Problems to solve for consolidating new material

Task No. 1. In how many ways can the 5 participants in the final be arranged?

race on 5 treadmills?

Solution: P 5 = 5!= 1 ∙2 ∙3 ∙4 ∙5 = 120 ways.

Task No. 2. How many three-digit numbers can be made from the digits 1,2,3, if each

does a digit appear in a number image only once?

Solution: The number of all permutations of three elements is equal to P 3 =3!, where 3!=1 * 2 * 3=6

This means that there are six three-digit numbers made up of the numbers 1,2,3.

Task No. 3. In how many ways can four young men invite four out of six

girls to dance?

Solution: two boys cannot invite the same girl at the same time. AND

options in which the same girls dance with different boys,

are considered different, therefore:

Problem No. 4. How many different three-digit numbers can be made from the numbers 1, 2, 3, 4, 5,

6, 7, 8, 9, provided that in writing the number each digit is used only

once?

Solution: In the problem statement it is proposed to count the number of possible combinations from

three digits taken from the assumed nine digits, and the order

the arrangement of numbers in a combination matters (for example, the number 132)

and 231 different). In other words, you need to find the number of placements out of nine

three elements each.

Using the formula for the number of placements we find:

Answer: 504 three-digit numbers.

Problem #5 In how many ways can a committee of 3 be selected from 7 people?

Solution: To consider all possible commissions, you need to consider all

possible 3-element subsets of a set consisting of 7

Human. The required number of ways is

Task No. 6. 12 teams participate in the competition. How many options are there?

distribution of prize (1, 2, 3) places?

Solution: A 12 3 = 12 ∙11 ∙10 = 1320 options for distribution of prize places.

Answer: 1320 options.

Task No. 7. At competitions athletics our school was represented by a team from

10 athletes. In how many ways can the coach determine which of them

will run in the 4x100m relay in the first, second, third and fourth stages?

Solution: Choice from 10 to 4, taking into account the order:
ways.

Answer: 5040 ways.

Task No. 8. In how many ways can red, black, blue and

green balls?

Solution: You can put any of the four balls in the first place (4 ways), on

second - any of the three remaining (3 methods), third place - any of

the remaining two (2 ways), for fourth place - the remaining last ball.

Total 4 · 3 · 2 · 1 = 24 ways.

P 4 = 4! = 1 · 2 · 3 · 4 = 24. Answer: 24 ways.

Problem No. 9. Students were given a list of 10 books that are recommended to be read in

vacation time. In how many ways can a student choose 6 books from them?

Solution: Choice 6 out of 10 without regard to order:
ways.

Answer: 210 ways.

Problem No. 10. There are 7 students in the 9th grade, 9 students in the 10th grade, and 8 students in the 11th grade. For

work on the school site, it is necessary to allocate two students from grade 9,

three out of 10, and one out of 11. How many ways are there to choose?

students to work in the school area?

Solution: Choice from three sets without regard to order, each choice from

first set (C 7 2) can be combined with each choice from

the second (C 9 3)) and with each choice of the third (C 8 1) according to the rule

multiplication we get:

Answer: 14,112 ways.

Task No. 11. Ninth-graders Zhenya, Seryozha, Kolya, Natasha and Olya ran to

recess to the tennis table, where the game was already in progress. How many

ways five ninth-graders running up to the table can take

queue for table tennis?

Solution: Any ninth grader could be first in line, and any of the students could be second.

the remaining three, the third - any of the remaining two and the fourth -

a ninth-grader who ran up second to last, and a fifth-grader who ran up last. By

The multiplication rule for five students has 5 4321=120 ways

METHODOLOGICAL DEVELOPMENT OF A PRACTICAL LESSON in the discipline: “MATHEMATICS”

Topic: “FUNDAMENTALS OF PROBABILITY THEORY AND MATHEMATICAL STATISTICS”

Example 1 . Calculate: a) ; b) ; V) .

Solution. A) .

b) Since , then we can put it out of brackets

Then we get

V) .

Example 2 . In how many ways can six different books be arranged on one shelf?

Solution. The required number of ways is equal to the number of permutations of 6 elements, i.e.

Example 3. How many options for distributing three vouchers to sanatoriums of various profiles can be compiled for five applicants?

Solution. The required number of options is equal to the number of placements of 5 elements of 3 elements, i.e.

.

