How many different options exist. Combinatorial tasks (class 5)

The combinatorial is called the Mathematics section, which studies the question of how many combinations of a certain type can be made from these items (elements).

Multiplication rule (basic formula combinatorics)

The total number of methods that can be selected by one element from each group and place them in a certain order (that is, to obtain an ordered set) is equal to:

Example 1.

The coin was thrown 3 times. How many different thumbnails can be expected?

Decision

The first coin has an alternative - either an eagle or a rush. For the second coin there are also alternatives, etc., i.e. .

The searched number of ways:

Completion rule

If any two groups do not have common elements, then choose one element or from, or from, ... or from can be carried out in ways.

Example 2.

On the shelf 30 books, of which are 20 mathematical, 6 technical and 4 economic. How many ways to choose one mathematical or one economic book.

Decision

The mathematical book can be selected by ways, economic - ways.

According to the rule, there is a way to choose a mathematical or economic book.

Accommodation and permutations

Accommodation - These are ordered set of elements that differ from each other or composition, or the order of elements.

Placement without repetitionWhen the selected element before the next selection is not returned to the general population. Such a choice is called a consistent choice without return, and its result is by placement without repetitions from the elements of software.

The number of different methods that can be sequentially selected without returning elements from the general population of volume, is:

Example 3.

The day schedule consists of 5 different lessons. Determine the number of schedule options when choosing from 11 disciplines.

Decision

Each version of the schedule is a set of 5 disciplines from 11, differing from other options of both the composition and the procedure for the following. so:

Rearranged - These are ordered sets, differing from each other only by the order of elements. The number of all the permutations of the set from the elements is equal

Example 4.

How many ways can you send 4 people at one table?

Decision

Each option of seating is different only by the order of participants, that is, it is a permutation of 4 elements:

Placement with repetitionWhen the selected item before selecting the next returned to the general population. Such a choice is called a consistent choice with a return, and its result is an accommodation with repetitions from software elements.

The total number of different methods that can be selected with the return of elements from the general population of volume, equal to

Example 5.

The elevator stops on 7 floors. How many ways can there be 6 passengers in the elevator cabin on these floors?

Decision

Each of the ways of the distribution of passengers on floors is a combination of 6 passengers in 7 floors, distinguished from other combinations of both the composition and their order. Since one floor can go out as one and several passengers, then the same passengers can repeat. Therefore, the number of such combinations is equal to the number of placements with repetitions of 7 elements of 6:

Combination

Combinations From n elements according to K are unordered sets, differing from each other at least one element.

Let several elements be taken from the general population (either the elements are taken consistently, but the order of their appearance is not taken into account). As a result of such a simultaneous disordered choice of elements from the general population of volume, combinations are obtained, which are called combinations without repetitions from elements of software.

The number of combinations from the elements is:

Example 6.

In the drawer 9 apples. How many ways can you choose 3 apples from the box?

Decision

Each option of choice consists of 3 apples and differs from other only composition, that is, it is a combination without repetitions of 9 elements:

The number of ways to choose 3 apples from 9:

Let the elements be selected from the general set of volume, one by one, with each selected element before the selection of the next returned to the general population. It is recorded, which elements appeared and how many times, however, the procedure for their appearance is not taken into account. The resulting aggregates are called combinations with repetitions from elements of software.

The number of combinations with repetitions from the following elements:

Example 7.

In the mail sell postcards 3 species. How many ways can I buy 6 postcards?

This is the task of finding the number of combinations with repetitions of 3 to 6:

Split a set in groups

Let the set of various elements be divided into groups so, then elements fall into the first group, in the second elements, a group of elements, and. This situation is called the splitting of the sets into groups.

The number of partitions on the groups when elements fall into the first, in the second elements, in k-Yu group - elements, equal:

Example 8.

A group of 16 people need to be divided into three subgroups, in the first of which there should be 5 people, in the second - 7 people, in the third - 4 people. How many ways can this be done?

The combinatorics is a section of mathematics, which studies questions about how many different combinations subordinate to those or other conditions can be made up of the specified objects. The basics of combinatorics are very important to assess the probabilities of random events, because It is they who allow you to calculate the fundamentally possible number of different options for the development of events.

Basic formula combinatorics

Let there be k groups of elements, and i-I group Consists of n i elements. Select one element from each group. Then total number N methods that can be made such a choice is determined by the relation N \u003d n 1 * N 2 * N 3 * ... * n k.

