6 1 2 4 0 2 0 0

In the Lehmer code for a permutation σ, the

number dn represents the choice made for the first

term σ1, the number dn−1 represents the choice made for

the second term σ1 among the remaining n − 1 elements

of the set, and so forth. More precisely, each dn+1−i gives

the number of remaining elements strictly less than the

term σi. Since those remaining elements are bound to

turn up as some later term σj, the digit dn+1−i counts

the inversions (i,j) involving i as smaller index (the

number of values j for which i < j and σi > σj).

Theinversion table for σ is quite similar, but

here dn+1−k counts the number of inversions (i,j)

where k = σjoccurs as the smaller of the two values

appearing in inverted order.[8] Both encodings can be

visualized by ann by n Rothe diagram[9] (named

after Heinrich August Rothe) in which dots at (i,σi) mark

the entries of the permutation, and a cross at (i,σj)

marks the inversion (i,j); by the definition of inversions a

cross appears in any square that comes both before the

dot (j,σj) in its column, and before the dot (i,σi) in its row.

The Lehmer code lists the numbers of crosses in

successive rows, while the inversion table lists the

numbers of crosses in successive columns; it is just the

Lehmer code for the inverse permutation, and vice

versa.

To effectively convert a Lehmer

code dn, dn−1, ..., d2, d1 into a permutation of an ordered

set S, one can start with a list of the elements of S in

increasing order, and for i increasing from 1 to n set σi to

the element in the list that is preceded by dn+1−i other

ones, and remove that element from the list. To convert

an inversion tabledn, dn−1, ..., d2, d1 into the

corresponding permutation, one can traverse the

numbers from d1 to dn while inserting the elements

of S from largest to smallest into an initially empty

sequence; at the step using the number d from the

inversion table, the element from S inserted into the

sequence at the point where it is preceded

by d elements already present. Alternatively one could

process the numbers from the inversion table and the

elements of S both in the opposite order, starting with a

row of n empty slots, and at each step place the

element from S into the empty slot that is preceded

by dother empty slots.

Converting successive natural numbers to the factorial

number system produces those sequences

in lexicographic order (as is the case with any mixed

radix number system), and further converting them to

permutations preserves the lexicographic ordering,

provided the Lehmer code interpretation is used (using

inversion tables, one gets a different ordering, where

one starts by comparing permutations by the place of

their entries 1 rather than by the value of their first

entries). The sum of the numbers in the factorial number

system representation gives the number of inversions of

the permutation, and the parity of that sum gives

the signature of the permutation. Moreover the positions

of the zeroes in the inversion table give the values of

left-to-right maxima of the permutation (in the example

6, 8, 9) while the positions of the zeroes in the Lehmer

code are the positions of the right-to-left minima (in the

example positions the 4, 8, 9 of the values 1, 2, 5); this

allows computing the distribution of such extrema

among all permutations. A permutation with Lehmer

code dn, dn−1, ..., d2, d1 has an ascent n − i if and only

if di ≥ di+1.

[edit]Algorithms to generate permutations

In computing it may be required to generate

permutations of a given sequence of values. The

methods best adapted to do this depend on whether

one wants some randomly chosen permutations, or all

permutations, and in the latter case if a specific ordering

is required. Another question is whether possible

equality among entries in the given sequence is to be

taken into account; if so, one should only generate

distinct multiset permutations of the sequence.

An obvious way to generate permutations of n is to

generate values for the Lehmer code (possibly using

the factorial number system representation of integers

up to n!), and convert those into the corresponding

permutations. However the latter step, while

straightforward, is hard to implement efficiently, because

it requires n operations each of selection from a

sequence and deletion from it, at an arbitrary position; of

the obvious representations of the sequence as

an array or a linked list, both require (for different

reasons) about n2/4 operations to perform the

conversion. With n likely to be rather small (especially if

generation of all permutations is needed) that is not too

much of a problem, but it turns out that both for random

and for systematic generation there are simple

alternatives that do considerably better. For this reason

it does not seem useful, although certainly possible, to

employ a special data structure that would allow

performing the conversion from Lehmer code to

permutation in O ( n   log   n )  time.

