Presentation on the topic "combinations". Presentation for a lesson on algebra and principles of analysis on the topic "Combinatorics: movements, permutations, combinations" Combinations with repetitions

“Combinatorics problems” - How many ways can one choose one book? In how many ways can a ship's crew be formed, consisting of a commander and an engineer? Combinatorics. Problem No. 2. K. Rule of addition Rule of multiplication. Sum rule. Solution: 30 + 40 = 70 (in ways). Task No. 1. Problem No. 3. I. Let there be three candidates for the post of commander and 2 for the post of engineer.

“Placement of elements” - Combinatorics. Accommodation. Placement and combination. Formulas: For any natural numbers n and k where n>k, the equalities are valid: For the number of choices of two elements from n data: Combination. In combinatorics, a combination of n through k is a set of k elements selected from given n elements.

“Statistical characteristics” - Mathematical statistics, etc. Statistical research. 5. What are statistics? 3. 9. Arithmetic mean Range Mode Median. Stages of research activities. 2. 14. “There are three types of lies: ordinary lies, blatant lies and statistical lies. "

“Combinations” - There are letters A, B, C, D. make all combinations of only two letters. Independent work consisted of 2 tasks. The problem was solved correctly by 13 students, and the example was 17. 3 students failed to complete the work. Combinatorial problems. Task No. 1. How many students successfully solved independent work. The work took 30 students to write.

“Permutations of elements” - Direct algorithm for lexicographic enumeration of permutations. Combinatorics. The greatest increasing subsequence problem. Numbering of the set. Formal description of the algorithm. Rearrangements. Theorem on lexicographic enumeration of permutations. Enumeration of permutations. Enumeration of permutations by elementary transpositions.

“Combinatorics 9th grade” - Out of 30 participants, the meeting must choose a chairman and secretary. Solution: a) 3! = 1 · 2 · 3 =6 b) 5! = 1 · 2 · 3 · 4 · 5 = 120. II. Designation: P n Formula for calculating permutations: P n = A6 10 =n ·(n -1) · (n-2) · … · 3 · 2 · 1=n! 2nd group. Designation: Formula for calculating combinations: *. Answers and solutions. 2nd group.

There are a total of 25 presentations in the topic

COMBINATORICS

Lesson objectives:

Find out what combinatorics studies
Find out how combinatorics arose
Study combinatorics formulas and learn how to apply them when solving problems

The birth of combinatorics as a branch of mathematics is associated with the works of Blaise Pascal and Pierre Fermat on the theory of gambling.

Blaise Pascal

Pierre Fermat

A great contribution to the development of combinatorial methods was made by G.V. Leibniz, J. Bernoulli and L. Euler.

G.V. Leibniz

L. Euler.

J. Bernoulli

Lemma. Let the set A have m elements, and the set B have n elements. Then the number of all distinct pairs (a,b), where a\in A,b\in B will be equal to mn. Proof. Indeed, with one element from the set A we can make n such different pairs, and in total there are m elements in the set A.

Placements, permutations, combinations Let us have a set of three elements a,b,c. In what ways can we select two of these elements? ab,ac,bc,ba,ca,cb.

Rearrangements We will rearrange them in all possible ways (the number of objects remains unchanged, only their order changes). The resulting combinations are called permutations, and their number is equal to Pn = n! =1 · 2 · 3 · ... · ( n-1) n

The symbol n! is called factorial and denotes the product of all integers from 1 to n. By definition, it is believed that 0!=1 1!=1 An example of all permutations of n=3 objects (different figures) is in the picture. According to the formula, there should be exactly P3=3!=1⋅2⋅3=6 , and this is what happens.

As the number of objects increases, the number of permutations grows very quickly and it becomes difficult to depict them clearly. For example, the number of permutations of 10 items is already 3628800 (more than 3 million!).