Example 4 . In a team of 25 people, you need to allocate four to work in a certain area. In how many ways can this be done?

Solution. Since the order of the four people selected does not matter, this can be doneways.

We find using the first formula

.

In addition, when solving problems, the following formulas are used, expressing the basic properties of combinations:

(by definition they assume and);

.

1.2. Solving combinatorial problems

Task 1. The faculty studies 16 subjects. You need to put 3 subjects on your schedule for Monday. In how many ways can this be done?

Solution. There are as many ways to schedule three items out of 16 as you can arrange placements of 16 items by 3.

Task 2. Out of 15 objects, 10 objects need to be selected. In how many ways can this be done?

Solution.

Task 3. Four teams took part in the competition. How many options for distributing seats between them are possible?

Solution.

.

Task 4. In how many ways can a patrol of three soldiers and one officer be formed if there are 80 soldiers and 3 officers?

Solution. You can choose a soldier on patrol

ways, and officers in ways. Since any officer can go with each team of soldiers, there are only so many ways.

Task 5. Find if it is known that .

Solution.

Since , we get

,

,

, .

By definition of a combination it follows that , . That. .

Answer: 9

1.3. The concept of a random event. Types of events. Probability of event

Example. The box contains 30 numbered balls. Determine which of the following events are impossible, reliable, or contrary:

took out a numbered ball(A);

got a ball with an even number(IN);

got a ball with an odd number(WITH);

got a ball without a number(D).

Which of them form a complete group?

Solution. A - reliable event;D - impossible event;

IN AndWITH - opposite events.

The complete group of events consists ofA AndD, V AndWITH .

Probability of event , is considered as a measure of the objective possibility of the occurrence of a random event.

1.4. Classic definition of probability

Task 1. In a lottery of 1000 tickets, there are 200 winning ones. One ticket is taken out at random. What is the probability that this ticket is a winner?

Solution. The total number of different outcomes isn =1000. The number of outcomes favorable to winning ism=200. According to the formula, we get

.

Task 2. In a batch of 18 parts there are 4 defective ones. 5 parts are selected at random. Find the probability that two of these 5 parts will be defective.

Solution. Number of all equally possible independent outcomesn equal to the number of combinations of 18 by 5 i.e.

Let's count the numberm, favorable to event A. Among 5 parts taken at random, there should be 3 high-quality and 2 defective. The number of ways to select two defective parts from 4 existing defective ones is equal to the number of combinations of 4 by 2:

The number of ways to select three quality parts from 14 available quality parts is equal to

.

Any group of good parts can be combined with any group of defective parts, so total number combinationsm amounts to

The desired probability of event A is equal to the ratio of the number of outcomesm, favorable to this event, to the numbernall equally possible independent outcomes:

.

1.5. Theorem for adding probabilities of incompatible events

Amount of a finite number of events is an event consisting of the occurrence of at least one of them.

The sum of two events is denoted by the symbol A+B, and the sumn events symbol A 1 +A 2 + … +A n .

Probability addition theorem.

Task 1. There are 100 lottery tickets. It is known that 5 tickets win 20,000 rubles each, 10 tickets win 15,000 rubles, 15 tickets win 10,000 rubles, 25 tickets win 2,000 rubles. and nothing for the rest. Find the probability that the purchased ticket will receive a winning of at least 10,000 rubles.

Solution. Let A, B, and C be events consisting in the fact that the purchased ticket receives a winning equal to 20,000, 15,000, and 10,000 rubles, respectively. since events A, B and C are incompatible, then

Task 2. On correspondence department technical school receives tests in mathematics from citiesA, B AndWITH . Probability of receiving a test paper from the cityA equal to 0.6, from the cityIN - 0.1. Find the probability that the next test will come from the cityWITH .

Solution. Events “the test came from the cityA ", "the test came from city B" and "the test came from city C" form complete system, so the sum of their probabilities is equal to one:

, i.e. .

Task 3. The probability that the day will be clear is . Find the probability that the day will be cloudy.

Solution. The events “clear day” and “cloudy day” are opposite, therefore

That is

1.6. Probability multiplication theorem independent events

Task 1. Calculate the probability that in a family where there is one baby boy, a second boy will be born.

Solution. Let the eventA is that there are two boys in the family, and the eventIN - that one boy.