Example 1. Let us explain this rule on a simple example. Let there be two groups of elements, and the first group consists of N 1 of the elements, and the second is from N 2 elements. How many different pairs of elements can be made up of these two groups, so that in a pair one element from each group? Suppose we took the first element from the first group and, without changing it, moved all possible pairs, changing only elements from the second group. Such pairs for this element can be made by N 2. Then we take the second element from the first group and also make up all possible couples for it. Such couples will also be N 2. Since in the first group of only N 1 element, all possible options will be N 1 * N 2.

Example 2. How many three-digit even numbers can be made from numbers 0, 1, 2, 3, 4, 5, 6, if the numbers can repeat?
Decision:n 1 \u003d 6 (because as the first digit can be taken any digit of 1, 2, 3, 4, 5, 6), N 2 \u003d 7 (because as a second digit can be taken any digit of 0 , 1, 2, 3, 4, 5, 6), N 3 \u003d 4 (because as a third digit, you can take any digit of 0, 2, 4, 6).
So, n \u003d n 1 * N 2 * N 3 \u003d 6 * 7 * 4 \u003d 168.

In the case when all groups consist of an identical number of items, i.e. N 1 \u003d N 2 \u003d ... n k \u003d n We can assume that each selection is made from the same group, and the element after the selection is returned to the group again. Then the number of all methods of choice is N K. This method of choice in combinatorics is called Sampling with return.

Example 3. How many of all four-digit numbers can be made from numbers 1, 5, 6, 7, 8?
Decision. For each discharge of the four-digit number there are five possibilities, which means N \u003d 5 * 5 * 5 * 5 \u003d 5 4 \u003d 625.

Consider a set consisting of n elements. This set in the combinatorics is called general Consideration.

Number of accommodation from n elements by m

Definition 1. Accommodation out n. Elements in m. in the combinatorics is called any ordered set of m. various elements selected from the general population in n. Elements.

Example 4.Various places of three elements (1, 2, 3) two will be kits (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2 ). The placement may differ from each other both elements and their order.

The number of accommodation in the combinatorics is denoted by A n M and is calculated by the formula:

Comment: n! \u003d 1 * 2 * 3 * ... * n (read: "en factorial"), in addition, it is believed that 0! \u003d 1.

Example 5.. How many two-digit numbers are there, in which the number of dozens and digit of units are different and odd?
Decision: Because odd figures five, namely 1, 3, 5, 7, 9, then this task is reduced to the selection and placement of two different positions of two of five different numbers, i.e. These numbers will be:

Definition 2. Combination of n. Elements in m. in the combinatorics is called any unordered set of m. various elements selected from the general population in n. Elements.

Example 6.. For the set (1, 2, 3), combinations are (1, 2), (1, 3), (2, 3).

The number of combinations from n elements by m

The number of combinations is denoted by C n M and is calculated by the formula:

Example 7.How many ways the reader can choose two books from six available?

Decision:The number of ways is equal to the number of combinations of six two books, i.e. equally:

Permutations from n elements

Definition 3. Perestanovka of n. Elements are called any ordered set These elements.

Example 7A. All sorts of permutations of the set consisting of three elements (1, 2, 3) are: (1, 2, 3), (1, 3, 2), (2, 3, 1), (2, 1, 3), ( 3, 2, 1), (3, 1, 2).

The number of different permutations from N elements is denoted by P n and is calculated by the formula P n \u003d n!.

Example 8. In how many ways, seven books of different authors can be placed on the shelf in one row?

Decision:this task is about the number of permutations of seven different books. There is P 7 \u003d 7! \u003d 1 * 2 * 3 * 4 * 5 * 6 * 7 \u003d 5040 ways to place books.

Discussion.We see that the number of possible combinations can be calculated by various rules (permutations, combinations, placement), and the result will be different, because The principle of calculation and the formulas themselves are different. Carefully looking at the definitions, it can be noted that the result depends on several factors at the same time.

First of all, from what amount of elements we can combine their sets (how big general aggregate elements).

Secondly, the result depends on what kind of element sets we need.

And the last, it is important to know whether it is essential for us the order of elements in the set. Let us explain the last factor in the following example.

Example 9.On the parent Meeting There are 20 people. How many different options are the composition of the parent committee, if 5 people should enter it?
Decision:In this example, we are not interested in the order of surnames in the list of Committee. If the same people will be among its composition, then in meaning for us it is the same option. Therefore, we can take advantage of the formula for counting the number Combinationsof the 20 elements of 5.

