Combinative

In Mathematical, the combinative , also called analyzes combinative , studies the configurations of finished collections of objects or the Combinaison S of finished units, and the enumerations.

In particular the combinative one is interested in the methods making it possible to count the elements in finished units (combinative énumérative) and in the research of the optima in the configurations like with their existence (combinative extreme).

The combinative one begins at the 17th century, at the same time as the calculation of the Probabilités. Initially, this part had as an aim the resolution of the problems of enumeration, coming from the study of the games of chance. It developed to a significant degree under the influence of the probability theory. Later, it bound to the Théorie of the numbers and the Graph theory.

Here some examples of situations giving place to questions of combinative analysis:

• arrangements of books on a rack;
• provisions of people around a roundtable;
• pullings with handing-over of a certain number of balls numbered in a ballot box;
• placements of tokens on a checkerwork.

Which is the number of possible schedulings of the charts of one pack of 52 cards?

This number is equal to 52! (it “! ” indicating the Factorial ). It can seem astonishing that this number, approximately 8,065 817.517.094 × 10⁶⁷, that is to say so large. They is approximately 8 follow-up of 67 zeros. It is, for example, larger than the Nombre of Avogadro, equal to 6,022 × 10²³ (many carbon 12 atoms in 12 grams of carbon 12).

Permutations (provisions, schedulings)

Permutations (without repetition) of discernible objects

Example of introduction, let us consider the number of provisions of six discernible objects in six consecutive boxes numbered with one and only one object by box. Each object can be placed in the first box, which gives six possibilities of occupying the first place. Once the first place occupied by one of the objects, he remains still five candidates for the second place, the second place being allotted, there remain only four candidates for the third place, and so on. For the penultimate place, it does not remain any more that two objects, and once one of both placed, the last place must be occupied by the last object.

There are thus 6 × 5 × 4 × 3 × 2 × 1 or 6! = 720 possibilities of laying out six discernible objects.

; Generalization:

We will see that the number of provisions of N discernible elements is equal to N !

A provision of the objects of a unit E of cardinal N , in N boxes with one and only one object by box, or a scheduling of the elements of E is represented by a bijection of {1, 2,…, N } in E or a Permutation of E . It is convenient to represent such a bijection by a '' N '' - uplet (or N - list) of elements of E , (x₁, x₂,…, x_n).

; Theorem: There is N ! permutations (without repetition) of N elements.

Indeed, to form a '' N '' - uplet of elements of E , we must choose an element of E for the first place of the '' N '' - uplet and there is not N possibilities, there is N - 1 possible choices of an element of E for the second place, N - 2 for the third etc It is not any more but one choice of element for the last place. Thus on the whole N × ( N -1) × ( N -2) ×… × 2 × 1 permutations.

This property is shown by Récurrence on N .

Permutations with repetition of discernible objects

To determine the number of the provisions possible of objects of several classes and mutually indistinguishable in each class, it is useful to consider the number of possible provisions of these discernible objects by supposing them all, and then to find how much these provisions are indistinguishable. The number of the possible provisions of these objects is equal to the number of possible provisions of the objects considered as discernible divided by the number of the indistinguishable provisions.

For example, if we must determine the full number of provisions of objects of which two are of a first class, three of a second class and five of a third class, then we calculate the full number of provisions of these objects considered as discernible, which gives (2 + 3 + 5)! , that is to say 3.628.800 possible provisions. But certain provisions remain unchanged when the indistinguishable objects of the same class are exchanged mutually, and there are 2! × 3! × 5! that is to say 1.440 ways of permuting the objects of each one of these classes.

We obtain 3.628.800 ÷ on the whole 1.440 = 2.520 different provisions. It is also about the number of permutations with repetition of 10 elements with 2,3 and 5 repetitions.

; Generalization:

The number of permutations of N elements, distributed in K classes whose N ₁ is of class 1, N ₂ are of class 2,…, N _k are of class K , indistinguishable in each class, or the number of permutations of N elements with N ₁, N ₂,…, N _k repetitions, with $\backslash \; left\; (\backslash \; sum^k\_\; \{i=1\}\; n\_i\; =\; N\; \backslash \; right)$, is equal to: $\backslash \; frac\; \{N!\}\{n\_1!\; n\_2!\; \backslash \; ldots\; n\_k!\}$.

