The number of different placements of a3 5 is equal. Combinatorics formulas

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Introduction. Sets and selections.

In this topic we will look at the basic concepts of combinatorics: permutations, combinations and placements. Let's find out their essence and formulas by which you can find their quantity.

To work, we need some auxiliary information. Let's start with such a fundamental mathematical concept as a set. The concept of a set was discussed in detail in the topic "The concept of a set. Methods of defining sets".

Very short story about sets: show\hide

In short: a set is a collection of objects. Write sets in curly braces. The order in which the elements are written does not matter; repetitions of elements are not allowed. For example, the set of digits of the number 11115555999 will be: $\(1,5,9\)$. The set of consonants in the word "tiger cub" is: $\(t, g, r, n, k\)$. The notation $5\in A$ means that element 5 belongs to the set $A=\(1,5,9 \)$. The number of elements in a finite set is called power of this set and denote $|A|$. For example, for a set $A=\(1,5,9 \)$ containing 3 elements, we have: $|A|=3$.

Consider a certain non-empty finite set $U$, whose cardinality is $n$, $|U|=n$ (i.e., the set $U$ has $n$ elements). Let's introduce a concept like sample(some authors call it a tuple). By a sample of volume $k$ from $n$ elements (abbreviated as $(n,k)$-sample) we mean a set of elements $(a_1, a_2,\ldots, a_k)$, where $a_i\in U$. A selection is called ordered if the order of its elements is specified. Two ordered samples that differ only in the order of the elements are different. If the order of the sample elements is not significant, then the sample is called unordered.

Note that the definition of a selection does not say anything about element repetitions. Unlike set elements, selection elements can be repeated.

For example, consider the set $U=\(a,b,c,d,e\)$. The set $U$ contains 5 elements, i.e. $|U|=5$. A sample without repetitions could be: $(a,b,c)$. This selection contains 3 elements, i.e. the size of this sample is 3. In other words, it is a $(5,3)$-sample.

A sample with repetitions can be like this: $(a,a,a,a,a,c,c,d)$. It contains 8 elements, i.e. its volume is 8. In other words, this is a $(5,8)$-sample.

Let's consider two more $(5,3)$-samples: $(a,b,b)$ and $(b,a,b)$. If we assume that our samples are unordered, then the sample $(a,b,b)$ is equal to the sample $(b,a,b)$, i.e. $(a,b,b)=(b,a,b)$. If we assume our samples are ordered, then $(a,b,b)\neq(b,a,b)$.

Let's look at another example, a little less abstract:) Suppose there are six candies in a basket, and they are all different. If we associate the first candy with the number 1, the second candy with the number 2, and so on, then the following set can be associated with the candies in the basket: $U=\(1,2,3,4,5,6\)$. Imagine that we randomly put our hand into a basket in order to pull out three candies. The pulled out candies are the selection. Since we take 3 candies out of 6, we get a (6,3)-sample. The order in which the candies are placed in the palm is completely irrelevant, so this sample is unordered. Well, and since all the candies are different, the selection is without repetition. So, in this situation we are talking about an unordered (6,3)-sample without repetitions.

Now let's approach from the other side. Let's imagine that we are in a candy production factory, and this factory produces four types of candy. The set $U$ in this situation is as follows: $U=\(1,2,3,4 \)$ (each number is responsible for its own type of candy). Now let’s imagine that all the candies are poured into a single chute, near which we are standing. And, placing our palms, we select 20 candies from this flow. A handful of sweets is a sample. Does the order in which the candies are placed in a handful matter? Naturally, no, so the sample is unordered. There are only 4 varieties of candies, and we select twenty pieces from the general flow - repetition of varieties is inevitable. At the same time, the samples can be very different: we may even have all the candies of the same type. Therefore, in this situation we are dealing with an unordered (4,20)-sample with repetitions.

