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Members: -Aicha Bakkali Kasmi. -Nicolás A. Peña Moreno. -Fco. David Pérez Rodríguez. -Victoria Prados Fernández. 4ºE//04-02-2013//GRUPO 3

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Members: -Aicha Bakkali Kasmi. -Nicolás A. Peña Moreno. -Fco. David Pérez Rodríguez. -Victoria Prados Fernández. 4ºE//04-02-2013//GRUPO 3


ORIGIN OF THE COMBINATORICS Combinatorics is the part of mathematics that is dedicated to find procedures and strategies for counting the elements of a set or how to group items in a set. We will learn how to make these clusters. The origin of combinatorics has been parallel to algebra, the number theory and probability. Since long time ago there are combinatorial problems that have caught the attention of mathematicians: - The magic squares problem: they are arrays of numbers with the property that the sum of the elements of any column, line or diagonal is the same number. - The binomial coefficients: integer coefficients, which are the development of ( + ) were known in the 12th century. - Pascal's triangle: it is a triangular disposal of the binomial coefficients that was developed in the 13th century. It’s thought that the combinatorics emerges in the 17th century with the work of Blaise Pascal and Pierre Fermat on gambling theory. This work, which laid the foundations of the probability theory, also contained the principles for determining the number of combinations of elements of a finite set, and so established the traditional connection between combinatorial and probability.

METHODS FOR COUNTING Counting methods are strategies used to determine the number of possible results to perform an experiment. Tree Diagram The tree diagram is a graphic method that consists of counting mark, like paths or branches of a tree. The number of possibilities is obtained by counting the final branches. For instance, if we want to know how many forms are to organize the numbers 1, 2 and 3 in groups of three numbers without repetition, you have to do this:

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The first event has three outcomes, so you draw three branches.

The second has two (without repetition) so at the end of each one of the first step you have to draw two branches.

You have to count the different solutions that you have obtained.

Method Product The method product is a method m of counting that consists of decomposing the experiment in other experiments simpler and multiplying the number of possibilities of each one. For example: A taco bar sells burritos with the following fillings: fillings: pork, grilled chicken, chicken mole, and beef. Burritos come small, medium, or large, with or without cheese, and with or without guacamole. How many different burritos can be ordered? Fillings: + + + 4. Weight: + + 3. Cheese: + 2. Guacamole: + 2. Experiments simpler: 1. 4 3 12 → . 2. 2 2 4 → . 3. 12 4 48 → ( ) ( ) ). Solution:: 48 different burritos can be ordered.

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FACTORIAL NUMBERS AND BINOMIAL COEFFICIENTS If we have n, a natural number greater than 1, factorial of n (n!), is the product of n by all natural numbers less than it. !=

·

−1 ·

− 2 …1

Considering that all products are at least two factors are meaningless symbols 0! and 1! But formulas apply to all cases defined factorial numbers 0 and 1 as 0! = 1 and 1! = 1. Combinatorial numbers are used to base groups in which the order does not matter and the elements cannot be repeated, namely to directly calculate combinations. Given two natural numbers m and n such that m ≥ n we define the combinatorial

numbers n on m and it is written like that:

*+ = , - =

! −

!

BINOMIAL THEOREM We use this theorem to calculate raise binomials to powers of the form (a +b) n, the called Newton’s binomial: + This formula is equal to: ±

=, 0

±, 1

23

= 0, ℎ

12 2

234

17

+, 2

18 8

± ⋯± , -

Let’s see how the powers of (a + b)n carry out: (a + b)1= a + b (a + b)2= a2 + 2ab + b2 (a + b)3= a3 + 3a2b + 3ab2 + b3 (a + b)4= a4 + 4a3b + 6a2b2 + 4ab3 + b3 Observing the power of (a + b)n, we notice that the powers of a, begins raised to n, reducing the value of n gradually until zero; while the exponents of b are increasing.

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Looking at the coefficient of each resultant polynomial, we can see that they follow the sequence known as the Tartaglia’s triangle, which is obtained writing in lines the binomial coefficient starting with n=0. This means that each number belongs to the values of one combinatorial number, like this:

n=0

1 n=1

1

1 n=2

1

2

1 n=3

1

3

1

3

n=4

4

1

6

4

1

Notice that each number is the sum of the two numbers which are above it. · Examples: 1. - Calculate those binomials: a)( + 2: ; = <;4= ; + <;7= > · 2: + <;8= 2:

;

· 2: 8 + <;?= 8 · 2: = ; + 10 > : + 40 ? : 8 + 80 8 : ? + 80 : > + 32: ; ?

= <>4=2> + <>7=2? · 3: + <>8=28 · 3: 216: 8 + 216: ? + 81: >

b) 2 + 3:

>

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8

+ <>?=2 · 3:

?

+ <;>= · 2:

>

?

+ <>>= · 3:

= 16 + 96: +

>

+ <;;= ·

Members: -Aicha Bakkali Kasmi. -Nicolás A. Peña Moreno. -Fco. David Pérez Rodríguez. -Victoria Prados Fernández. 4ºE//04-02-2013//GRUPO 3


Does the order affect? YES

NO

Do all the elements enter?