Otherwise, things will be faced if each member of the Committee is initially responsible for a certain direction of work. Then, with the same list of the Committee, it is possible 5! Options rearrangedthat matter. The number of different (and in composition, and on the scope of responsibility) is determined in this case by the number accommodation Of the 20 elements of 5.

Tasks for self-test
1. How many three-digit even numbers can be made from numbers 0, 1, 2, 3, 4, 5, 6, if the numbers can repeat?

2. How many five-digit numbers exist, which are equally read from left to right and right left?

3. In the class ten items and five lessons per day. How many ways can be a schedule for one day?

4. how many ways can you choose 4 delegates to the conference if in a group of 20 people?

5. How many ways can you decompose eight different letters on eight different envelopes if only one letter is put in each envelope?

6. Of the three mathematicians and ten economists, it is necessary to compile a commission consisting of two mathematicians and six economists. How many ways can this be done?

Abstract on the topic:

Performed a student of grade 10 "B"

high School №53

Glukhov Mikhail Alexandrovich

naberezhnye Chelny

2002
Content

From the history of the combinatorics _________________________________________	3
The rule of the amount ___________________________________________________	4
	-
Rule of work _____________________________________________	4
Examples of tasks ____________________________________________________	-
Intersecting sets ________________________________________	5
Examples of tasks ____________________________________________________	-
Euler circles _________________________________________________	-
Placement without repetitions ________________________________________	6
Examples of tasks ____________________________________________________	-
Permutations without repetitions _______________________________________	7
Examples of tasks ____________________________________________________	-
Combinations without repetitions __________________________________________	8
Examples of tasks ____________________________________________________	-
Placement and combinations without repetitions ______________________________	9
Examples of tasks ____________________________________________________	-
Rearrangements with repetitions ___________________________________	9
Examples of tasks ____________________________________________________	-
Tasks for self-decide________________________________	10
Bibliography___________________________________	11

From the history of combinatorics

Combinatorics is engaged of various types compounds that can be formed from the elements of the final set. Some elements of the combinatorics were known in India in the II century. BC e. Nidice could calculate the numbers that are now called "combinations". In the XII century Bhaskar calculated some types of combinations and permutations. It is assumed that Indian scientists studied compounds in connection with the use of them in the poetics, science on the structure of verse and poetic works. For example, due to the calculation of possible combinations of drums (long) and unstressed (short) syllables of the foot from N syllables. As a scientific discipline, the combinatorics was formed in the XVII century. In the book "Theory and Practice of Arithmetic" (1656), French author A. also devotes combinations and permutations a whole chapter.
B. Pascal in the "Treatise on an arithmetic triangle" and in the "treatise on numerical orders" (1665) outlined the doctrine of binomial coefficients. P. Farm knew about the bonds of mathematical squares and curly numbers with the theory of compounds. The term "combinatorics" began to be used after the published by the Leibnian in 1665, the work "Correction on the combinatorial art", in which the first given was given scientific justification Theories of combinations and permutations. The study of accommodation was first engaged in J. Bernoulli in the second part of his book "ARS Conjectandi" (the art of prevailing) in 1713. Modern symbolism of combinations was proposed by various authors of educational guidelines only in the XIX century.

All variety of combinatorial formulas can be derived from two main statements concerning finite sets - The rule of the amount and the rule of the work.

Rule sum

If the final sets do not intersect, the number of elements x U y (or) is equal to the sum of the number of elements of the set X and the number of elements of the set Y.

That is, if there are X books on the first shelf, and on the second Y, then choose a book from the first or second shelf, you can x + y in ways.

Examples of tasks

The student must perform practical work in mathematics. He was offered to choose from 17 by algebra and 13 by geometry. How many ways he can choose one topic for practical work?

Solution: x \u003d 17, y \u003d 13

According to the rule of the sum x U y \u003d 17 + 13 \u003d 30.

There are 5 monetary lottery tickets, 6 tickets of Sports and 10 tickets of the automaker. How many ways can you choose one ticket from Sportloto or Auto Maker?

Solution: Since the monetary lottery does not participate in the choice, then only 6 + 10 \u003d 16 options.

Rule of work

If the X element can be selected by k methods, and item Y-M ways That pair (x, y) can be chosen K * M methods.

That is, if there are 5 books on the first shelf, and on the second 10, then select one book from the first shelf and one with the second can be 5 * 10 \u003d 50 ways.

Examples of tasks

The rebooter must intertwine 12 different books in red, green and brown bounds. How many ways can he do it?