Let's look at a couple more examples. Let different 7 letters be written on the cubes: k, o, n, f, e, t, a. These letters form the set $U=\(k,o,n,f,e,t,a\)$. Let's say that from these cubes we want to make “words” of 5 letters. The letters of these words (for example, “confe”, “tenko” and so on) form (7,5)-selections: $(k,o,n,f,e)$, $(t,e,n,k ,o)$, etc. Obviously, the order of the letters in such a sample is important. For example, the words “nokft” and “kfton” are different (although they consist of the same letters), because the order of the letters in them does not match. There are no repetitions of letters in such “words”, because there are only seven cubes. So, the set of letters of each word is an ordered (7,5)-sample without repetitions.

Another example: we make all kinds of eight-digit numbers from four digits 1, 5, 7, 8. For example, 11111111, 15518877, 88881111 and so on. The set $U$ is: $U=\(1,5,7,8\)$. The digits of each composed number form a (4,8)-sample. The order of the digits in a number is important, i.e. the sample is ordered. Repetitions are allowed, so here we are dealing with an ordered (4,8)-sample with repetitions.

Placements without repetitions of $n$ elements by $k$

Placement without repetitions of $n$ elements by $k$ - ordered $(n,k)$-selection without repetitions.

Since the elements in the sample under consideration cannot be repeated, we cannot select more elements into the sample than are in the original set. Therefore, for such samples the following inequality is true: $n≥ k$. The number of placements without repetitions of $n$ elements by $k$ is determined by the following formula:

\begin(equation)A_(n)^(k)=\frac(n{(n-k)!} \end{equation} !}

What does the "!" sign mean?: show\hide

Recording "n!" (read "en factorial") denotes the product of all numbers from 1 to n, i.e.

$$ n!=1\cdot2\cdot 3\cdot \ldots\cdot n $$

By definition, it is assumed that $0!=1!=1$. For example, let's find 5!:

$$ 5!=1\cdot 2\cdot 3\cdot 4\cdot 5=120. $$

Example No. 1

The alphabet consists of a set of symbols $E=\(+,*,0,1,f\)$. Let us determine the number of such three-character words in this alphabet that do not contain repeating letters.

By three-character words we mean expressions like “+*0” or “0f1”. The set $E$ has five elements, so the letters of three-character words form (5,3)-selections. The first question is: are these samples ordered or not? Words that differ only in the order of their letters are considered different, so the order of the elements in the sample is important. This means that the sample is ordered. Second question: are repetitions allowed or not? The answer to this question is given by the condition: words should not contain repeating letters. To summarize: the letters of each word that satisfies the conditions of the problem form an ordered (5,3)-sample without repetitions. In other words, the letters of each word form a placement without repetition of 5 elements of 3. Here are examples of such placements:

$$ (+,*,f), \; (*,+,f), \; (1,+,0) $$

We are interested in total quantity these placements. According to formula (1), the number of placements without repetitions of 5 elements of 3 will be as follows:

$$ A_(5)^(3)=\frac(5{(5-3)!}=\frac{5!}{2!}=60. $$ !}

Those. you can make 60 three-character words, the letters of which will not be repeated.

Answer: 60.

Placements with repetitions of $n$ elements of $k$

Placement with repetitions of $n$ elements by $k$ - ordered $(n,k)$-selection with repetitions.

The number of placements with repetitions of $n$ elements of $k$ is determined by the following formula:

\begin(equation)\bar(A)_(n)^(k)=n^k \end(equation)

Example No. 2

How many five-digit numbers can be made from the set of digits $\(5,7,2\)$?

From this set of numbers you can make five-digit numbers 55555, 75222, and so on. The digits of each such number form a (3,5)-sample: $(5,5,5,5,5)$, $(7,5,2,2,2)$. Let us ask ourselves: what kind of samples are these? First, digits in numbers can be repeated, so we are dealing with samples with repetitions. Secondly, the order of the digits in a number is important. For example, 27755 and 77255 - different numbers. Consequently, we are dealing with ordered (3,5)-samples with repetitions. We find the total number of such samples (i.e. the total number of required five-digit numbers) using formula (2):

$$ \bar(A)_(3)^(5)=3^5=243. $$

Therefore, 243 five-digit numbers can be made from the given digits.