Placements Let there be n different objects. We will select m objects from them and rearrange them in all possible ways (that is, both the composition of the selected objects and their order change). The resulting combinations are called placements of n objects by m, and their number is Aⁿm =n!(n−m)!=n⋅(n−1)⋅...⋅(n−m+1)

Definition. By placing a set of n different elements into m elements (m ≤ n) are called combinations , which are composed of given n elements by m elements and differ either in the elements themselves or in the order of the elements.

Combinations Let there be n different objects. We will select m objects from them in every possible way (that is, the composition of the selected objects changes, but the order is not important). The resulting combinations are called combinations of n objects by m, and their number is Cmn=n!(n−m)!⋅m!

An example of all combinations of n=3 objects (different figures) by m=2 is in the picture below. According to the formula, there should be exactly C23=3!(3−2)!⋅2!:3!=3. It is clear that there are always fewer combinations than placements (since the order is important for placements, but not for combinations), and specifically in m! times, that is, the connection formula is correct: Amn=Cmn⋅Pm.

Method 1. There are 2 people participating in one game, therefore, you need to calculate how many ways you can select 2 people out of 15, and the order in such pairs is not important. Let's use the formula to find the number of combinations (samples that differ only in composition) of n different elements of m elements each

n!= 1⋅ 2 ⋅3⋅...⋅ n , with n=2, m=13.

Method 2. The first player played 14 games (2nd, 3rd, 4th, and so on until 15th), the 2nd player played 13 games (3rd, 4th, etc. until 15th). well, we exclude the fact that there was already a game with the first), 3rd player - 12 games, 4th - 11 games, 5 - 10 games, 6 - 9 games, 7 - 8 games, 8 - 7 games,

and the 15th one has already played with everyone.

Total: 14+13+12+11+10+9+8+7+6+5+4+3+2+1=105 games

ANSWER. 105 games.

Mathematics teacher Svetlana Valerievna Aksenova

Bugrovskaya secondary school, Vsevolozhsk district, Leningrad region

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Slide captions:

Combinations

Combinations The number of all selections of n elements from m data without taking into account the order is called the number of combinations of m elements by n. All combinations differ from each other in at least one element; The order of the elements is not significant here; The difference between a combination and an arrangement is that if you rearrange the elements in an arrangement, you will get a different arrangement, but the combination does not depend on the order of the elements included in it.

Combinations The number of all selections of n elements from m data without taking into account the order is called the number of combinations of m elements by n. Find: Number of combinations from 6 to 3: Number of combinations from 4 to 4:

Task No. 1 Out of 20 students, you need to choose two duty officers. In how many ways can this be done? Solution: We need to choose two people out of 20. It is clear that nothing depends on the order of choice, that is, Ivanov - Petrov or Petrov - Ivanov are the same pair of duty officers. Therefore, these will be combinations of 20 by 2.

Task No. 2. The Minotaur has 25 prisoners languishing in the labyrinth. a) In how many ways can he choose three of them for breakfast, lunch and dinner? b) How many ways are there to release three captives to freedom? Solution: A) Order is important. B) The order is not important

Task No. 3 There are 27 students in the class, three of them need to be selected. In how many ways can this be done if: a) the first student must solve the problem, the second must go for chalk, the third must go on duty in the dining room; b) should they sing in chorus? 6

In how many different ways can a team of two be formed from seven members of a math club to participate in the Olympiad? Task No. 4

Task No. 5 The department has 5 leading and 8 senior employees. Two leading and two senior researchers should be sent on a business trip. In how many ways can a choice be made?

From a shuffled deck of 36 cards, 4 cards are drawn at random. What is the probability that all cards drawn are aces? Problem #6

Problem No. 7 In a batch of 50 parts, there are 10 defective ones. Four parts are taken out from the batch at random. Determine the probability that all 4 parts will be defective. Total outcomes: Favorable outcomes: Probability.

Rearrangements Placements Combinations Probability

Municipal educational institution secondary school No. 30, Volgograd

Mathematics teacher Skleinova N.I.

Factorial

Definition 1

The factorial is the product of the first n natural numbers

n! = 1*2*2*…(n-2)(n-1)n

2!=1*2=2

3!=1*2*3=6

4!= 1*2*3*4=24

5!=1*2*3*4*5=120

Rearrangements

Definition 2

A permutation of n elements is each arrangement of these elements in a certain order P=n!