Let's consider everything possible outcomes: boy and boy; boy and girl; girl and boy; girl and girl.

Then, and using the formula we find

.

Task 2. The first urn contains 6 black and 4 white balls, the second urn contains 5 black and 7 white balls. One ball is drawn from each urn. What is the probability that both balls will be white?

Solution. Let - a white ball is drawn from the first urn; - a white ball is drawn from the second urn. It is obvious that the events are independent.

Because , , then using the formula we find

.

Task 3. The device consists of two elements that work independently. The probability of failure of the first element is 0.2; the probability of failure of the second element is 0.3. Find the probability that: a) both elements will fail; b) both elements will work.

Solution. Let the eventA - failure of the first element, eventIN - the output of their structure of the second element. These events are independent (by condition).

a) Simultaneous appearanceA AndIN there is an eventAB . Hence,

b) If the first element works, then an event occurs (opposite to the eventA - failure of this element); if the second element works - eventIN. Let's find the probabilities of events and:

Then the event that both elements will work is and, therefore,

II . RANDOM VARIABLE, ITS DISTRIBUTION FUNCTION

2.1. Random variable, methods of specifying it

Random is a quantity that, as a result of testing, can take one or another numeric value, and it is not known in advance which one.

If for any quantity its measurement is repeated many times under practically identical conditions, you will find that each time you obtain slightly different results. This is the influence of two types of causes: 1) basic ones, determining the main meaning of the result; 2) secondary ones, causing their divergence.

With the combined action of these causes, the concepts of necessity and chance are closely related to each other, but the necessary prevails over the chance.

Thus, the possible values ​​of random variables belong to some numerical sets.

What is random is that on these sets the quantities can take on any value, but which one cannot be said in advance.

A random variable is associated with a random event.

If a random event -quality characteristic tests, then the random variable is itsquantitative characteristic .

Random variables are denoted in capitals in Latin letters and their meaning is in capitals - .

The probability that a random variable will take a value is denoted by:

etc.

Random variables are specified by distribution laws.

Law of distribution random variable is the correspondence established between the possible values ​​of a random variable and their probabilities.

Distribution laws can be specified in three ways: tabular, graphical, analytical. The method of setting depends on the type of random variable.

There are two main types of random variables:discrete and continuously distributed random variables.

2.2. Discrete and continuous random variables

If the values ​​that a given random variable can take form a discrete (finite or infinite) series of numbers, then the random variable itself is calleddiscrete.

If the values ​​that a given random variable can take fill a finite or infinite interval (a, b) of the numerical axisOh, then the random variable is calledcontinuous.

Each value of a random variable of a discrete type corresponds to a certain probability; Each interval (a, b) from the range of values ​​of a random variable of continuous type also corresponds to a certain probability that the value taken by the random variable falls into this interval.

2.3. Distribution law of a random variable

A relationship that establishes in one way or another a connection between the possible values ​​of a random variable and their probabilities is calledlaw of distribution random variable.

The distribution law of a discrete random variable is usually givennext distribution:

At the same time, where the summation extends to the entire (finite or infinite) set possible values given random variable.

It is convenient to specify the distribution law of a continuous random variable usingprobability density function .

The probability that the value taken by the random variable will fall into the interval (a, b) is determined by the equality

.

The graph of the function is calleddistribution curve . Geometrically, the probability of a random variable falling into the interval (a, b) is equal to the area of ​​the corresponding curved trapezoid, limited by the distribution curve, axisOh and straightx=a, x=b.

Task 1. The probabilities of random variable values ​​are given: value 10 has a probability of 0.3; value 2 – probability 0.4; value 8 – probability 0.1; value 4 – probability 0.2. Construct a distribution series of a random variable.

Solution. By arranging the values ​​of the random variable in ascending order, we obtain the distribution series:

Let's take it on a planechorus points (2; 0.4), (4; 0.2), (8; 0.1) and (10; 0.3). By connecting successive points with straight line segments, we getpolygon (orpolygon ) distribution of a random variable

X

Task 2. Two items worth 5,000 rubles each and one item worth 30,000 rubles are up for grabs. Draw up a law of distribution of winnings for a person who bought one ticket out of 50.

Solution. The desired random variable is a gain and can take three values: 0, 5000 and 30000 rubles. The first result is favored by 47 cases, the second result by two cases and the third by one case. Let's find their probabilities:

; ; .