Answer: 243.

Permutations without repetitions of $n$ elements

A permutation without repetitions of $n$ elements is an ordered $(n,n)$-selection without repetitions.

In essence, permutation without repetition is a special case of placement without repetition, when the sample size is equal to the cardinality of the original set. The number of permutations without repetition of $n$ elements is determined by the following formula:

\begin(equation)P_(n)=n! \end(equation)

By the way, this formula is easy to obtain if you consider that $P_n=A_(n)^(n)$. Then we get:

$$ P_n=A_(n)^(n)=\frac(n{(n-n)!}=\frac{n!}{0!}=\frac{n!}{1}=n! $$ !}

Example No. 3

There are five servings of ice cream from different companies in the freezer. In how many ways can you choose the order in which they are eaten?

Let the number 1 correspond to the first ice cream, the number 2 to the second, and so on. We will get the set $U=\(1,2,3,4,5\)$, which will represent the contents of the freezer. The order of eating can be as follows: $(2,1,3,5,4)$ or as follows: $(5,4,3,1,2)$. Each such set is a (5,5)-sample. It will be orderly and without repetition. In other words, each such sample is a permutation of 5 elements of the original set. According to formula (3), the total number of these permutations is as follows:

$$ P_5=5!=120. $$

Consequently, there are 120 orders for choosing the order of eating.

Answer: 120.

Permutations with repetitions

Permutation with repetitions is an ordered $(n,k)$-sample with repetitions, in which element $a_1$ is repeated $k_1$ times, $a_2$ is repeated $k_2$ times, and so on, until the last element $a_r$, which is repeated $ k_r$ times. In this case, $k_1+k_2+\ldots+k_r=k$.

The total number of permutations with repetitions is determined by the formula:

\begin(equation)P_(k)(k_1,k_2,\ldots,k_r)=\frac(k{k_1!\cdot k_2!\cdot \ldots \cdot k_r!} \end{equation} !}

Example No. 4

Words are composed based on the alphabet $U=\(a,b,d\)$. How many different words can be composed of seven characters if in these words the letter “a” must be repeated 2 times; the letter "b" - 1 time, and the letter "d" - 4 times?

Here are examples of search words: "aabdddd", "daddabd" and so on. The letters of each word form a (3,7)-sample with repetitions: $(a,a,b,d,d,d,d)$, $(d,a,d,d,a,b,d)$ and etc. Each such selection consists of two elements "a", one element "b" and four elements "d". In other words, $k_1=2$, $k_2=1$, $k_3=4$. The total number of repetitions of all symbols, naturally, is equal to the sample size, i.e. $k=k_1+k_2+k_3=7$. Substituting these data into formula (4), we will have:

$$ P_7(2,1,4)=\frac(7{2!\cdot 1!\cdot 4!}=105. $$ !}

Therefore, the total number of search words is 105.

Answer: 105.

Combinations without repetitions of $n$ elements of $k$ each

A combination without repetitions of $n$ elements by $k$ is an unordered $(n,k)$-sample without repetitions.

The total number of combinations without repetitions of $n$ elements of $k$ is determined by the formula:

\begin(equation)C_(n)^(k)=\frac(n{(n-k)!\cdot k!} \end{equation} !}

Example No. 5

The basket contains cards with integers written on them from 1 to 10. 4 cards are taken out of the basket and the numbers written on them are added up. How many different sets of cards can be drawn from the basket?