[edit]Random generation of permutations

Main article: Fisher–Yates shuffle

For generating random permutations of a given

sequence of n values, it makes no difference whether

one means apply a randomly selected permutation

of n to the sequence, or choose a random element from

the set of distinct (multiset) permutations of the

sequence. This is because, even though in case of

repeated values there can be many distinct

permutations of n that result in the same permuted

sequence, the number of such permutations is the same

for each possible result. Unlike for systematic

generation, which becomes unfeasible for large n due to

the growth of the number n!, there is no reason to

assume that n will be small for random generation.

The basic idea to generate a random permutation is to

generate at random one of the n! sequences of

integers d1,d2,...,dn satisfying 0 ≤ di < i (since d1 is

always zero it may be omitted) and to convert it to a

permutation through a bijective correspondence. For the

latter correspondence one could interpret the (reverse)

sequence as a Lehmer code, and this gives a

generation method first published in 1938 by Ronald A.

Fisher and Frank Yates.[10] While at the time computer

implementation was not an issue, this method suffers

from the difficulty sketched above to convert from

Lehmer code to permutation efficiently. This can be

remedied by using a different bijective correspondence:

after using di to select an element among i remaining

elements of the sequence (for decreasing values of i),

rather than removing the element and compacting the

sequence by shifting down further elements one place,

one swaps the element with the final remaining element.

Thus the elements remaining for selection form a

consecutive range at each point in time, even though

they may not occur in the same order as they did in the

original sequence. The mapping from sequence of

integers to permutations is somewhat complicated, but it

can be seen to produce each permutation in exactly one

way, by an immediate induction. When the selected

element happens to be the final remaining element, the

swap operation can be omitted. This does not occur

sufficiently often to warrant testing for the condition, but

the final element must be included among the

candidates of the selection, to guarantee that all

permutations can be generated.

The resulting algorithm for generating a random

permutation of a[0], a[1], ..., a[n − 1] can be described

as follows in pseudocode:

for i from n downto 2

do   di ← random element of { 0, ..., i − 1 }

swap a[di] and a[i − 1]

This can be combined with the initialization of

the array a[i] = i as follows:

for i from 0 to n−1

do   di+1 ← random element of { 0, ..., i }

a[i] ← a[di+1]

a[di+1] ← i

If di+1 = i, the first assignment will copy

an uninitialized value, but the second

will overwrite it with the correct

value i.

[edit]Generation in

lexicographic order

There are many ways to

systematically generate all

permutations of a given

sequence[citation needed]. One classical

algorithm, which is both simple and

flexible, is based on finding the next

permutation in lexicographic ordering,

if it exists. It can handle repeated

values, for which case it generates

the distinct multiset permutations

each once. Even for ordinary

permutations it is significantly more

efficient than generating values for

the Lehmer code in lexicographic

order (possibly using the factorial

number system) and converting those

to permutations. To use it, one starts

by sorting the sequence in

(weakly) increasing order (which

gives its lexicographically minimal

permutation), and then repeats

advancing to the next permutation as

long as one is found. The method

goes back to Narayana Pandita in

14th century India, and has been

frequently rediscovered ever since.[11]

The following algorithm generates the

next permutation lexicographically

after a given permutation. It changes

the given permutation in-place.

1. Find the largest index k such

that a[k] < a[k + 1]. If no

such index exists, the

permutation is the last

permutation.

2. Find the largest index l such

that a[k] < a[l]. Since k + 1 is

such an index, l is well

defined and satisfies k < l.

3. Swap a[k] with a[l].

4. Reverse the sequence

from a[k + 1] up to and

including the final

element a[n].