The distribution law of a random variable has the form:

As a check we will find

Task 3. The random variable is subject to a distribution law with density , and

Required: 1) Find coefficient a; 2) build a density distribution graph; 3) find the probability of falling into the interval (1; 2).

Solution. 1) Since all values ​​of a given random variable are contained on the segment , then

, where

, or

Those. .

2) The graph of a function in the interval is a parabola, and outside this interval the x-axis itself serves as the graph.

X

) The probability of a random variable falling into the interval (1; 2) can be found from the equality

2.4. Binomial distribution

Let a certain number be producedn independent experiments, and in each of them some event can occur with the same probabilityR . Consider a random variable representing the number of occurrences of eventsA Vn experiments. The law of its distribution has the form

Where, is calculated using Bernoulli's formula.

The distribution law, which is characterized by such a table, is calledbinomial .

Task. The coin is tossed 5 times. Draw up a law of distribution of a random variable - the number of the coat of arms.

Solution. The following values ​​of the random variable are possible: 0, 1, 2, 3, 4, 5. Knowing that the probability of a coat of arms falling out in one trial is equal to , we will find the probabilities of the values ​​of the random variable using the Bernoulli formula:

The distribution law has the form

Let's check:

III . MATHEMATICAL EXPECTATION AND VARIANCE OF A RANDOM VARIABLE

3.1. Expectation of a discrete random variable

Example 1 . Find the mathematical expectation of a random variable, knowing the law of its distribution

Solution.

Properties of mathematical expectation.

1. The constant factor can be taken out of the mathematical expectation sign:

2. Mathematical expectation of a constant valueWITH equal to this value itself:

3. The mathematical expectation of the sum of two random variables is equal to the sum of their mathematical expectations:

4. The mathematical expectation of the product of independent random variables is equal to the product of the mathematical expectations of these variables:

3.2. Standard deviation and variance of a random variable.

Example 2. Let's find the mathematical expectation of random variables and , knowing the laws of their distribution

2)

Solution:

P

We got an interesting result: the laws of distribution of quantities and are different, but their mathematical expectations are the same.

b)

From the drawingb it is clear that the value of the quantity is more concentrated around the mathematical expectation than the values ​​of the quantity that are scattered (scattered) relative to its mathematical expectation (FigureA ).

The main numerical characteristic of the degree of dispersion of the values ​​of a random variable relative to its mathematical expectation is dispersion, which is denoted by .

In how many ways can two students be selected for a conference if there are 33 people in the group?

Solve equations

A) . b) .

How many four-digit numbers divisible by 5 can be made from the digits 0, 1, 2, 5, 7, if each number must not contain the same digits?

From a group of 15 people, a foreman and 4 team members should be selected. In how many ways can this be done?

Morse code letters are made up of symbols (dots and dashes). How many letters can be drawn if you require that each letter contain no more than five characters?

In how many ways can four-color ribbons be made from seven ribbons of different colors?

In how many ways can four persons be selected from nine candidates for four different positions?

In how many ways can you choose 3 out of 6 cards?

Before graduation, a group of 30 students exchanged photos. How many photo cards were distributed?

In how many ways can 10 guests be seated in ten places at a festive table?

How many games should 20 football teams play in a one-round championship?

In how many ways can 12 people be distributed among teams if each team has 6 people?

Probability theory

The urn contains 7 red and 6 blue balls. Two balls are drawn from the urn at the same time. What is the probability that both balls are red (event A)?

Nine different books are arranged at random on one shelf. Find the probability that four specific books will be placed next to each other (event C).

Out of 10 tickets, 2 are winning. Determine the probability that among 5 tickets taken at random, one is winning.

3 cards are drawn at random from a deck of cards (52 cards). Find the probability that it is a three, a seven, an ace.

A child plays with the five letters of the split alphabet A, K, R, Sh, Y. What is the probability that if the letters are randomly arranged in a row, he will get the word “Roof”.

There are 6 white and 4 red balls in the box. Two balls are taken at random. What is the probability that they will be the same color?

The first urn contains 6 black and 4 white balls, the second urn contains 5 black and 7 white balls. One ball is drawn from each urn. What is the probability that both balls are white?

Random variable, mathematical expectation and variance of a random variable

Draw up a distribution law for the number of hits on a target with six shots, if the probability of a hit with one shot is 0.4.