So, in this problem the initial set is: $U=\(1,2,3,4,5,6,7,8,9,10\)$. From this set we select four elements (i.e., four cards from the basket). The numbers of the pulled out elements form a (10,4)-selection. Repetitions in this selection are not allowed, since the numbers of all cards are different. The question is: does the order in which cards are selected matter or not? That is, for example, are the samples $(1,2,7,10)$ and $(10,2,1,7)$ equal or not equal? Here you need to turn to the conditions of the problem. The cards are taken out in order to later find the sum of the elements. This means that the order of the cards is not important, since changing the places of the terms will not change the sum. For example, the sample $(1,2,7,10)$ and the sample $(10,2,1,7)$ will correspond to the same number $1+2+7+10=10+2+1+7= $20. Conclusion: from the conditions of the problem it follows that we are dealing with unordered samples. Those. we need to find the total number of unordered (10,4)-samples without repetitions. In other words, we need to find the number of combinations of 10 elements of 4. We use formula (5) for this:

$$ C_(10)^(4)=\frac(10{(10-4)!\cdot 4!}=\frac{10!}{6!\cdot 4!}=210. $$ !}

Therefore, the total number of searched sets is 210.

Answer: 210.

Combinations with repetitions of $n$ elements of $k$ each

A combination with repetitions of $n$ elements of $k$ is an unordered $(n,k)$-sample with repetitions.

The total number of combinations with repetitions of $n$ elements of $k$ is determined by the formula:

\begin(equation)\bar(C)_(n)^(k)=\frac((n+k-1){(n-1)!\cdot k!} \end{equation} !}

Example No. 6

Imagine that we are at a candy factory, right next to a conveyor along which four types of candies move. We put our hands into this stream and pull out twenty pieces. How many different "candy combinations" can there be in a handful?

If we assume that the first type corresponds to the number 1, the second type - the number 2, and so on, then the initial set in our problem is as follows: $U=\(1,2,3,4\)$. From this set we select 20 elements (i.e., those same 20 candies from the assembly line). A handful of sweets forms a (4,20)-sample. Naturally, there will be repetitions of varieties. The question is, does the order of the elements in the sample matter or not? From the conditions of the problem it follows that the order in which the elements are arranged does not matter. It makes no difference to us whether the handful contains first 15 lollipops, and then 4 chocolate candies, or first 4 chocolate candies, and only then 15 lollipops. So, we are dealing with an unordered (4,20) sample with repetitions. To find the total number of these samples we use formula (6):

$$ \bar(C)_(4)^(20)=\frac((4+20-1){(4-1)!\cdot 20!}=\frac{23!}{3!\cdot 20!}=1771. $$ !}

Therefore, the total number of searched combinations is 1771.

Consider the problem of counting the number of samples from a given set in general view. Let there be some set N, consisting of n elements. Any subset consisting of m elements can be considered without taking into account their order, or taking it into account, i.e. when changing the order, move to another m– sampling.

Let us formulate the following definitions:

Placements without repetition

Placement without repetition ofn elements bym Ncontainingmvarious elements.

From the definition it follows that the two arrangements differ from each other, both in their elements and in their order, even if the elements are the same.

Theorem 3. The number of placements without repetition is equal to the product m factors, the largest of which is the number n . Write down:

Permutations without repetition

Permutations fromn elements are called different orderings of the setN.

From this definition it follows that the two permutations differ only in the order of the elements and they can be considered as a special case of placements.

Theorem 4. The number of different permutations without repetition is calculated by the formula

Combinations without repetitions

A combination without repetition ofn elements bym any unordered subset of a set is calledNcontainingm various elements.

From the definition it follows that the two combinations differ only in elements; the order is not important.

Theorem 5. The number of combinations without repetitions is calculated using one of the following formulas:

Example 1. There are 5 chairs in the room. In how many ways can you place them on them?

a) 7 people; b) 5 people; c) 3 people?

Solution: a) First of all, you need to select 5 people out of 7 to sit on chairs. It can be done
way. With each choice of a specific five, you can produce
rearrangements. According to the multiplication theorem, the required number of landing methods is equal.