After step 1, one knows that all of the

elements strictly after position k form

a weakly decreasing sequence, so no

permutation of these elements will

make it advance in lexicographic

order; to advance one must

increase a[k]. Step 2 finds the

smallest value a[l] to replace a[k] by,

and swapping them in step 3 leaves

the sequence after position k in

weakly decreasing order. Reversing

this sequence in step 4 then produces

its lexicographically minimal

permutation, and the lexicographic

successor of the initial state for the

whole sequence.

[edit]Generation with minimal

changes

Main article: Steinhaus–Johnson–

Trotter algorithm

An alternative to the above algorithm,

the Steinhaus–Johnson–Trotter

algorithm, generates an ordering on

all the permutations of a given

sequence with the property that any

two consecutive permutations in its

output differ by swapping two

adjacent values. This ordering on the

permutations was known to 17th-

century English bell ringers, among

whom it was known as "plain

changes". One advantage of this

method is that the small amount of

change from one permutation to the

next allows the method to be

implemented in constant time per

permutation. The same can also

easily generate the subset of even

permutations, again in constant time

per permutation, by skipping every

other output permutation.[11]

[edit]Software implementations

[edit]Calculator functions

Many scientific calculators and

computing software have a built-in

function for calculating the number

of k-permutations of n.

Casio and TI calculators: nPr

HP calculators: PERM[12]

Mathematica: FallingFactorial

[edit]Spreadsheet functions

Most spreadsheet software also

provides a built-in function for

calculating the number of k-

permutations of n, called PERMUT in

many popular spreadsheets.

[edit]Applications

Permutations are used in

the interleaver component of the error

detection and correction algorithms,

such as turbo codes, for

example 3GPP Long Term

Evolution mobile telecommunication

standard uses these ideas (see 3GPP

technical specification 36.212 [13]).

Such applications raise the question

of fast generation of permutation

satisfying certain desirable properties.

One of the methods is based on

the permutation polynomials.

[edit]See also

Mathematics portal

Alternating permutation

Binomial coefficient

Combination

Combinatorics

Convolution

Cyclic order

Cyclic permutation

Even and odd permutations

Factorial number system

Superpattern

Josephus permutation

List of permutation topics

Levi-Civita symbol

Permutation group

Permutation pattern

Permutation polynomial

Probability

Random permutation

Rencontres numbers

Sorting network

Substitution cipher

Symmetric group

Twelvefold way

Weak order of permutations

Wikimedia Commons has 

media related 

to: Permutations

[edit]Notes

1. ̂  N. L. Biggs, The roots of

combinatorics, Historia Math.

6 (1979) 109−136

2. ̂  Charalambides, Ch A.

(2002). Enumerative

Combinatorics. CRC Press.

p. 42. ISBN 978-1-58488-290-

9.

3. ̂  Humphreys (1996), p. 84

4. ̂  Combinatorics of

Permutations, ISBN 1-58488-

434-7, M. Bona, 2004, p. 3

5. ̂  Combinatorics of

Permutations, ISBN 1-58488-

434-7, M. Bona, 2004, p. 4f

6. ̂  Combinatorics of

Permutations, ISBN 1-58488-

434-7, M. Bona, 2004, p. 43

7. ̂  Combinatorics of

Permutations, ISBN 1-58488-

434-7, M. Bona, 2004, p. 43ff

8. ^ a b D. E. Knuth, The Art of

Computer Programming, Vol

3, Sorting and Searching,

Addison-Wesley (1973), p. 12.

This book mentions the

Lehmer code (without using

that name) as a

variant C1,...,Cnof inversion

tables in exercise 5.1.1−7

(p. 19), together with two other

variants.

9. ̂  H. A. Rothe, Sammlung

combinatorisch-analytischer

Abhandlungen 2 (Leipzig,

1800), 263−305. Cited in,

[8] p. 14.

10. ̂  Fisher, R.A.; Yates, F.

(1948) [1938]. Statistical

tables for biological,

agricultural and medical

research (3rd ed.). London:

Oliver & Boyd. pp. 26–

27. OCLC 14222135.

11. ^ a b Knuth, D. E. (2005).

