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. Examples of 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 primary school student can easily handle: 15 + 10. After counting, 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, so let’s move on to consider a simple example of applying 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 two officers on duty can be elected in six hundred ways. We obtained this number 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). The two balls can be placed in two 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? It would take a very long time to list all the possible options, 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 best friends in the squad - 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 get a completely laconic formula: A = n! / (n - m)!
Combination
We reviewed almost all the basic combinatorics formulas with examples. Now let's move on to the final stage of considering the basic combinatorics course - getting to know combinations. Now we will choose m items from the n we have, and we will choose everything in every possible way. How then is this different from placement? We will not take into account the order. This unordered set will be the 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:
- Ask yourself: is the order in which the elements are placed taken into account in the text of the problem?
- If the answer is no, then use the combination formula (C = n! / (m! * (n - m)!)).
- If the answer is no, then another question needs to be answered: are all the elements included in the combination?
- If the answer is yes, then use the permutation formula (P = n!).
- 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 the 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.
Combinatorics is a branch of mathematics that studies questions about how many different combinations, subject to certain conditions, can be made from given objects. The basics of combinatorics are very important for estimating the probabilities of random events, because It is they that allow us to calculate the fundamentally possible number of different scenarios for the development of events.
Basic formula of combinatorics
Let there be k groups of elements, and the i-th group consists of n i elements. Let's select one element from each group. Then the total number N of ways in which such a choice can be made is determined by the relation N=n 1 *n 2 *n 3 *...*n k .
Example 1. Let us explain this rule with a simple example. Let there be two groups of elements, and the first group consists of n 1 elements, and the second - of n 2 elements. How many different pairs of elements can be made from these two groups, such that the pair contains one element from each group? Let's say we took the first element from the first group and, without changing it, went through all possible pairs, changing only the elements from the second group. There can be n 2 such pairs for this element. Then we take the second element from the first group and also make all possible pairs for it. There will also be n 2 such pairs. Since there are only n 1 elements in the first group, the total possible options will be n 1 *n 2 .
Example 2. How many three-digit even numbers can be made from the digits 0, 1, 2, 3, 4, 5, 6, if the digits can be repeated?
Solution: n 1 =6 (because you can take any number from 1, 2, 3, 4, 5, 6 as the first digit), n 2 =7 (because you can take any number from 0 as the second digit , 1, 2, 3, 4, 5, 6), n 3 =4 (since any number from 0, 2, 4, 6 can be taken as the third digit).
So, N=n 1 *n 2 *n 3 =6*7*4=168.
In the case when all groups consist of the same number of elements, i.e. n 1 =n 2 =...n k =n we can assume that each selection is made from the same group, and the element after selection is returned to the group. Then the number of all selection methods is n k . This method of selection in combinatorics is called samples with return.
Example 3. How many four-digit numbers can be made from the digits 1, 5, 6, 7, 8?
Solution. For each digit of a four-digit number there are five possibilities, which means N=5*5*5*5=5 4 =625.
Consider a set consisting of n elements. In combinatorics this set is called general population.
Number of placements of n elements by m
Definition 1. Accommodation from n elements by m in combinatorics any ordered set from m various elements selected from the population in n elements.
Example 4. Different arrangements of three elements (1, 2, 3) by two will be the sets (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2 ). Placements may differ from each other both in elements and in their order.
The number of placements in combinatorics is denoted by A n m and is calculated by the formula:
Comment: n!=1*2*3*...*n (read: “en factorial”), in addition, it is assumed that 0!=1.
Example 5. How many two-digit numbers are there in which the tens digit and the units digit are distinct and odd?
Solution: because If there are five odd digits, namely 1, 3, 5, 7, 9, then this task comes down to selecting and placing two of the five different digits in two different positions, i.e. the indicated numbers will be:
Definition 2. Combination from n elements by m in combinatorics any unordered set from m various elements selected from the population in n elements.
Example 6. For the set (1, 2, 3), the combinations are (1, 2), (1, 3), (2, 3).
Number of combinations of n elements, m each
The number of combinations is denoted by C n m and is calculated by the formula:
Example 7. In how many ways can a reader choose two books out of six available?
Solution: The number of methods is equal to the number of combinations of six books of two, i.e. equals:
Permutations of n elements
Definition 3. Permutation from n elements are called any ordered set these elements.
Example 7a. All possible permutations of a 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 of n elements is denoted by P n and is calculated by the formula P n =n!.
Example 8. In how many ways can seven books by different authors be arranged in one row on a shelf?
Solution: This problem is about the number of permutations of seven different books. There are P 7 =7!=1*2*3*4*5*6*7=5040 ways to arrange the books.
Discussion. We see that the number of possible combinations can be calculated according to different rules (permutations, combinations, placements) and the result will be different, because The calculation principle and the formulas themselves are different. Looking carefully at the definitions, you will notice that the result depends on several factors simultaneously.
Firstly, from how many elements we can combine their sets (how large is the totality of elements).
Secondly, the result depends on the size of the sets of elements we need.
Finally, it is important to know whether the order of the elements in the set is significant to us. Let us explain the last factor using the following example.