Example 1

In how many ways can the 8 participants in the final race be placed on eight treadmills?

R 8 =8!=1*2*3*4*5*6*7*8= 40320(ways)

Placements

Definition 3

An arrangement of n elements by k (k≤ n) is any set consisting of any k elements taken in a certain order from the given n elements

Example 2

Second grade students study 8 subjects. In how many ways can you create a schedule for one day so that it includes 4 different subjects?

A 8 4 =8*7*6*5= 1680 (ways)

A n k =

Combinations

Definition 4

A combination of n elements of k is any set composed of k elements selected from the given n elements

WITH n k =

Example 3

Of the 15 members of the tourist group, three duty officers must be selected. In how many ways can this choice be made?

WITH 15 3 =15!/(3!*12!)=(13*14*15)/(1*2*3)= 455(ways)

Probability

Definition 5

The probability of an event A is the ratio of the number of favorable outcomes N(A) of a test to the number of all equally possible outcomes N

P(A)= N(A)/N

Example 4

Out of 25 exam papers in geometry, the student prepared 11 first and 8 last papers. What is the probability that in the exam he will receive a ticket that he did not prepare?

P(A)=(25-11-8)/25= 0,24

Addition of probabilities

Definition 6

If event C means that one of two incompatible events occurs: A or B, then the probability of event C is equal to the sum of the probabilities of events A and B

P(C)=P(A)+P(B)

The sum of the probabilities of opposite events is 1

P(A)+P( A )=1

Multiplying Probabilities

Definition 7

If event C means the joint occurrence of two independent events A and B, then the probability of event C is equal to the product of the probabilities of events A and B

P(C)=P(A)*P(B)

Probability

Sum of probabilities

The sum of the probabilities of two events is equal to the sum of the probability of the product of these events and the probability of the sum of these events

P(A)+P(B)= P(A*B) +P(A+B)

Probability of the amount

The probability of the sum of two events is equal to the difference between the sum of the probabilities of these events and the product of the probabilities of these events

P(A+B)=P(A)+P(B)-P(A)*P(B)

Problem 1

Solution

Condition

Probability of each hits equal to 0,8.

A biathlete shoots at targets 5 times. The probability of hitting the target with one shot is 0.8. Find the probability that the biathlete hit the targets the first 3 times and missed the last 2 times. Round the result to hundredths.

Probability of each miss equal to 1-0.8= 0,2 .

Using the probability multiplication formula, we get

P(A )=0,8*0,8*0,8*0,2*0,2

P(A )= 0,02048 0,02

Answer: 0.02

Problem 2

Condition

Solution

In Fairytale Land there are two types of weather: good and excellent, and the weather, once established in the morning, remains unchanged all day. It is known that with probability 0.6 the weather tomorrow will be the same as today. Today is September 18th, the weather in Fairytale Land is good. Find the probability that the weather will be great in Fairyland on September 21.

Since the weather is good on September 18, then on September 19 the weather is good with a probability of 0.6, and excellent with a probability of 0.4.

If the weather is good on September 19, then on September 20 the probability of good weather is 0.6*0.6=0.36

The probability of excellent weather is 0.6*0.4=0.24

Similarly, if the weather is excellent on September 19, then with a probability of 0.4 * 0.6 = 0.24 it will be excellent on September 20. The probability of good weather on September 20 is 0.4*0.4=0.16.

Reasoning similarly, we find that the probability of excellent weather on September 21 will be equal to the probability of the sum: 0.6*0.24+ +0.6*0.24+0.4*0.16+0.6*0.24= 0,496

Problem 3

Condition

Solution

An automatic line produces batteries. The probability that a finished battery is faulty is 0.02. Before packaging, each battery goes through a control system. The probability that the system will block a faulty battery is 0.98. The probability that the system will mistakenly block a working battery is 0.03. Find the probability that a randomly selected manufactured battery will be blocked by the control system.