"Generating All Tuples and

Permutations". The Art of

Computer Programming. 4,

Fascicle 2. Addison-Wesley.

pp. 1–26. ISBN 0-201-85393-

0.

12. ̂  http://

h20331.www2.hp.com/

Hpsub/downloads/

50gProbability-

Rearranging_items.pdf

13. ̂  3GPP TS 36.212

[edit]References

Miklos Bona . "Combinatorics of

Permutations", Chapman Hall-

CRC, 2004. ISBN 1-58488-434-

7.

Donald Knuth . The Art of

Computer Programming,

Volume 4: Generating All Tuples

and Permutations, Fascicle 2,

first printing. Addison-Wesley,

2005. ISBN 0-201-85393-0.

Donald Knuth. The Art of

Computer Programming,

Volume 3: Sorting and

Searching, Second Edition.

Addison-Wesley, 1998. ISBN 0-

201-89685-0. Section 5.1:

Combinatorial Properties of

Permutations, pp. 11–72.

Humphreys, J. F.. A course in

group theory. Oxford University

Press, 1996. ISBN 978-0-19-

853459-4

CombinationFrom Wikipedia, the free encyclopedia

"Combin" redirects here. For the mountain massif, see Grand Combin.

For other uses, see Combination (disambiguation).

In mathematics a combination is a way of selecting several things out of a larger group, where

(unlike permutations) order does not matter. In smaller cases it is possible to count the number of

combinations. For example given three fruit, say an apple, orange and pear, there are three combinations of

two that can be drawn from this set: an apple and a pear; an apple and an orange; or a pear and an orange.

More formally a k-combination of a set S is a subset of k distinct elements of S. If the set has n elements the

number of k-combinations is equal to the binomial coefficient

which can be written using factorials as   whenever  , and which is zero

when  . The set of all k-combinations of a set S is sometimes denoted by  .

Combinations can refer to the combination of n things taken k at a time without or with repetitions.[1] In the

above example repetitions were not allowed. If however it was possible to have two of any one kind of fruit

there would be 3 more combinations: one with two apples, one with two oranges, and one with two pears.

With large sets, it becomes necessary to use more sophisticated mathematics to find the number of

combinations. For example, a poker hand can be described as a 5-combination (k = 5) of cards from a 52

card deck (n = 52). The 5 cards of the hand are all distinct, and the order of cards in the hand

does not matter. There are 2,598,960 such combinations, and the chance of drawing any one

hand at random is 1 / 2,598,960.

Contents

  [hide] 

1     Number of      k  -combinations   

o 1.1      Example of counting    

combinations

o 1.2      Enumerating      k  -combinations   

2     Number of combinations with repetition   

o 2.1      Example of counting    

multicombinations

3     Number of      k  -combinations for all      k  

4     Probability: sampling a random    

combination

5     See also   

6     References   

7     External links   

[edit]Number of k-combinations

3-element subsets of a 5-element set

Main article: Binomial coefficient

The number of   k -combinations  from a given set S of n elements is often denoted in elementary

combinatorics texts by C(n, k), or by a variation such as  ,  ,   or even   (the latter form is

standard in French, Russian, and Polish texts[citation needed]). The same number however occurs in many other

mathematical contexts, where it is denoted by   (often read as "n choose k"); notably it occurs as

coefficient in the binomial formula, hence its name binomial coefficient. One can define   for all

natural numbers k at once by the relation

from which it is clear that   and   for k > n. To see that these coefficients

count k-combinations from S, one can first consider a collection of n distinct variables Xs labeled by

the elements s of S, and expand the product over all elements of S:

it has 2n distinct terms corresponding to all the subsets of S, each subset giving the product of the

corresponding variables Xs. Now setting all of the Xsequal to the unlabeled variable X, so that the

product becomes (1 + X)n, the term for each k-combination from S becomes Xk, so that the

coefficient of that power in the result equals the number of such k-combinations.