Example 9. There are 20 people present at the parent meeting. How many different options are there for the composition of the parent committee if it must include 5 people?
Solution: In this example, we are not interested in the order of names on the committee list. If, as a result, the same people turn out to be part of it, then in meaning for us this is the same option. Therefore, we can use the formula to calculate the number combinations of 20 elements 5 each.
Things will be different if each committee member is initially responsible for a specific area of work. Then, with the same list composition of the committee, there are possibly 5 within it! options permutations that matter. The number of different (both in composition and area of responsibility) options is determined in this case by the number placements of 20 elements 5 each.
Self-test tasks
1. How many three-digit even numbers can be made from the digits 0, 1, 2, 3, 4, 5, 6, if the digits can be repeated?
2. How many five-digit numbers are there that are read the same from left to right and from right to left?
3. There are ten subjects in the class and five lessons a day. In how many ways can you create a schedule for one day?
4. In how many ways can 4 delegates be selected for a conference if there are 20 people in the group?
5. In how many ways can eight different letters be placed in eight different envelopes, if only one letter is placed in each envelope?
6. A commission consisting of two mathematicians and six economists should be composed of three mathematicians and ten economists. In how many ways can this be done?
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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We sometimes make a choice from many without regard to order. This choice is called combination . If you play cards, for example, you know that in most situations the order in which you hold the cards doesn't matter.
Example 1 Find all combinations of 3 letters taken from a set of 5 letters (A, B, C, D, E).
Solution These combinations are as follows:
(A, B, C), (A, B, D),
(A, B, E), (A, C, D),
(A, C, E), (A, D, E),
(B, C, D), (B, C, E),
(B, D, E), (C, D, E).
There are 10 combinations of three letters chosen from five letters.
When we find all combinations from a set with 5 objects, if we take 3 objects at a time, we find all 3-element subsets. In this case, the order of objects is not considered. Then,
(A, C, B) is called the same set as (A, B, C).
Subset
A set A is a subset of B, which means that A is a subset of and/or the same as B if every element of A is an element of B.
The elements of the subset are not ordered. When combinations are considered, order is not considered!
Combination
Combination,
containing k objects is a subset consisting of k objects.
We want to write down a formula for calculating the number of combinations of n objects if k objects are taken at the same time.
Combination designations
The number of combinations of n objects, if k objects are taken simultaneously, is denoted n C k .
We call n C k number of combinations . We want to write down a general formula for n C k for any k ≤ n. First, it is true that n C n = 1, because a set with n elements has only one subset with n elements, which is the set itself. Second, n C 1 = n because a set with n elements has only n subsets with 1 element each. Finally, n C 0 = 1 because a set with n elements has only one subset with 0 elements, that is, the empty set ∅. To look at other combinations, let's go back to Example 1 and compare the number of combinations with the number of permutations.
Please note that each combination of 3 elements has 6, or 3!, permutations.
3! . 5 C 3 = 60 = 5 P 3 = 5. 4 . 3,
so 
.
In general, the number of combinations of k elements selected from n objects, n C k times permutations of these elements k!, must be equal to the number of permutations of n elements by k elements:
k!. n C k = n P k
n C k = n P k /k!
n C k = (1/k!). n P k
n C k = 
Combinations of k objects from n objects
The total number of combinations of k elements from n objects is denoted by n C k , determined by
(1) n C k = ,
or
(2) n C k = 
Another type of notation for n C k is binomial coefficient . The reason for this terminology will be clear below.
Binominal coefficient
Example 2 Calculate using formulas (1) and (2).
Solution
a) According to (1), 
.
b) According to (2), 
Keep in mind that n/k does not mean.
Example 3 Calculate and .
Solution We use formula (1) for the first expression and formula (2) for the second. Then 
,
using (1), and 
,
using formula (2).
note that 
,
and using the result of example 2 gives us
.
It follows that the number of a 5-element subset of a set of 7 elements is the same as the number of a 2-element subset of a set of 7 elements. When 5 elements are selected from a set, they do not include 2 elements. To see this, consider the set (A, B, C, D, E, F, G): 
In general, we have the following. This result provides an alternative way to calculate the combination.
Subsets of size k and size
and n C k = n C n-k
The number of subsets of size k of a set with n objects is the same as the number of subsets of size n - k. The number of combinations of k objects from a set of n objects is the same as the number of combinations of n objects taken at the same time.
Now we will solve problems with combinations.
Example 4 Michigan Lottery. Michigan's twice-weekly lottery WINFALL has a jackpot of at least $2 million. For one dollar, a player can cross out any 6 numbers from 1 to 49. If these numbers match those drawn in the lottery, the player wins. (
The first place in a row can be any of N elements, therefore, there are N options. In second place - any, except 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 an engineering calculator. That is, such a calculator on the keyboard has symbols of mathematical functions (cos, sin, √). Type the original number into the calculator, and then click the factorial button. Usually a button like “n!” or similarly (instead of “n” there can be “N” or “x”, but the exclamation point “!” in the designation of the factorial 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).
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