The probability that a student will find the book he needs in the library is 0.3. Draw up a distribution law for the number of libraries he will visit if there are four libraries in the city.

The hunter shoots at the game until the first hit, but manages to fire no more than four shots. Find the variance of the number of misses if the probability of hitting the target with one shot is 0.7.

Find the mathematical expectation of a random variableX, if the law of its distribution is given by the table:

The plant operates four automatic lines. The probability that during a work shift the first line will not require adjustment is 0.9, the second – 0.8, the third – 0.75, the fourth – 0.7. find the mathematical expectation of the number of lines that will not require adjustment during a work shift.

Find the variance of the random variable X, knowing the law of its distribution: 5. REFERENCES

Main:

1. Bogomolov N.V. Practical lessons in mathematics. – M.: graduate School, 1990. – 495 p.

   Soloveychik I.L. Collection of problems in mathematics for technical schools / I.L. Soloveychik, V.T. Lisichkin. – M.: Onyx 21st century, 2003. – 464 p.

   Valutse I.I. Mathematics for technical schools / I.I. Valuta, G.D. Diligul. - M.: Nauka, 1989. – 575 p.

   Danko P.E. Higher mathematics in exercises and problems. In two parts. PartII/ P.E. Danko, A.G. Popov, T.Ya. Kozhevnikova. – M.: Higher School, 1986. – 415 p.

   Vygodsky M.Ya. Handbook of higher mathematics. – M.: Nauka, 1975. – 872 p.

Additional:

Griguletsky V.G. Mathematics for students of economic specialties. Part 2 / V.G. Griguletsky, I.V. Lukyanova, I.A. Petunina. – Krasnodar, 2002. – 348 p.

Malykhin V.I. Mathematics in Economics. – M.: Infra-M, 1999. – 356 p.

Gusak A.A. Higher mathematics. In 2 volumes, T.2. – training manual for university students. – M.: TetraSystems, 1988. – 448 p.

Griguletsky V.G. Higher mathematics / V.G. Griguletsky, Z.V. Yashchenko. – Krasnodar, 1998.-186 p.

Gmurman V.E. A guide to solving problems in probability theory and mathematical statistics. – M.: Higher School, 2000. – 400 p.

Methods for solving combinatorial problems

Enumeration of possible options

Simple problems are solved by an ordinary exhaustive search of possible options without drawing up various tables and diagrams.

Task 1.
Which double figures can it be made from the numbers 1, 2, 3, 4, 5?

Answer: 11, 12, 13, 14, 15, 21, 22, 23, 24, 25, 31, 32, 33, 34, 35, 41, 42, 43, 44, 45, 51, 52, 53, 54, 55.

Task 2.
Ivanov, Gromov and Orlov are participating in the final 100 m race. Name possible options distribution of prizes.

Answer:
Option1: 1) Ivanov, 2) Gromov, 3) Orlov.
Option2: 1) Ivanov, 2) Orlov, 3) Gromov.
Option 3: 1) Orlov, 2) Ivanov, 3) Gromov.
Option4: 1) Orlov, 2) Gromov, 3) Ivanov.
Option5: 1) Gromov, 2) Orlov, 3) Ivanov.
Option6: 1) Gromov, 2) Ivanov, 3) Orlov.

Task 3.
Petya, Kolya, Vitya, Oleg, Tanya, Olya, Natasha, Sveta signed up for the ballroom dance club. What dance pairs of a girl and a boy can form?

Answer:
1) Tanya - Petya, 2) Tanya - Kolya, 3) Tanya - Vitya, 4) Tanya - Oleg, 5) Olya - Petya, 6) Olya - Kolya, 7) Olya - Vitya, 8) Olya - Oleg, 9) Natasha - Petya, 10) Natasha - Kolya, 11) Natasha - Vitya, 12) Natasha - Oleg, 13) Sveta - Petya, 14) Sveta - Kolya, 15) Sveta - Vitya, 16) Sveta - Oleg.

Tree of possible options

A variety of combinatorial problems are solved by drawing up special circuits. Outwardly, this scheme resembles a tree, hence the name of the method - tree of possible options.

Task 4.
Which three digit numbers can it be made from the numbers 0, 2, 4?

Solution.Let's build a tree of possible options, taking into account that 0 cannot be the first digit in the number.