Comment: The problem can be solved using only the product theorem, reasoning as follows: for seating on the 1st chair there are 7 options, on the 2nd chair there are 6 options, on the 3rd -5, on the 4th -4 and on 5- th -3. Then the number of ways to seat 7 people on 5 chairs is . The solutions by both methods are consistent, since

b) The solution is obvious -

V) - number of elections of occupied chairs.

- the number of seats for three people on three selected chairs.

The total number of elections is .

It's not hard to check the formulas
;

;

The number of all subsets of a set consisting of n elements.

Repeat placements

By placing with repetition fromn elements bym every ordered subset of a set is calledN, consisting ofm elements so that any element can be included in this subset from 1 tomtimes, or be absent from it altogether.

The number of placements with repetition is denoted by and calculated using the formula, which is a consequence of the multiplication theorem:

Example 2. Let N = (a, b, c) be a set of three letters. Let's call any set of letters included in this set a word. Find the number of words of length 2 that can be made from these letters:
.

Comment: Obviously, placements with repetition can also be considered when
.

Example 3. You need to use the letters (a, b) to create all possible words of length 3. In how many ways can this be done?

Answer:

The first place in a row can be any of N elements, therefore, there are N options. In second place - any, except for the one that has already been used for first place. Therefore, for each of the N options already found, there are (N - 1) second place options, and the total number of combinations becomes N*(N - 1).
The same can be repeated for the remaining elements of the series. For the very last place, there is only one option left - the last remaining element. For the penultimate one there are two options, and so on.
Therefore, for a series of N non-repeating elements, the possible permutations are equal to the product of all integers from 1 to N. This product is called N and N! (read “en factorial”).

In the previous case, the number of possible elements and the number of places in the row coincided, and their number was equal to N. But a situation is possible when there are fewer places in the row than there are possible elements. In other words, the number of elements in the sample is equal to a certain number M, and M< N. В этом случае задача определения возможных комбинаций может иметь два различных варианта.
First, you may need to count the total number of possible ways in which M elements out of N can be arranged in a row. Such ways are by arrangement.
Secondly, the researcher may be interested in the number of ways in which M elements can be selected from N. In this case, the order of the elements is no longer important, but any two options must differ from each other by at least one element. Such methods are called combinations.

To find the number of placements of M elements out of N, you can resort to the same method of reasoning as in the case of permutations. There can still be N elements in the first place, N - 1 in the second place, and so on. But for the last place, the number of possible options is not equal to one, but (N - M + 1), since when the placement is completed, there will still be (N - M) unused elements.
Thus, the number of placements of M elements from N is equal to the product of all integers from (N - M + 1) to N, or, what is the same, the quotient N!/(N - M)!.

Obviously, the number of combinations of M elements from N will be less than the number of placements. For every possible combination there is an M! possible placements depending on the order of the elements of this combination. Therefore, to find this quantity, you need to divide the number of placements of M elements from N by N!. In other words, the number of combinations of M elements from N is equal to N!/(M!*(N - M)!).

Sources:

• number of combinations

Factorial a natural number is the product of all previous natural numbers, including the number itself. Factorial zero is equal to one. It seems that calculating the factorial of a number is very simple - just multiply all natural numbers not exceeding the given one. However, the value of the factorial increases so quickly that some calculators cannot cope with this task.

You will need

• calculator, computer

Instructions

To calculate the factorial of a natural number, multiply all , not exceeding the given one. Each number is counted only once. In the form of a formula, this can be written as follows: n! = 1*2*3*4*5*…*(n-2)*(n-1)*n, where n is a natural number whose factorial needs to be calculated.
0! is taken to be equal to one (0!=1). As the argument increases, the value of the factorial increases very quickly, so the usual (accounting) one, already for a factorial of 15, may give an error instead of a result.