Let the event A = (the battery will be blocked), then the probability of the occurrence of this event can be found as the union of intersections of events.

P(A)=0.02*0.98+0.98*0.03

P(A)=0.98(0.02+0.03)

P(A)=0.98*0.05= 0,049

Answer: 0.049

Literature

Makarychev Yu.N. Algebra: elements of statistics and probability theory: textbook. manual for general education students. Institutions. Publishing house "Prosveshcheniye", 2003

Mordkovich A.G., Semenov P.V. Algebra and beginning of mathematical analysis. Part 1. Textbook for general education organizations. Publishing house "Mnemosyne", 2015

Lysenko F.F., Kulabukhova S.Yu. Mathematics. Preparation for the Unified State Exam 2016. Publishing house "Legion" LLC, 2015

Vysotsky I.R., Yashchenko I.V. Unified State Exam 2016. Mathematics. Probability theory. Workbook. Publishing house MCNMO, 2016

1. Organizational moment
Greeting students, communicating the topic and purpose of the lesson
2. Repetition and consolidation of the material covered
· Answers to questions on homework (analysis of unsolved problems).
· Monitoring the assimilation of the material (written survey).
Option 1
1. A reliable event and its probability.
2. a) In a random experiment, two dice are thrown. Find the probability that the total will be 7 points. Round the result to hundredths.
b) 40 athletes are participating in the gymnastics championship: 12 from Argentina, 9 from Brazil, the rest from Paraguay. The order in which the gymnasts perform is determined by lot. Find the probability that the athlete competing first is from Paraguay.
c) On average, out of 500 garden pumps sold, 4 leak. Find the probability that one pump randomly selected for control does not leak.
Option 2
1. Impossible event and its probability.
2. a) In a random experiment, two dice are thrown. Find the probability that the total will be 9 points. Round the result to hundredths.
b) 64 athletes are participating in the gymnastics championship: 20 from Japan, 28 from China, the rest from Korea. The order in which the gymnasts perform is determined by lot. Find the probability that the athlete competing first is from Korea.
c) The factory produces bags. On average, for every 170 quality bags, there are six bags with hidden defects. Find the probability that the purchased bag will be of high quality. Round the result to hundredths.

Answer: option 1. 2. a) 0.17; b) 0.475; c) 0.992.
option 2. 2. a) 0.11; b) 0.25; c) 0.97.
3. Learning new material
The class was divided into groups that collected information, compiled and presented the results of their work in class (students presenting the results of their work).
1 group(find information about what factors (reasons) contributed to the emergence of the science of combinatorics, which scientists were at the very origins of its emergence).
2nd group(find information about whether combinatorics exists in real life, and if so, in which industries it is used).

3 group (find information about what problems are called combinatorial and how they can be solved, consider each solution method and make a selection of several problems solved by a specific method).
3.1. 1 group.
Representatives of a wide variety of specialties have to solve problems that involve certain combinations made up of letters, numbers and other objects.
When considering the simplest probabilistic problems, we had to count the number of different outcomes (combinations). For a small number of elements such calculations are easy to do. Otherwise, such a task poses a significant difficulty. (slide 1)

Combinatoricsis a branch of mathematics that studies questions about the number of different combinations (satisfying certain conditions) that can be made from given elements.
Combinatorics- a branch of mathematics in which the simplest “connections” are studied. Permutations are compounds that can be made up of n objects, changing their order in all possible ways; the number of their placements - compounds containing m objects from the number n given, differing either in the order of the objects or in the objects themselves; their number Combinations - compounds containing m items out of n, differing from each other in at least one item (in the modern explanatory dictionary, published by “Big Soviet Encyclopedia”).
People faced problems in which they had to choose certain objects, arrange them in a certain order and find the best among different arrangements back in the prehistoric era, choosing the best position for hunters during a hunt, warriors during a battle, tools during work. . (slide 2)