Solution: There are 12 books and 3 colors, it means on the rules of the work it is possible 12 * 3 \u003d 36 of the binding options.

How many five-digit numbers exist, which are equally read from left to right and right to left?

Solution: In such numbers, the last figure will be the same as the first, and the penultimate - as well as the second. The third digit will be any. This can be represented as Xyzyx.where y and z-thigh numbers, and x - not zero. It means according to the rule of the work the number of numbers equally ready both from left to right and right to left equal to 9 * 10 * 10 \u003d 900 options.

Intersecting sets

But it happens that the sets X and Y intersect, then use the formula

where X and Y are sets, and the area of \u200b\u200bintersection. Examples of tasks

20 people are 10 - German, abundant 5 I know Intiangali, Inherets. How many times?

Answer: 10 + 20-5 \u003d 25 people.

Also often for visual solution The tasks are used circles Euler. For example:

Of the 100 tourists going to the passage trip, german language Own 30 people, English - 28, French - 42. English and German at the same time own 8 people, English and French - 10, German and French - 5, by all three languages \u200b\u200b- 3. How many tourists do not own any language?

Decision: Express the condition of this task graphically. Denote by the circle of those who know English, another round - those who know French, and the third round - those who know German.

Three tourists own all three languages, it means that in the total parts of the circles enter the number 3. English and french language Owned 10 people, and 3 of them also own German. Consequently, only English and French own 10-3 \u003d 7 people.

Similarly, we obtain that only English and German own 8-3 \u003d 5 people, and German and French 5-3 \u003d 2 tourists. We introduce this data to the appropriate parts.

We define now how many people own only one of the listed languages. German knows 30 people, but 5 + 3 + 2 \u003d 10 of them own other languages, therefore, only German know 20 people. Similarly, we get that one English is owned by 13 people, and one French is 30 people.

Under the condition of the task of only 100 tourists. 20 + 13 + 30 + 5 + 7 + 2 + 3 \u003d 80 Tourists know at least one language, therefore, 20 people do not own any of these languages.

Placement without repetitions.

How much can phone numbers from 6 digits each, so that all the numbers are different?

This is an example of the task of placement without repetitions. 10 numbers are placed here 6. And options in which the same numbers are in different order are considered different.

If the X-set consisting of N elements, M≤n, then by placing without repetitions from n elements of the set x by m is called an ordered set X, which contains m elements is called an ordered set X containing M elements.

The number of all accommodation from N elements by m is denoted

n! - N-factorial (Factorial Ang. Fancy) The product of the numbers of a natural row from 1 to any number n A task

How many ways are 4 young men who can invite four of six girls to dance?

Decision: Two young men can simultaneously invite the same girl. And options at which the same girls dance with different young men are considered different, so:

Perhaps 360 options.

Rearrangements without repetitions

In the case of n \u003d m (see placement without repetitions) from N elements by m is called the permutation of the set x.

The number of all permutations from n elements is denoted by P n.

Really at n \u003d m:

Examples of tasks

How many different six-digit numbers can be made from numbers 0, 1, 2, 3, 4.5, if the numbers are among the number they do not repeat?

1) We find the amount of all permutations from these numbers: P 6 \u003d 6! \u003d 720

2) 0 cannot stand ahead of the number, therefore, from this number it is necessary to take away the number of permutations, at which 0 is in front. And this p 5 \u003d 5! \u003d 120.

P 6 -p 5 \u003d 720-120 \u003d 600

Martushka lean

Yes Kosolapy Mishka

Speed \u200b\u200bto play quartet

Stop, brothers stand! -

Screams Marty, - Wait!

How to go music?

After all, you are not so sitting ...

And so, and the point was transplanted - again the music on the way does not go.

In many combinatorial tasks, the immediate foundation of the number of options of interest is difficult. However, with some change, the conditions of the task can find the number of options that are superior to the initial number of times. This reception is called method of multiple counting.

1. How many anagrams have the word class?

The difficulty is that in this word two identical letters S. We will temporarily consider them different and denoted from 1 and C 2. Then the number of anagrams will be equal to 5! \u003d 120. But those words that differ from each other only with permutation of letters from 1 and C 2, are the same and the same analog! Therefore, 120 anagrams are divided into pairs of the same, i.e. The desired number of anagrams is 120/2 \u003d 60.