Answer: 200, 202, 204, 220, 222, 224, 240, 242, 244, 400, 402, 404, 420, 422, 424, 440, 442, 444.

Task 5.
School tourists decided to take a trip to a mountain lake. The first stage of the journey can be covered by train or bus. The second stage is by kayaks, bicycles or on foot. And the third stage of the journey is on foot or using a cable car. What possible travel options do school tourists have?

Solution.Let's build a tree of possible options, denoting travel by train P, by bus - A, by kayak - B, by bicycle - B, on foot - X, by cable car - K.

Answer:The figure lists all 12 possible travel options for school tourists.

Task 6.
Write down all possible options for scheduling five lessons per day from the subjects: mathematics, Russian, history, English language, physical education, and mathematics should be the second lesson.

Solution.Let's build a tree of possible options, denoting M - mathematics, R - Russian, I - history, A - English, F - physical education.

Answer:There are 24 possible options in total:

R M AND A F

R M AND F A

R M A AND F

R M A F AND

R M F AND A

R M F A AND

AND M R A F

AND M R F A

AND M A R F

AND M A F R

AND M F R A

AND M F A R

A M R AND F

A M R F AND

A M AND R F

A M AND F R

A M F R AND

A M F AND R

F M R AND A

F M R A AND

F M AND R A

F M AND A R

F M A R AND

F M A AND R

Task 7.
Sasha goes to school in trousers or jeans; he wears gray, blue, green or checkered shirts with them, and takes shoes or sneakers as a change of shoes.
a) How many days will Sasha be able to look new?
b) How many days will he wear sneakers?
c) How many days will he wear a checkered shirt and jeans?

Solution.Let's build a tree of possible options, denoting B - trousers, D - jeans, C - gray shirt, G - blue shirt, Z - green shirt, P - checkered shirt, T - shoes, K - sneakers.

Answer:a) 16 days; b) 8 days; c) 2 days.

Compiling tables

You can solve combinatorial problems using tables. They, like the tree of possible options, clearly represent the solution to such problems.

Task 8.
How many odd two-digit numbers can be made from the digits 1, 3, 4, 6, 7, 8, 9?

Solution.Let's make a table: the first column on the left is the first digits of the required numbers, the first row at the top is the second digits.

Answer: 28.

Task 9.
Masha, Olya, Vera, Ira, Andrey, Misha and Igor were preparing to become presenters at New Year's holiday. Name possible options if only one girl and one boy can lead.

Solution.Let's make a table: the first column on the left is the names of girls, the first row at the top is the names of boys.

Answer:All possible options are listed in the rows and columns of the table.

Multiplication rule

This method of solving combinatorial problems is used when it is not necessary to list all possible options, but you need to answer the question - how many of them exist.

Problem 10.
Several teams participate in the football tournament. It turned out that they all used white, red, blue and green colors, and all possible options were presented. How many teams participated in the tournament?

Solution.
Briefs can be white, red, blue or green, i.e. there are 4 options. Each of these options has 4 jersey color options.

4 x 4 = 16.

Answer: 16 teams.

Problem 11.
6 students take a test in mathematics. In how many ways can they be arranged in the list?

Solution.
The first on the list can be any of the 6 students,
the second on the list can be any of the remaining 5 students,
third - any of the remaining 4 students,
fourth - any of the remaining 3 students,
fifth - any of the remaining 2 students,
sixth - the last 1 student.

6 x 5 x 4 x 3 x 2 x 1 = 720.

Answer: 720 ways.

Problem 12.
How many even two-digit numbers can be made from the digits 0, 2, 3, 4, 6, 7?

Solution.
The first in a two-digit number can be 5 digits (digit 0 cannot be the first in the number), the second in a two-digit number can be 4 digits (0, 2, 4, 6, since the number must be even).
5 x 4 = 20.

Answer: 20 numbers.

Problem 12. Of the students attending the math club, in which there are 5 girls and 3 boys, two need to be sent to the Olympiad: one girl and one boy. How many different couples are there that can be sent to the Olympics?

Solution: A girl from the circle can be selected in five ways, and a boy in three. A couple (a girl with a boy) can be selected in fifteen different ways

5 3 = 15 ways.

Answer: 15 ways.

Problem 13. 12 teams participate in the competition. How many options are there for the distribution of prize (1, 2, 3) places?