Binomial coefficients can be computed explicitly in various ways. To get all of them for the

expansions up to (1 + X)n, one can use (in addition to the basic cases already given) the

recursion relation

which follows from (1 + X)n = (1 + X)n − 1(1 + X); this leads to the construction of Pascal's

triangle.

For determining an individual binomial coefficient, it is more practical to use the formula

The numerator gives the number of k -permutations  of n, i.e., of sequences of k distinct

elements of S, while the denominator gives the number of such k-permutations that give

the same k-combination when the order is ignored.

When k exceeds n/2, the above formula contains factors common to the numerator and

the denominator, and canceling them out gives the relation

This expresses a symmetry that is evident from the binomial formula, and can also

be understood in terms of k-combinations by taking the complement of such a

combination, which is an (n − k)-combination.

Finally there is a formula which exhibits this symmetry directly, and has the merit of

being easy to remember:

where n! denotes the factorial of n. It is obtained from the previous formula by

multiplying denominator and numerator by (n − k)!, so it is certainly inferior as a

method of computation to that formula.

The last formula can be understood directly, by considering the n! permutations

of all the elements of S. Each such permutation gives a k-combination by

selecting its first k elements. There are many duplicate selections: any

combined permutation of the first k elements among each other, and of the

final (n − k) elements among each other produces the same combination; this

explains the division in the formula.

From the above formulas follow relations between adjacent numbers in

Pascal's triangle in all three directions:

,

,

.

Together with the basic cases  , these allow

successive computation of respectively all numbers of

combinations from the same set (a row in Pascal's triangle), of k-

combinations of sets of growing sizes, and of combinations with a

complement of fixed size n − k.

[edit]Example of counting combinations

As a concrete example, one can compute the number of five-card

hands possible from a standard fifty-two card deck as:

Alternatively one may use the formula in terms of factorials

and cancel the factors in the numerator against parts of the

factors in the denominator, after which only multiplication of

the remaining factors is required:

Another alternative computation, almost equivalent to

the first, is based on writing

which gives

When evaluated as 52 ÷ 1 × 51 ÷ 2 × 50 ÷ 3

× 49 ÷ 4 × 48 ÷ 5, this can be computed

using only integer arithmetic. The reason that

all divisions are without remainder is that the

intermediate results they produce are

themselves binomial coefficients.

Using the symmetric formula in terms of

factorials without performing simplifications

gives a rather extensive calculation:

[edit]Enumerating k-combinations

One can enumerate all k-combinations of

a given set S of n elements in some fixed

order, which establishes a bijection from

an interval of   integers with the set of

those k-combinations. Assuming S is itself

ordered, for instance S = {1,2, ...,n},

there are two natural possibilities for

ordering its k-combinations: by comparing

their smallest elements first (as in the

illustrations above) or by comparing their

largest elements first. The latter option has

the advantage that adding a new largest

element to S will not change the initial part

of the enumeration, but just add the

new k-combinations of the larger set after

the previous ones. Repeating this process,

the enumeration can be extended

indefinitely with k-combinations of ever

larger sets. If moreover the intervals of the

integers are taken to start at 0, then the k-

combination at a given place i in the

enumeration can be computed easily

from i, and the bijection so obtained is

known as the combinatorial number

system. It is also known as

"rank"/"ranking" and "unranking" in

computational mathematics.[2][3]

[edit]Number of combinations with repetition

See also: Multiset coefficient

Bijection between

3-element multisets with elements from a 5-

element set (on the right)

and 3-element subsets of a 7-element set (on

the left)

A k-combination with repetitions, or k-

multicombination, or multiset of

size k from a set S is given by a sequence

of knot necessarily distinct elements of S,

where order is not taken into account: two

sequences of which one can be obtained

from the other by permuting the terms

define the same multiset. In other words,

the number of ways to sample k elements

from a set of n elements allowing for

duplicates (i.e., with replacement) but

disregarding different orderings (e.g.