2. How many anagrams have the word charade?

Considering three letters a varied letters a 1, and 2, and 3, we get 6! Anagram. But words that are obtained from each other only by permutation of letters a 1, and 2, and 3, in fact are the same analog. As there are 3! permutations of letters a 1, and 2, and 3, obtained initially 6! Anagrams are divided into groups of 3! The same, and the number of different anagrams turns out to be equal to 6! / 3! \u003d 120.

3. How many four-digit numbers exist, in which there are at least one even digit?

We find the number of "unnecessary" four-digit numbers, in which only odd numbers are present in the record. Such numbers 5 4 \u003d 625. But only four-digit numbers 9000, so the desired number of "necessary" numbers is 9000 - 625 \u003d 8375.

Find the number of anagrams from the words Heather, Balagan, city.
Find the number of anagrams from the words of Baobab, Ballad, Trouble, Anagram, Mathematics, Combinatorics, Defense capability.
How many ways can I settle 7 visitors to three hotel rooms: Single, Double and Quadruple?
In the refrigerator lie two apples, three pears and four oranges. Every day, for nine days in a row, Pet gives one of some kind of fruit. How many ways can it be done?
From the seven best skiers of the school you need to select the team of three to participate in urban competitions. How many ways can this be done?
Before the exam, the professor promised to put two half examinations. 20 students came to the exam. How many ways can he fulfill the promise?
How many words can be made up of five letters a and no more than three letters b?
There is a chocolate, strawberry and dairy ice cream on sale. How many ways can you buy three ice creams?
When preparing pizza to cheese, different components are added, providing one or another taste. At the disposal of Bill there are onions, mushrooms, tomatoes, peppers and anchovies, all this, in his opinion, can be added to cheese. How many types of pizza can cook Bill?
The witness of the criminally disassembly remembered that the criminals were hidden on Mercedes, the number of which contained the letters t, s, y and numbers 3 and 7 (the number is the line in which three letters go first, and then three numbers). How many such numbers exist?
How many diagonals in the convex n.-Golf?
How much is there exist n.-Quality numbers?
How many decade-digit numbers exist, in which there are at least two identical numbers?
The cube throws three times. Among all sorts of sequences, there are those in which six at least once fell. How many of them?
How many five-digit numbers have numbers in their record 1?
How many ways can be placed on chessboard White and black king so that they do not beat each other?
How many divisors have 10800?

Dedicated to solving problems of selection and location of elements of some, usually finite, sets in accordance with the specified rules. For example, how many ways you can choose 6 cards from a deck consisting of 36 cards, or how many ways can be a queue consisting of 10 people, etc. Each rule in the combinatorics determines the method of constructing some design composed of elements of the original set and called combination. The main goal of the combinatorics is counting the number of combinations that can be made from the elements of the source set in accordance with the specified rule. The simplest examples of combinatorial structures are permutations, accommodation and combinations.

Birth combinatorics associated with the works B. Pascal And P. Farm about gambling, leibyans, Bernoulli, Euler contributed great contribution. Currently, interest in the combinatorics is associated with the development of computers. Combinatorics will be interested in the possibility of determining quantitatively different subsets of finite sets to calculate the likelihood by a classic method.

To determine the power of a set that corresponds to a particular event, it is useful to deal with the two rules of the combinatorics: the rule of the work and the rule rule (sometimes they are called the principles of multiplication and addition, respectively).

Deference rule: Let from some finite set

1st object can be chosen k. 1 ways

2nd object - k. 2 ways

n.object - k N. ways. (1.1)

Then the arbitrary set listed n. objects from this set can be chosen k. 1 K. 2 , ..., k n ways.

Example 1. How many three-digit numbers with different numbers?

Decision. In the decimal calculus system, ten digits: 0.1,2,3,4,5,6,7,8,9. In the first place can stand any of nine digits (except zero). In second place - any of the remaining 9 digits, except for the selected one. On the last place Any of the remaining 8 digits.

According to the rule, 9 · 9 · 8 \u003d 648 three-digit numbers have different numbers.

Example 2. From paragraph 3 roads lead to item, and from the point to paragraph - 4 roads. In how many ways you can travel from in through?

Decision. In point there are 3 ways to choose the road to the item, and in the point there are 4 ways to get to the item. According to the principle of multiplication, there are 3 × 4 \u003d 12 methods to get from the point to the item.

Rule amount: When performing conditions (1.1), any of the objects can be selected k. 1 + K. 2 + ... + k n ways.

Example 3. How many ways to choose one pencil out of a box containing 5 red, 7 blue, 3 green pencils.