Solution: A 12 3 = 12 11 10 = 1320 options for distribution of prize places.

Answer: 1320 options.

Problem 14. At the athletics competition, our school was represented by a team of 10 athletes. In how many ways can a coach determine which of them will run in the 4100 m relay in the first, second, third and fourth legs?

Solution: Choice from 10 to 4, taking into account the order: methods.

Answer: 5040 ways.

Problem 15. In how many ways can red, black, blue and green balls be placed in a row?

Solution: You can place any of the four balls in the first place (4 ways), any of the remaining three balls in the second place (3 ways), any of the remaining two balls in the third place (2 ways), and the last remaining ball in the fourth place. Total 4 · 3 · 2 · 1 = 24 ways.

R 4 = 4! = 1 · 2 · 3 · 4 = 24.

Answer: 24 ways.

Problem 16 . The students were given a list of 10 books to read during the holidays. In how many ways can a student choose 6 books from them?

Solution: Selecting 6 out of 10 without taking into account the order: methods.

Answer: 210 ways.

Problem 17 . Volodya goes to his classmates’ birthday party, twins Yulia and Ira. He wants to give each of them a ball. There are only 3 balls left for sale in the store different colors: white, black and striped. In how many ways can Volodya give gifts to his sisters by purchasing 2 balls?

Solution: According to the conditions of the problem, two sequential choices are provided: first, Volodya chooses 2 balls out of three available in the store, and then decides which of the twin brothers to give each of the purchased balls. Two out of three balls can be selected in three ways. After that, each selected pair can be gifted in two ways (methods) (the order is important). Then, according to the multiplication rule, the required number of ways is equal to the ways.

Answer: 6 ways.

Problem 18 . There are 7 students in the 9th grade, 9 students in the 10th grade, and 8 students in the 11th grade. To work in a school site, two students from grade 9 must be selected, three from grade 10, and one from grade 11. How many ways are there for selecting students to work in a school site?

Solution: Choice from three sets without taking into account the order, each choice from the first set (C 7 2) can be combined with each choice from the second (C 9 3) and with each choice from the third (C 8 1) using the multiplication rule we obtain:

S 7 2 · S 9 3 · S 8 1 =------ · -------- · ---- = 14,112 ways for students to choose.

Answer: 14,112 ways.

Problem 19. Ninth-graders Zhenya, Seryozha, Kolya, Natasha and Olya ran during recess to the tennis table, where a game was already underway. In how many ways can five ninth-graders running up to the table take a turn to play table tennis?

Solution: Any ninth grader could be the first to stand in line, the second - any of the remaining three, the third - any of the remaining two, and the fourth - the ninth grader who ran up next to last, and the fifth-last. According to the multiplication rule, five students have

5· 4321=120 ways to get in line.

Answer: 120 ways.

Basic concepts of combinatorics

In the branch of mathematics, which is called combinatorics, problems related to the consideration of sets and the composition of various combinations of elements of these sets are solved. For example, if we take 10 different numbers 0, 1, 2, 3, ..., 9 and make combinations of them, then we will get various numbers, for example 345, 534, 1036, 45, etc.

We see that some of these combinations differ only in the order of the digits (for example, 345 and 534), others in the digits they contain (for example, 1036 and 5671), and others also differ in the number of digits (for example, 345 and 45).

Thus, the resulting combinations satisfy different conditions. Depending on the rules of composition, three types of combinations can be distinguished: permutations, placements, combinations. Let's consider them separately. However, first familiarize yourself with the concept of factorial.

1. The concept of factorial

Product of all natural numbers from 1 to n inclusively called n-factorial and write n! = 1 · 2 · 3 · ... · (n – 1) · n.

Example 1. Calculate:

a) 3!; b) 7! – 5!; V)

Solution. a) 3! = 1 · 2 · 3 = 6.

b) Since 7! = 1 2 3 4 5 6 7 and 5! = 1 · 2 · 3 · 4 · 5, then we can put 5 out of brackets!. Then we get 5! (6 · 7 – 1) = 5! · 41 = 120 · 41 = 4920.

V)

Example 2. Simplify:

Solution. a) Considering that (n + 1)! = 1 · 2 · 3 · … · n · (n + 1), and n! = 1 · 2 · 3 ... · n, reduce the fraction;

b) Since (n + 1)! = 1 · 2 · 3 · ... · (n – 1) · n · (n + 1), then after reduction we get

(n+1)! = 1 · 2 · 3 · ... · n · (n + 1), n! = 1 · 2 · 3 · ... · n.