To calculate the factorial of a large natural number, take engineering calculator. That is, such a calculator on the keyboard has symbols mathematical functions(cos, sin, √). Type the original number into the calculator, and then click the factorial button. Usually a button like “n!” or similar (instead of “n” there can be “N” or “x”, but exclamation mark"!" the factorial designation must be present in any case).
For large values ​​of the argument, the calculation results begin to be displayed in “exponential” (exponential) form. So, for example, a factorial of 50 would be represented in the form: 3.0414093201713378043612608166065e+64 (or similar). To get the result of calculations in the usual form, add as many zeros to the number shown before the symbol “e” as are indicated after “e+” (if, of course, there is enough space).

In this article we will talk about a special branch of mathematics called combinatorics. Formulas, rules, examples of problem solving - you can find all this here by reading the article to the very end.

So what is this section? Combinatorics deals with the issue of counting any objects. But in this case, the objects are not plums, pears or apples, but something else. Combinatorics helps us find the probability of an event. For example, when playing cards - what is the probability that the opponent has a trump card? Or this example: what is the probability that you will get a white one from a bag of twenty marbles? It is for this kind of problem that we need to know at least the basics of this branch of mathematics.

Combinatorial configurations

Considering the issue of basic concepts and formulas of combinatorics, we cannot help but pay attention to combinatorial configurations. They are used not only to formulate, but also to solve various examples such models are:

• accommodation;
• rearrangement;
• combination;
• number composition;
• splitting a number.

We will talk about the first three in more detail later, but we will pay attention to composition and partitioning in this section. When they talk about the composition of a certain number (for example, a), they mean representing the number a as an ordered sum of certain positive numbers. And a partition is an unordered sum.

Sections

Before we move directly to the formulas of combinatorics and consideration of problems, it is worth paying attention to the fact that combinatorics, like other branches of mathematics, has its own subsections. These include:

• enumerative;
• structural;
• extreme;
• Ramsey theory;
• probabilistic;
• topological;
• infinitary.

In the first case, we are talking about calculative combinatorics; the problems consider enumeration or counting of different configurations that are formed by elements of sets. As a rule, some restrictions are imposed on these sets (distinctiveness, indistinguishability, possibility of repetition, and so on). And the number of these configurations is calculated using the rules of addition or multiplication, which we will talk about a little later. Structural combinatorics includes the theories of graphs and matroids. An example of an extremal combinatorics problem is what is the largest dimension of a graph that satisfies the following properties... In the fourth paragraph, we mentioned Ramsey theory, which studies the presence of regular structures in random configurations. Probabilistic combinatorics is able to answer the question - what is the probability that a given set has a certain property. As you might guess, topological combinatorics applies methods in topology. And finally, the seventh point - infinitary combinatorics studies the application of combinatorics methods to infinite sets.

Addition rule

Among the combinatorics formulas you can find quite simple ones, with which we have been familiar for quite a long time. An example is the sum rule. Suppose that we are given two actions (C and E), if they are mutually exclusive, action C can be performed in several ways (for example, a), and action E can be performed in b-ways, then any of them (C or E) can be performed in a + b ways .

In theory, this is quite difficult to understand; we will try to convey the whole point using a simple example. Let's take the average number of students in one class - let's say it's twenty-five. Among them are fifteen girls and ten boys. One person on duty is assigned to each class every day. How many ways are there to appoint a class monitor today? The solution to the problem is quite simple; we will resort to the addition rule. The text of the problem does not say that only boys or only girls can be on duty. Therefore, it could be any of the fifteen girls or any of the ten boys. Applying the sum rule, we get a fairly simple example that a schoolchild can easily handle primary classes: 15 + 10. Having counted, we get the answer: twenty-five. That is, there are only twenty-five ways to assign a class on duty for today.