Do all the elements enter?

YES

NO

NO

Can they be repeated?

Can they be repeated?

Can they be repeated?

YES

NO

YES

NO

YES

NO

Permutations with repetition

Permutations without repetition

Variations with repetition

Variations without repetition

Combinations with repetition

Combinations without repetition

Pn=n(n-1)(n-2)...3·2·1=n!

VRm,n=mn

Vm,n=m(m-1)(m-2)...(m-n+1)

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VARIATIONS Variations without repetitions of m elements taken n by n (n<m) are the different groups that could be formed with the m elements, so: • In each group enters n different elements. • Two groups are different if they differ in some element or in the position of their elements. The number of these variations of m elements taken n by n is represented by B+ , and it is equal to: !

B+

( − )!

The variations with repetition of n numbers taken m by m,BC groups that could be formed with the m elements so:

+

, the different

BC+

· Examples: 1. The final of the long jump will be celebrated in a championship. 8 people take part. How many ways could be the three medals (golden, silver, bronze) shared out? Available elements: 8 athlete, m= 8. Elements each group: 3 medals, n= 3 1. - Does the order of elements’ position affect? Yes, because it’s different to get golden, silver or bronze. 2. - Do we catch all the available elements? m= 8, n= 3 No, we just catch 3 of them, so they are variations. 3. - Could the elements be repeated? No, because nobody can take more than one medal. Variations without repetitions of 8 elements, m, taken 3 by 3, n. BD?

8 8 − 1 8 − 2 … 8 − 3 + 1 = 8 · 7 · 6 = 336

.

Solution: The medals can be shared out in 336 forms. 2. The system of registers consists of 4 numbers followed by 3 consonant letters. How many registers could we have with a particular block of letters? We have 22 consonants, m= 22. We form groups of 3 letters, n= 3 1. Does the order of elements’ position affect? Yes, if we change the order we’ll get new registers.

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2. Do we catch all the available elements? m= 22, n= 3. No, we just ust take 3 of them. 3. Could the elements be repeated? Yes, they could. Variations with repetitions of 22 elements, m, taken 3 by 3. ? BC88

22?

10648.

Solution: We can have 10648 different registers. 3. From a team of 5 racers, A, B, C, D, E, we choose 3 racers. The order of the selected selec racers has importance, so it is a variation of 5 elements where we choose 3 of them. a) The number ber of such choices isGIH . B;? 5 % 4 % 3 60. wit the four letters a, b, c, d? b) How many words of two letters can you make with BC>8

48

16

.

c) How many three-digit digit numbers can be formed with the digits 1, 2, 3, 4, 5? BC;?

5?

125

.

PERMUTATIONS

1

We call permutation of n elements to the different associations of these n elements, where all the elements enter and the order is important. ·Permutations Permutations without repetition of n elements PERMUTATION RESULT are the different kinds that they could be ABC formed, so: ACB · In each group there are the n elements. BAC · One group iss different to another one just BCA in the position on order of the elements. CAB The number of these permutations of n CBA elements is represented like Pn Pn, and this is:

F

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( − 1)( − 2) … 3 % 2 % 1

!

Meembers: -Aiicha Bakkali Kasmi. -N Nicolás A. Peña Moreno. -Fcco. David Pérez Rodríguez. -Vi Victoria Prados Fernández. 4ººE//04-02-2013//GRUPO 3


Permutations with repetition of n elements, with one of them repeated a times, and other b times, and so until the last one, which is repeated k times (a + b + … + k = n), are the different groups that can be formed with these n elements, in such way that one group is different to another one just in the position order of its elements.

FC Q,R,…S =

! ! !… !

·Problems: 1. In fact there is an easy way to work out how many ways "1 2 3" could be placed in order, and we have already talked about it. The answer is: F? 3! = 3 · 2 · 1 = 6 2. How many different ways can sit 8 people in a theatre seat line? - All the elements enter, because the 8 people must sit. - The order affects. - The elements can´t be repeated, because a person can´t be repeated. FD = 8! = 40320 3. With the numbers 2, 2, 2, 3, 3, 3, 3, 4, 4; how many numbers of 9 digits can be obtained? a=3 b=4 c=2 m=a+b+c=9 -All the elements enter. -The order is important. -The elements can be repeated. 9! FCN?,>,8 = = 1260 3! · 4! · 2! 4. What order could 16 pool balls be in? After choosing, say, number "14" you can't choose it again. So, your first choice would have 16 possibilites, and your next choice would then have 15 possibilities, then 14, 13, etc. And the total permutations would be: P7P = 16! = 16 · 15 · 14 . . .· 3 · 2 · 1

20,922,789,888,000.

COMBINATIONS without repetition The ordinary combinations or without repetition of m elements n by n (n ≤ m) are the different groups that can be formed with the m elements, so that: -In each group there are n different elements. -Two groups are different if they have some different element. The number of ordinary combinations of m elements taken n by n is represented by *+ , and is equal to:

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*+

, -

! ! % ( − )!