Arrangements (choice by holding suspense account)

Arrangements without repetition

We have N discernible objects and we want to place K of it, by holding suspense account, in K boxes numbered of 1 with K with one and only one object by box. The number of provisions is then equal to the number of K - distinct lists formed starting from these objects. Instead of constituting a '' N '' - uplet, starting from N discernible objects, we form here '' K '' - uplet S with K ≤ N , ( X ₁, X ₂,…, X _K )--> $k\; \backslash \; N\; \backslash \; text\; \{,\}\; \backslash \; left\; (x\_1,\; x\_2,\; \backslash \; ldots,\; x\_k\; \backslash \; right)$ starting from these N objects such as for $i\; \backslash \; neq\; j$, one has $x\_i\; \backslash \; neq\; x\_j$. Such a '' K '' - uplet is called a Arrangement without repetition of N elements taken K with K .

; Theorem: The number of arrangements without repetition of N elements taken K with K is equal to $A\_n^k$ (equal to $\backslash \; frac\; \{N!\}\{(n-k)!\}$ if K ≤ N and to 0 if not). Indeed, There is possible N choice object which occupies the first place of the '' K '' - uplet, N -1 choice for the object of the 2nd place; for the K E, there remain nothing any more but N - ( K -1) objects and thus N - K +1 choices possible. The product N ( N -1)… ( N - K +1)--> $n\; \backslash \; cdot\; (n-1)\; \backslash \; ldots\; (n-k+1)$ is written well in the form: $\backslash \; frac\; \{N!\}\{(n-k)!\}$.

The case N = K then obliges us to divide by (0)! that one defines as being worth 1.

Arrangements with repetition

When we want to place objects taken among N discernible objects in K sites by holding suspense account, these objects being able to appear several times, the number of provisions is then equal to the number of '' K '' - uplets formed starting from these N objects. Such a '' K '' - uplet, with K ≤ N , ( X ₁, X ₂,…, X _K ) formed starting from these N objects is called a Arrangement with repetition of N elements taken K with K .

As each site can be indifferently occupied by any of these N objects, there is of it on the whole N ^K .

When we draw 11 times one from 3 numbers by holding account about appearance we obtain 3¹¹ on the whole = 177  147 different pullings. Like example drawn from the genetics, we can give the full number of Codon S basic (formed triplets of four codes): 4³= 64.

Combinations (choice without holding suspense account)

Contrary to arrangements, the combinations are provisions of objects which do not hold account about placement of these objects. For example, if has , B and C is balls drawn from a ballot box, ABC and acb corresponds to the same pulling. There are thus less combinations than of arrangements.

Combinations without repetition

If we draw without handing-over K objects among N discernible objects, and lay out we them without holding account about appearance, we can represent these K objects by a part with K elements of a whole with N elements. They are combinations without repetition of N elements taken K with K .

To determine the number of these provisions, we can determine the number of arrangements of K objects and divide by the number of provisions obtained the ones from the others by a permutation. There is $\{N\; \backslash \; choose\; K\}\; =\; \backslash \; frac\; \{A\_n^k\}\; \{K!\}$ (for the notation, to also see the article on the binomial Coefficient).

With the play of the Lotto, we must choose among 49 different numbers, 6 numbers without holding suspense account, and there is $\{49\; \backslash \; choose\; 6\}\; =\; \backslash \; frac\; \{49!\}\{6!\; \backslash \; cdot\; 43!\}=13\; \backslash ,\; 983\; \backslash ,\; 816$ possible choices.

On the same principle, the play “euro-million” requires to choose 5 different numbers between 1 and 50 and 2 numbers between 1 and 9, that is to say $\{50\; \backslash \; choose\; 5\}\; \backslash \; times\; \{9\; \backslash \; choose\; 2\}\; =76\; \backslash ,\; 275\; \backslash ,\; 360$ possibilities.

Combinations with repetition

If we draw with handing-over K objects among N discernible objects, and we them let us lay out without holding account about appearance; these objects can appear several times and we can represent them neither with a part with K elements, nor with a '' K '' - uplet since their order of placement does not intervene. It is however possible to represent such provisions with applications called combinations with repetition.

The number of combinations with repetition of N elements taken K with K is equal to: $\backslash \; Gamma\_n^k=\; \{n+k-1\; \backslash \; choose\; K\}$.

Let us give the example of the play of domino. The parts are manufactured by laying out side by side two elements of the white {unit, 1,2,3,4,5,6}. If we turn over a domino, we change the order of the two elements, but the domino remains identical. We have a combination with repetition of 8 elements taken 2 to 2, and on the whole it there a: $\backslash \; Gamma\_7^2=\; \{8\; \backslash \; choose\; 2\}\; =28$ dominos in a play.

Function of counting

That is to say Ϭ _N the whole of the permutations of {1, 2,…, N }. We can consider the function which with N associates the number of permutations. This function is the factorial function and is used to count the permutations.