Multiplication rule

The basic formulas of combinatorics also include the multiplication rule. Let's start with the theory. Let's say we need to perform several actions (a): the first action is performed in 1 ways, the second - in 2 ways, the third - in 3 ways, and so on until the last a-action, performed in 3 ways. Then all these actions (of which we have a total) can be performed in N ways. How to calculate unknown N? The formula will help us with this: N = c1 * c2 * c3 *…* ca.

Again, nothing is clear in theory, let’s move on to consideration simple example to apply the multiplication rule. Let's take the same class of twenty-five people, in which there are fifteen girls and ten boys. Only this time we need to choose two people on duty. They can be just boys or girls, or a boy and a girl. Let's move on to the elementary solution of the problem. We choose the first person on duty, as we decided in the last paragraph, we get twenty-five possible options. The second person on duty can be any of the remaining people. We had twenty-five students, we chose one, which means the second person on duty could be any of the remaining twenty-four people. Finally, we apply the multiplication rule and find that the two officers on duty can be elected in six hundred ways. We given number obtained by multiplying twenty-five and twenty-four.

Rearrangement

Now we will look at another combinatorics formula. In this section of the article we will talk about permutations. We propose to immediately consider the problem using an example. Let's take billiard balls, we have nth number of them. We need to count how many options there are to arrange them in a row, that is, to create an ordered set.

Let's start, if we don't have balls, then we also have zero options for placement. And if we have one ball, then the arrangement is also the same (mathematically this can be written as follows: P1 = 1). Two balls can be placed in two in different ways: 1.2 and 2.1. Therefore, P2 = 2. Three balls can be arranged in six ways (P3 = 6): 1,2,3; 1,3,2; 2,1,3; 2,3,1; 3,2,1; 3,1,2. What if there are not three such balls, but ten or fifteen? List everything possible options for a very long time, then combinatorics comes to our aid. The permutation formula will help us find the answer to the question that interests us. Pn = n *P (n-1). If we try to simplify the formula, we get: Pn = n* (n - 1) *…* 2 * 1. And this is the product of the first natural numbers. This number is called factorial, and is denoted as n!

Let's consider the problem. Every morning the counselor lines up his squad (twenty people). There are three in the squad best friend- Kostya, Sasha and Lesha. What is the probability that they will stand next to each other? To find the answer to the question, you need to divide the probability of a “good” outcome by the total number of outcomes. The total number of permutations is 20! = 2.5 quintillion. How to count the number of “good” outcomes? Let's assume that Kostya, Sasha and Lesha are one superman. Then we have only eighteen subjects. The number of permutations in this case is 18 = 6.5 quadrillion. With all this, Kostya, Sasha and Lesha can arbitrarily move among themselves in their indivisible three, and that’s 3 more! = 6 options. This means that we have 18 “good” arrangements in total! * 3! All we have to do is find the desired probability: (18! * 3!) / 20! Which equals approximately 0.016. If converted into percentages, it turns out to be only 1.6%.

Accommodation

Now we will look at another very important and necessary combinatorics formula. Placement is our next issue, which we invite you to consider in this section of the article. We are going for complications. Suppose we want to consider possible permutations, not from the entire set (n), but from a smaller one (m). That is, we are considering permutations of n items by m.

The basic formulas of combinatorics should not only be memorized, but understood. Even though they become more complicated, since we have not one parameter, but two. Suppose that m = 1, then A = 1, m = 2, then A = n * (n - 1). If we further simplify the formula and switch to notation using factorials, we will get a completely laconic formula: A = n! / (n - m)!

Combination

We have reviewed almost all the basic combinatorics formulas with examples. Now let's move on to final stage consideration basic course combinatorics - introduction to combination. Now we will choose m items from the n we have, and we will choose everyone possible ways. How then is this different from placement? We will not take into account the order. This unordered set will be a combination.