·Problems: 1. Out of a collection of twenty books, you take four books to read them at home. We'll calculate the number of choices. The order of the selected books has no importance. The number of such choices is: 84! 84%7N%7D%7T > *84 <84 = 4845 . > >!%(841>)!

>%?%8%7

2. With the numbers 1, 2, 3, 4, 5, 6 and 7 how many different products can I get if I take 2 in two? *T8 =

7! 7 · 6 · 5! 7 · 6 = = = 21 2! 5! 2 · 5! 2

.

3. How many ways are there to mix the seven colors of the rainbow taking them three at a time?

*T? =

7! 7·6·5 = = 35 3! · 4! 3·2

: .

4. In a class of 10 pupils they are going to distribute three prizes. In how many ways can they do it if the prizes are equal? ? *74 =

10! 10! 10 · 9 · 8 = = = 120 10 − 3 ! · 3! 7! · 3! 3·2·1

U .

SOME PROBLEMS SOLVED WITH THE HELP OF COMBINATORICS 1) We have 3 books: one of math (A), one of biology (B), and another of history (C), and we want to see how many ways there are to tidy them in a book self. At the beginning we can choose which of them take first: 1a 2a 3a A B C Once chosen the first, still 2 more to locate.

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Now, there are 3×2 2 different ways to tidy the books. But once arranged the two first we know which the last is. So the number of possible ways to tidy the 3 books could be calculate as: 3×2×1=6 2) With the numbers 3, 5, 8, 9; how many different numbers of three numbers could cou be formed which are higher than 600? FIRST: We examine the result which achieves the condition. If the digit of 3 digit that we form must be higher than 600, it has to be begin with 8 or 9. So the numbers searched are going to be: 8ab-------------- where a and b could be 3, 5 or 9 9ab-------------- where a and b could be 3, 5 or 8 SECOND: We calculate the possibilities. In both cases, the order affects, so they are variations. In both cases there are 3 elements elements which are grouped in two too.

B?8

3! 3&2 !

3! 1!

6

Solution: there will be 6 numbers which begin with 8 and another 6 numbers which begin with 9.

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APPLICATIONS OF THE COMBINATORICS IN EVERYDAY LIFE The combinatorics is used to know how many ways there are to combine a number of things. Some examples: How many ways are there to combine the teams in a football tournament doing that each team plays with all the others? How many cars can you register with the same system of combination? Which 5 people out of 50 could be chosen to form a jury? How many possible combinations are there to the winning lottery ticket? Is a game of chance profitable? Combinatorics, the mathematics of finite (or countable) structures is often used in computer science and communications. Also give you powerful methods for modeling and solving large scheduling and management problems, such as the optimization of flight schedules, the planning of equipment maintenance, and the scheduling of staff. Besides, it’s used in areas as diverse as cryptography, electronic security, planning the efficient layout of a factory floor, synchronizing traffic lights, and improving telecommunications signals. Also, we can use them for example if we go to a travel and we will visit six cities and we want to know how many forms of do that we have, we use them (FP 6! = 720 V ). There are thousands and thousands of applications more could help us resolve some of everyday problems.

·Glossary: - Cluster: a number of things growing, fastened, or occurring close together. Synonym= group. Translation: grupo. - Array: an impressive display or collection. Synonym= formation. Traduction: formación. - Gambling theory: it’s a theory who studies the possibilities of a game of chance. .Translation: teoría de los juegos de azar. - Namely: Translation: es decir. - Register: Translation: matrícula. (Registration number: matrícula de coche)

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· Bibliography: Newton’s binomial ·http://platea.pntic.mec.es/anunezca/ayudas/newton/binomio_de_newton.htm ·http://www.vitutor.com/pro/1/a_11.html

Combinatorics ·http://recursostic.educacion.es/descartes/web/materiales_didacticos/Combinatoria/vari acionescon.htm

Variations ·http://www.vadenumeros.es/sociales/variaciones.htm

Permutations ·http://www.disfrutalasmatematicas.com/combinatoria/combinacionespermutaciones.html ·http://guiamath.sytes.net/ejercicios_resueltos/01_04_05_01Tec_Conteo_Permutaciones/0_permutaciones.html ·http://descartes.cnice.mecd.es/Algebra/combinatoria_jjce/combinatoria_2.htm

Problems ·http://www.unlu.edu.ar/~dcb/matemat/combina1.htm ·http://webpages.ull.es/users/jjsalaza/curriculum/books/GOBCAN02.pdf

·ANZOLA, Máximo, DE LOS SANTOS, Isabel, CARLOS HERVÁS, Juan, RAMÓN VIZMANOS, José, Libro de Matemáticas de 4º ESO, opción A, Editorial SM, Madrid, 2008. ·ÁLVAREZ, María Dolores, GAZTELU, Ana María, GONZÁLEZ, Augusto..., Libro de Matemáticas de 4º ESO, opción B, Editorial Santillana, Madrid, 2008.

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Combinatorics