· The term “combinatorics” was introduced into mathematical use by Leibniz, who in 1666 published his work “Discourses on the Art of Combination”. (slide 3)
· Combinatorics originally arose in XVI in connection with the spread of various gambling games. (slide 4)
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3.1. 2nd group.(slide 1)
It is remarkable that a science which began with the consideration of gambling promises to become the most important object of human knowledge. After all, most of life's questions are actually problems from probability theory.
P. Laplace

Areas of application of combinatorics:
. educational institutions (scheduling) (slide 2)
. catering industry (menu planning)
. linguistics (considering options for letter combinations)
. geography (map coloring) (slide 3)
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3.1. 3 group
Problems that involve certain combinations of objects are called combinatorial.(slide 1)
Addition rule: if some object A can be selected m ways, and another object B can be selected n ways, then the choice “either A or B” can be made m + n ways.
(slide 2)
For example:
· There are 5 apples and 4 oranges on a plate. In how many ways can you choose one fruit?
According to the conditions of the problem, an apple can be chosen in five ways, an orange in four. Since the problem is about choosing “either an apple or an orange,” then, according to the addition rule, it can be done in 5 + 4 = 9 ways.
· Let's look at this problem: how many two-digit numbers can be made from the digits 1,4,7, using each of them no more than once in writing the number? (slide 3)
· Solution: in order not to miss or repeat any of the numbers, we will write them in ascending order. First we write down the numbers starting with the number 1, then with the number 4, and finally with the number 7:
14, 17, 41, 47, 71, 74.
Answer: 6.
This method is called enumeration of options. Thus, from these three digits, a total of 6 different two-digit numbers can be formed.
This problem can be solved in another way. His name - tree of possible options. A special circuit has been built for this task. (slide 4) (slide 5)
We put an asterisk. It will indicate the number of possible options.
Next, we take 3 segments from the asterisk. In the problem statement, 3 numbers are given - 1, 4, 7.
We put these numbers at the ends of the segments. They will indicate the number of tens in a given number.
Next, we draw 2 segments from each number.
At the ends of these segments we also write the numbers 1, 4, 7. They will indicate the number of ones.
Let's look at the numbers we got: 14, 17, 41, 47, 71, 74. That is, we got 6 numbers in total.
Answer: 6.

This diagram really looks like a tree, albeit “upside down” and without a trunk.
Multiplication rule: if object A can be selected m ways and if after each such choice object B can be selected in n ways, then the choice of the pair (A, B) in the specified order can be carried out m∙ n ways. (slide 6)
· How many two-digit numbers can be made from the digits 1,4,7, using each of them no more than once?
This problem can be solved differently and much faster, without building a tree of possible options. Let's think like this. The first digit of a two-digit number can be selected in three ways. Since after selecting the first digit there will be two left, the second digit can be selected from the remaining digits in two ways. Therefore, the total number of required three-digit numbers is equal to the product 3∙2, i.e. 6.
· How many five-digit numbers can be made from the numbers 5, 9, 0, 6?

According to the multiplication rule we get: 4∙4∙4∙4=256 numbers.
(slide 7)
Rearrangements -connections, each of which contains n various elements taken in a certain order. (slide 8)
Pn=n! = 1 · 2 · 3 · … · (n -2) · (n -1) · n
Task.(slide 9)
In how many ways can seven different books be arranged on a shelf?
Solution:
The number of such ways is equal to the number of permutations of seven elements,
those. P 7 = 7! = 1 · 2 · 3 · … · 7 = 5040.
Answer: 5040.
Task.(slide 10)
There are 10 different books, three of which are reference books. In how many ways
Is it possible to arrange these books on a shelf so that all the reference books are next to each other?
Solution:
Because If the reference books should stand side by side, then we will consider them as one book. Then you need to arrange 10 books on the shelf - 3+1=8 books. It can be done P 8 ways. For each of the resulting combinations you can do P 3 directory rearrangements.
Therefore, the number of ways to arrange books on a shelf is equal to the product:
P 8 P 3 = 8! · 3! = 40320 · 6 =241920.
Answer: 241920.
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