Let's bring the fraction to a common denominator, for which we take (n + 1)!. Then we get

1 – 3. Calculate:

1. 2. 3.

4 – 9. Simplify the expressions:

4. 6. 8.

5. 7. 9. -

2. Rearrangements

Let three letters A, B, C be given. Let's make all possible combinations of these letters: ABC, ASV, BSA, BAC, CAB, CBA (6 combinations in total). We see that they differ from each other only in the order of the letters.

Combinations of n elements that differ from each other only in the order of the elements are called permutations.

Permutations are denoted by the symbol Рn, where n is the number of elements included in each permutation.

The number of permutations can be calculated using the formula

Рn = n (n – 1) (n – 2) · ... · 3 · 2 · 1 (1)

or using factorial:

Pn = n!. (2)

Thus, the number of permutations of three elements according to formula (2) is

P3 = 3! = 3 · 2 · 1 = 6, which coincides with the result of the example discussed above.

Indeed, three letters can be placed in first place in a combination (permutation). Only two of the three letters can be placed in second place (one took first place), and only one of the remaining ones will be in third place. This means 3 · 2 · 1 = 6 = P3.

10. How many different five-digit numbers can be made from the digits 1, 2, 3, 4, 5, provided that not a single digit is repeated in the number?

11. Four teams took part in the competition. How many options for distributing seats between them are possible?

12 – 14. Calculate:

12. 13. 14.

3. Placements

Let there be four letters A, B, C, D. Composing all combinations of only two letters, we get:

We see that all the resulting combinations differ either in letters or in their order (combinations BA and AB are considered different).

Combinations of m elements of n elements that differ from each other or by the elements themselves are called arrangements.

Placements are indicated by the symbol A, where m is the number of all available elements, n is the number of elements in each combination. In this case, it is assumed that n m. The number of placements can be calculated using the formula

n factors

A = (3)

that is, the number of all possible arrangements of m elements by n is equal to the product of n consecutive integers, of which the largest is m.

So, A = 4 · 3 = 12, which coincides with the result of the above example: since the number of rows corresponds to the number of all available letters, i.e. m = 4, and the number of columns is 3, there are 12 different combinations in total.

Example 3. Calculate: a) A; b)

Solution. a) A = 6 5 4 = 120.

b) Since A = 15 14 13, A = 15 14 13 12, A = 15 14 13 12 11, then

Example 4. How many two-digit numbers can be made from five digits 1, 2, 3, 4, 5, provided that none of them are repeated?

Solution. Since two-digit numbers differ from each other either in the numbers themselves or in their order, the required quantity is equal to the number of placements of five elements in twos: A = 5 · 4 = 20. So, you can make 20 different two-digit numbers.

When finding the number of placements, we multiply n successively decreasing integers, i.e., there are not enough (m – n) successively decreasing integer factors to reach the full factorial.

m factors

Therefore, the formula for the number of placements can be written as

A =

Hence, taking into account that the numerator is equal to m!, and the denominator is equal to (m – n)!, we write this formula in factorial form:

A = (4)

Example 5. Calculate A in factorial form.

Solution. A =

15-20. Calculate in any way:

15. A; 16. A; 17. A; 18. A; 19. A; 20.

21. How many options are there for distributing three prizes if 7 teams participate in the draw?

22. How many different four-digit numbers can be made from the digits 0, 1, 2, ..., 8, 9?

23. How many schedule options can be created for one day if there are 8 in total? educational subjects, and only three of them can be included in the daily schedule?

24. How many options for distributing three vouchers to sanatoriums of various profiles can be compiled for five applicants?

4. Combinations

Combinations are all combinations of m elements of n that differ from each other by at least one element (here m and n are natural numbers, and n m).

So, from four different letters A, B, C, D, you can make the following combinations that differ from each other in at least one element: AB, AC, AD, BC, BD, CD. This means that the number of combinations of four elements of two is 6. This is briefly written as follows: C = 6.

In each combination we will rearrange the elements:

AB, AC, AD, BC, BD, CD;

BA, CA, DA, CB, DB, DC.

As a result, we received an arrangement of four elements, two each. Therefore, CP2 = A, whence C =