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The set of formulas for combination, alignment, and permutation always makes it easy for high school students to confuse the definitions and exact calculation formulas. During the learning process, you must clearly distinguish these three types of formulas to do exercises and exams effectively. The following article will introduce more clearly the formulas for alignment, combination, and permutation for you to learn properly.

Permutations in mathematics and common forms

First we will learn about permutations and the most common types of permutations:

Definition of permutation

If you explain each word, you can consider the word “swap” to mean “exchange” and the word “position” to mean “position”. For example, we have a set of numbers X consisting of a number of “n” different elements (the condition is n ≥ 0). Thus, each arrangement of elements of set X in a specific order is considered a permutation of “n”. Often the number of permutations of “n” will be written with the short notation “Pn”.

How many common types of permutations are there?

Currently there are 3 common types:

Repeated permutation

This is a form of permutation when we are given “n” number of objects and among those objects there are “ni” objects of type “i” and have identical structural types. To put it simply, for each type of arrangement of n elements, there will be about “n1” elements will be “a1, n2” elements will be “a2″… and there will be “nk” elements will be ” ak” (n1 + n2 + n3 +…..+ nk = n) is arranged in random order and is considered a repeating permutation of level “n” with type (n1, n2,…, nk) belongs to the “k” section death.

Each type of arrangement that includes the order of “n” given elements is called a repeating permutation of “n”.

The formula used to calculate the iterative permutation type will be:

In it, we have:

• “Pn” is an iterative permutation of level “n” with type (n1,n2,…, nk) of “k” number of elements.
• n = n1 + n2 +…..+ nk: Number of elements.
• “n1” is the number of elements “a1” with the same structure
• “n2” is the number of “a2” elements with the same structure.
• “nk” is the number of “ak” elements with the same structure.

Circular permutation

This type of permutation includes elements inside the permutation that can create a round with the number of elements k greater than 1 and “k” must be an integer. The formula for calculating circular permutation is:

Q(n) = (n-1)!

Permutation of uniform form

Or many people also call it permutation, this is a type of permutation with the first element with the first element, element 2 with element 2… Meaning that there is actually no exchange between these elements. .

What is combination and harmony?

Before learning the combination and alignment formulas, everyone must understand the following concepts:

Definition of combination

Combination is a method by which we select elements in a large group without having to differentiate in terms of ordering. In some cases, you can count the number of combinations. The combination has the convolutional form “k” of the number of “n” elements, which means the number of groups consisting of “k” elements taken from the group “n” elements and between which there are only differences in the elements in structure without considering the order of the elements.

For each subset consisting of a number of “k” elements in the large set consisting of a number of “n” elements (with n>0), it is considered a combination of “k” convolutions of a number of “n”. element.

Definition of alignment

This is a method where we select elements in a larger set while still distinguishing the sort order. This is the opposite of the combinatorial form in that it does not require distinction in terms of order.

We have a convolution consisting of the convolution “k” of “n” which will be a subset of a large set S including a number of “n” elements. This subset will include a number of “k” elements belonging to the set S and follows a sort order.

The most complete set of combination, concordance, and permutation formulas with examples

The following is the most complete set of combination and permutation formulas in Mathematics:

Conformity calculation formula

According to the concept mentioned above, we have the number of convolutions “k” of a set consisting of “n” elements (1≤ k ≤ n) which will be calculated according to the formula:

Conformity calculation formula

Illustrative example 1: How many ways do we have to arrange 3 friends Hoang, Hieu, Hung into 2 available seats?

Answer:

Illustrative example 2: There are several numbers consisting of 04 completely different digits established from the digits 1,2,3,4,5,6,7.

Answer:

We will have each number consisting of four completely different digits established by taking the four digits of the set A consisting of elements 1, 2, 3, 4, 5, 6, 7 and then arranging it. in specific orders. Each of these numbers will be considered a concordance with convolution 04 of 07 elements.

It follows that the number of natural numbers generated from the above set is 840 numbers.

Combination calculation formula

When it comes to combinatorial and composite formulas, we will have a formula for calculating combinations consisting of “k” convolutions of “n” number of elements (1≤ k ≤ n) as:

Combination calculation formula

In which “kn” has an answer of 0 when k > n.

Illustrative example: Mr. B played with a total of 11 people. But Mr. B wants to send an invitation to 5 of them to eat together. Among those 11 friends, 2 people did not want to meet. So how many ways does Mr. B have to invite them to eat?

