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)
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