{2,1,2} = {1,2,2}). If S has n elements, the

number of such k-multicombinations is

also given by a binomial coefficient,

namely by

(the case where both n and k are zero

is special; the correct value 1 (for the

empty 0-multicombination) is given by

left hand side  , but not by the

right hand side  ).

[edit]Example of counting multicombinations

For example, if you have ten types of

donuts (n = 10) on a menu to choose

from and you want three donuts

(k = 3), the number of ways to choose

can be calculated as

The analogy with the k-

combination case can be

stressed by writing the numerator

as a rising power

There is an easy way to

understand the above result.

Label the elements of S with

numbers 0, 1, ..., n − 1, and

choose a k-combination from

the set of numbers { 1,

2, ..., n + k − 1 } (so that

there are n −

1 unchosen numbers). Now

change this k-combination

into a k-multicombination

of S by replacing every

(chosen) number x in the k-

combination by the element

of Slabeled by the number of

unchosen numbers less

than x. This is always a

number in the range of the

labels, and it is easy to see

that every k-

multicombination of S is

obtained for one choice of

a k-combination.

A concrete example may be

helpful. Suppose there are 4

types of fruits (apple,

orange, pear, banana) at a

grocery store, and you want

to buy 12 pieces of fruit.

So n = 4 and k = 12. Use

label 0 for apples, 1 for

oranges, 2 for pears, and 3

for bananas. A selection of

12 fruits can be translated

into a selection of 12 distinct

numbers in the range 1,...,15

by selecting as many

consecutive numbers

starting from 1 as there are

apples in the selection, then

skip a number, continue

choosing as many

consecutive numbers as

there are oranges selected,

again skip a number, then

again for pears, skip one

again, and finally choose the

remaining numbers (as

many as there are bananas

selected). For instance for 2

apples, 7 oranges, 0 pears

and 3 bananas, the numbers

chosen will be 1, 2, 4, 5, 6,

7, 8, 9, 10, 13, 14, 15. To

recover the fruits, the

numbers 1, 2 (not preceded

by any unchosen numbers)

are replaced by apples, the

numbers 4, 5, ..., 10

(preceded by one unchosen

number: 3) by oranges, and

the numbers 13, 14, 15

(preceded by three

unchosen numbers: 3, 11,

and 12) by bananas; there

are no chosen numbers

preceded by exactly 2

unchosen numbers, and

therefore no pears in the

selection. The total number

of possible selections is

[edit]Number of k-combinations for all k

See also: Binomial

coefficient#Sum of

coefficients row

The number of k-

combinations for

all k, 

, is the sum of the nth

row (counting from 0) of

the binomial

coefficients. These

combinations are

enumerated by the 1

digits of the set of base

2 numbers counting

from 0 to  ,

where each digit

position is an item from

the set of n.

[edit]Probability: sampling a random combination

There are

various algorithms to

pick out a random

combination from a

given set or

list. Rejection

sampling is extremely

slow for large sample

sizes. One way to select

a k-combination

efficiently from a

population of size n is to

iterate across each

element of the

population, and at each

step pick that element

with a dynamically

changing probability

of 

.

[edit]See also

Combinatorial

number system

Combinatorics

Multiset

Binomial coefficient

Permutation

List of permutation

topics

Subset

Probability

Pascal's Triangle

[edit]References

1. ̂  Erwin

Kreyszig, Adva

nced

Engineering

Mathematics,

John Wiley &

Sons, INC,

1999

2. ̂  http://

www.site.uotta

wa.ca/~lucia/

courses/5165-

09/

GenCombObj.p

df

3. ̂  http://

www.sagemath.

org/doc/

reference/

sage/

combinat/

subset.html

[edit]External links

C code to generate

all combinations of

n elements chosen

as k

Many Common

types of

permutation and

combination math

problems, with

detailed solutions

The Unknown

Formula For

combinations when

choices can be

repeated and order

does NOT matter

[1]  Combinations

with repetitions (by:

Akshatha AG and

Smitha B)