Being given an infinite collection of finished units E _N / N ∈ℕ}--> $\backslash \; left\; \backslash \; \{E\_n\; |\; N\; \backslash \; in\; \backslash \; mathbb\; \{NR\}\; \backslash \; right\; \backslash \}$ indexed by the whole of the natural entireties, a function of counting is a function which with an entirety N associates the number of elements of E _N . A function of counting F thus makes it possible to count the objects of E _N for any N . The elements of E _N have usually a relatively simple combinative description and an additional structure, often making it possible to determine F .

Certain functions of counting, are given by “closed” formulas, and can be expressed like made up of elementary functions such as factorials, powers, and so on.

This approach can not be entirely satisfactory (or practical) for certain combinative problems. For example, that is to say F ( N ) the number of subsets distinct from integers in the interval '' N '' which do not contain two consecutive integers. For example, with N = 4, we obtain ∅, {1}, {2}, {3}, {4}, {1,3}, {1,4}, {2,4}, and thus F (4) = 8. It proves that F ( N ) is N ème number of Fibonacci, which can be expressed in the “closed” form following:

$f\; (N)\; =\; \backslash \; frac\; \{\backslash \; phi^n\}\; \{\backslash \; sqrt\; \{5\}\}\; -\; \backslash \; frac\; \{(1\; \backslash \; phi)\; ^n\}\; \{\backslash \; sqrt\; \{5\}\}$

where φ = (1 + √5) /2, is the golden section. However, since we consider whole of integers, the presence of the √5 in the result can be regarded as unaesthetic from a combinative point of view. Also F ( N ) can it be expressed by a relation of recurrence:

F ( N ) = F ( N - 1) + F ( N - 2)

what can be more satisfactory (from a purely combinative point of view), since the relation shows more clearly how the result was found.

In certain cases, an asymptotic equivalent G of F ,

F ( N ) ~ G ( N ) when N tends towards infinite the

where G is a “familiar” function, makes it possible to obtain a good approximation of F . A simple asymptotic function can be preferable with a “closed” formula extremely complicated and which informs little on the behavior of the number of objects. In the example above, an asymptotic equivalent would be:

$f\; (N)\; \backslash \; sim\; \backslash \; frac\; \{\backslash \; phi^n\}\; \{\backslash \; sqrt\; \{5\}\}$

when N becomes large.

Another approach is that of the whole series. F ( N ) can be expressed by a formal whole series, called generating function of F , which can be most usually:

• the ordinary generating function

$\backslash \; sum\_\; \{N\; >\; 0\}\; F\; (N)\; \backslash \; cdot\; x^n$

• or the exponential generating function

$\backslash \; sum\_\; \{N\; >\; 0\}\; F\; (N)\; \backslash \; cdot\; \backslash \; frac\; \{x^n\}\; \{N!\}$ Once determined, the generating function can make it possible to obtain all the furnished informations by the preceding approaches. Moreover, the various usual operations like the addition, the multiplication, derivation, etc, have a combinative significance; and this makes it possible to prolong results of a combinative problem in order to solve other problems.

Some results

A theorem, due to Franck P. Ramsey, gives a surprising result. With one evening to which at least six people go, there are at least three people who know each other mutually or at least three which is foreign the ones with the others.

; Demonstration: either $P\_1$ an unspecified person presents to the evening. On the n-1 others, either she knows some to more both, or she knows at least three of them. Let us suppose that one is in the second case, and are $P\_2,\; \backslash ,\; P\_3\; \backslash \; text\; \{and\}\; P\_4$ three people known of $P\_1$. If two of enter they know, let us put $P\_2\; \backslash \; text\; \{and\}\; P\_3$, then $P\_1,\; \backslash ,\; P\_2\; \backslash \; text\; \{and\}\; P\_3$ know themselves all three. If not, $P\_2,\; \backslash ,\; P\_3\; \backslash \; text\; \{and\}\; P\_4$ do not know themselves at all, and the announced result is still right. In the other case of figure ($P\_1$ knows with more the two people of the group), the same reasoning, reversed, functions with the three people that $P\_1$ does not know.

(It is a particular case of the theorem of Ramsey.)

The idea to find an order in random configurations leads to the theory of Ramsey. Primarily, this theory indicates that any sufficiently large configuration will contain at least another type of configuration.

See too

Related articles

• Mathematical discrete.
• Word without square factor.
• Principle of inclusion-exclusion of Moivre.
• binomial Coefficient.

External bonds

• history of the economic thinking: in connection with Frank P. Ramsey.

References

• Richard P. Stanley, Enumerative Combinatorics , volumes 1 and 2, Cambridge University Near, resp. 1997 (ISBN 0-521-55309-1) and 1999 (ISBN 0-521-56069-1).

• Giacomo Lorenzoni, The solution off has combinatorial problem (file pdf, considering the 12/13/2006).

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