Let us immediately introduce the notation: C. We take the placement of m balls out of n. We stop paying attention to order and end up with repeating combinations. To get the number of combinations we need to divide the number of placements by m! (m factorial). That is, C = A / m! Thus, there are only a few ways to select from n balls, which is approximately equal to the number of ways to select almost all of them. There is a logical expression for this: choosing a little is the same as throwing out almost everything. It is also important to mention at this point that the maximum number of combinations can be achieved when trying to select half of the items.

How to choose a formula to solve a problem?

We examined in detail the basic formulas of combinatorics: placement, permutation and combination. Now our task is to facilitate the selection of the necessary formula for solving a combinatorics problem. You can use the following fairly simple scheme:

1. Ask yourself: is the order in which the elements are placed taken into account in the text of the problem?
2. If the answer is no, then use the combination formula (C = n! / (m! * (n - m)!)).
3. If the answer is no, then another question needs to be answered: are all the elements included in the combination?
4. If the answer is yes, then use the permutation formula (P = n!).
5. If the answer is no, then use the placement formula (A = n! / (n - m)!).

Example

We looked at elements of combinatorics, formulas and some other issues. Now let's move on to consider real problem. Imagine that you have a kiwi, an orange and a banana in front of you.

Question one: in how many ways can they be rearranged? To do this, we will use the permutation formula: P = 3! = 6 ways.

Question two: in how many ways can you choose one fruit? This is obvious, we have only three options - choose kiwi, orange or banana, but let's apply the combination formula: C = 3! / (2! * 1!) = 3.

Question three: in how many ways can you choose two fruits? What options do we even have? Kiwi and orange; kiwi and banana; orange and banana. That is, there are three options, but this is easy to check using the combination formula: C = 3! / (1! * 2!) = 3

Question four: in how many ways can you choose three fruits? As you can see, there is only one way to choose three fruits: take kiwi, orange and banana. C = 3! / (0! * 3!) = 1.

Question five: in how many ways can you choose at least one fruit? This condition means that we can take one, two or all three fruits. Therefore, we add C1 + C2 + C3 = 3 + 3 + 1 = 7. That is, we have seven ways to take at least one fruit from the table.

Number of combinations

Combination from n By k called a set k elements selected from data n elements. Sets that differ only in the order of the elements (but not in composition) are considered identical; this is why combinations differ from placements.

Explicit formulas

Number of combinations of n By k equal to the binomial coefficient

For a fixed value n generating function of numbers of combinations with repetitions from n By k is:

The two-dimensional generating function of numbers of combinations with repetitions is:

Links

• R. Stanley Enumerative combinatorics. - M.: Mir, 1990.
• Calculate the number of combinations online

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See what “Number of combinations” is in other dictionaries:

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In combinatorics, a combination of by is a set of elements selected from a given set containing different elements. Sets that differ only in the order of the elements (but not in composition) are considered identical, these combinations ... ... Wikipedia

Engaged in the study of events the occurrence of which is not known with certainty. It allows us to judge the reasonableness of expecting the occurrence of some events compared to others, although assigning numerical values ​​to the probabilities of events is often unnecessary... ... Collier's Encyclopedia

1) the same as mathematical combinatorial analysis. 2) A section of elementary mathematics associated with the study of the number of combinations, subject to certain conditions, that can be composed from a given finite set of objects... ... Big Soviet encyclopedia

- (Greek paradoxos unexpected, strange) in a broad sense: a statement that sharply diverges from generally accepted, established opinion, a denial of what seems “unconditionally correct”; in a narrower sense, two opposing statements, for... ... Philosophical Encyclopedia

- (or the principle of inclusions and exclusions) a combinatorial formula that allows you to determine the power of the union of a finite number of finite sets, which in general case can intersect with each other... Wikipedia

Mathematical theory concerned with the determination of number in various ways distribution of these items in a known order; is especially important in the theory of equations and probability theory. The simplest tasks of this kind are... ... Encyclopedic Dictionary F. Brockhaus and I.A. Ephron

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