Decision. One pencil, according to rule, you can choose 5 + 7 + 3 \u003d 15 methods.

Example 4. Let from the city the city can be reached by one airplane, two rail routes and three bus routes. How many ways can be reached from the city in town ?

Decision. All conditions of the principle of addition here are completed here, therefore, in accordance with this principle, we obtain 1 + 2 + 3 \u003d 6 methods.

Consider an example illustrating the difference between the principles of multiplication and addition.

Example 5. Three brands of televisions and two types of video recorders are sold in the electronics store. The buyer has the opportunity to purchase either a TV or a VCR. How many ways can he make one purchase? How many different sets containing a TV and a tape recorder can be purchased in this store if the buyer is going to purchase a pair and TV and a VCR?

Decision. One TV can be chosen in three ways, and the tape recorder is the other two ways. Then the TV or the tape recorder can be bought 3 + 2 \u003d 5 ways.

In the second case, one TV can be chosen in three ways, after that the VCR can be selected in two ways. Consequently, due to the principle of multiplication, you can buy a TV and a video recorder 3 × 2 \u003d 6 methods.

We now consider examples in which both rules of combinatorics are applied: and the principle of multiplication, and the principle of addition.

Example 6. 12 apples and 10 oranges lie in the basket. Vanya chooses either an apple or an orange. After that, Nadya chooses from the remaining fruits and an apple and an orange. How many possible elections are possible?

Decision. Vanya can choose an apple 12 ways, orange - 10 ways. If Vanya chooses an apple, then Nadia can choose an apple 11 ways, and an orange - 10 ways. If Vanya chooses an orange, then Nadia can choose an apple 12 ways, and orange - 9 ways. Thus, Vanya and Nadia can make their choice in ways.

Example 7. There are 3 letters, each of which can be sent to 6 addresses. How many ways can this be done?

Decision. In this task, we must consider three cases:

a) all letters are sent at different addresses;

b) all letters are sent at one address;

c) Only two letters are sent at one address.

If all letters are sent at different addresses, the number of such methods is easily located from the principle of multiplication: n. 1 \u003d 6 × 5 × 4 \u003d 120 methods. If all letters are sent at one address, there will be such ways n. 2 \u003d 6. Thus, it remains only to consider the third case when only 2 letters are sent at one address. We can choose any letter in 3 ways, and send it to any selected address can 6 ways. We can send the remaining two letters to the remaining addresses 5 ways. Consequently, send only two letters to one address we can n. 3 \u003d 3 × 6 × 5 \u003d 90 methods. Thus, sent 3 letters to 6 addresses in line with the principle of addition

ways.

Usually in the combinatorics there is an idealized experiment on choosing k. Elements from n.. At the same time, items: a) are not returned back (selection circuit without returns); b) Return back (selecting a selection circuit).

1. Selection Scheme without returns

Accommodationof n. Elements in k. call any ordered set of k. Elements belonging n. - element set. Various accommodations are different from each other or order of elements, or composition.

Number of accommodation out n. Elements in k. denotes and calculated by the formula

(1.2)

where n.! \u003d 1 × 2 × 3 × ... × n., 1! = 1, 0! = 1.

Example 8. 10 people participate in competitions, three of them will take 1, 2, 3rd place. How many different options are there?

Decision. In this case, the procedure for distributing places is important. The number of different options is equal

Permutationof n. elements call accommodation from n. Elements in n. The number of permutations is n. Elements are denoted P N. and calculate the formula

(1.3)

Example 9. How many ways to arrange 10 books on the shelf?

Decision. The total number of ways of arrangement is defined as the number of permutations (1.3) out of 10 elements and equal R 10 = 10! = 3628 800.

2. Selection Scheme with Returies

If when choosing k. Elements from n., items are returned back and streamlined, they say it accommodation with reasons .

The number of placements with repetitions:

Example 11. The hotel has 10 rooms, each of which can accommodate four people. How many accommodation options arrived by four guests?

Decision. Each next guest out of 4 can be placed in any of the 10 rooms, as idealized experience is considered, therefore the total number of accommodations, by the formula of repetitions (1.5), is equal

If when choosing k. Elements from n. items are returned back without subsequent ordering, they say it combines with replacements. The number of combinations with repetitions from n. Elements in k. Determined:

Example 12. The store sells 10 types of cakes. Another buyer knocked on three cake. Considering that any set of products is equal to, determine the number of possible orders.

Decision. The number of equal orders by formula (1.6) is equal