Answer:

Mr. B can only invite one of two friends and he will invite about 4 more friends out of the other 9 friends. Thus we have:

Mr. B did not send invitations to the other two friends but only sent invitations to 5 of the nine friends. Thus we have:

In total, Mr. B will have 328 ways to invite.

Permutation calculation formula

The formula to calculate permutations is quite simple. If we have a set consisting of “n” number of elements (condition n> 0), then we have the formula to calculate the permutation of “n” number of elements as below:

Pn=n!

Illustrative exercise 1: Given a set named A consisting of 5 numbers 3, 4, 5, 6, 7. Based on this set, you can establish a number of natural numbers consisting of 5 other digits. each other?

Answer: You can apply the permutation formula Pn=n!. So we have P5 = 5! and the answer is 120 numbers.

Illustrative exercise 2: Please calculate the number of ways to arrange 10 people in a vertical row?

Answer:

Each arrangement of 10 people in a vertical row is a permutation of the 10 elements. Thus, the number of ways to arrange 10 people in a vertical row is P10 = 10!

The relationship of conformity, combination, and permutation in mathematics

Through the concepts and formulas of combination, alignment, and permutation above, we can see that they are closely related to each other. Specifically, a convolution with “k” convolutions of “n” elements will be established by performing the following two steps:

• Step 1: You take a combination with “k” convolutions of “n” elements.
• Step 2: You perform permutation for “k” elements.

Therefore, you will have the formula used to express the relationship between combination, concordance and permutation as follows:

The formula represents the relationship of union, combination, and permutation in mathematics

Rules for counting correct combinations, combinations, and permutations

In addition to combination and permutation formulas, students should also grasp the rules for counting correct combinations, combinations, and permutations as follows:

Rules used to count combinations

We have a set A consisting of n elements with the condition that n > 0. Thus, a random “k” convolutional combination of elements in set A will be a subset with about “k” elements. belongs to A, where 0 ⩽ k ⩽ n ; k ∈ N.

Thus the number of combinations will be calculated based on this formula: n!(nk)!

Rules used to count conformations

We have a set A consisting of a number of “n” elements with n⩾1. Thus, a concordance has a convolution of “k” distinct elements belonging to A. In which there are 1⩽k⩽n, k ∈ N.

Thus the number of conformations is calculated based on this formula: n!k!(nk)!

The rule used to count permutations

With a set consisting of a number of “n” distinct elements, we will create a permutation of “r” elements taken from the set as follows:

• Taking the first element, we will have a total of n ways;
• Taking the second element, we will have n-1 permutation arrangements;
• ….

Similarly, when we take the rth element in the set, we have r-1 permutation arrangements:

• In the case where r = n, we will have a formula to count the number of different permutations formed from n elements. That is: P(n) = n!
• In case r < n, the number of permutations will be counted according to this formula: P(n,r)= n!(nr)!

Exercises illustrate the formulas for combination, alignment, and permutation

After understanding the formulas for combinations of combinations and permutations, you can see a few illustrative exercises as follows:

Exercise 1

The 12th grade math exam at a high school includes two types of multiple-choice questions and essay questions. Each student taking the exam must take 2 exams including 1 multiple choice and 1 essay. There are 12 essay questions and 15 multiple-choice questions. So how many ways will each student choose to take the exam?

Prize:

We will have the number of ways to choose an essay question: 12 ways and the number of ways to choose a multiple choice question will be 15 ways. Therefore, a student must complete both topics in parallel. So there will be a total of 12 x 15 = 180 ways to choose exam questions.

Exercise 2

We have a set A consisting of the digits 1, 2, 3, 5, 7, 9:

a. From the above set, it is possible to establish several natural numbers consisting of 4 different digits.

b. From the above set, it is possible to establish several even natural numbers consisting of 5 different digits.

Prize:

a. We call the 4-digit natural number n = a1a2a3a4. To get such a number n, we must choose the numbers a1, a2, a3, a4 in parallel. In it we have:

• a1 has a total of 6 ways to choose from.
• a2 has a total of 5 ways to choose from.
• a3 has a total of 4 ways to choose from.
• a4 has a total of 3 ways to choose from.

So we have a total of: 6 x 5 x 4 x 3 = 360 numbers n we want to find.

Example illustrating combinatorial and conformal formulas

b. We call an even natural number consisting of 05 numbers n = a1a2a3a4a5. In it we have:

• a5 has exactly 1 choice: 2.
• a1 has a total of 5 ways to choose from.
• a2 has a total of 4 ways to choose from.
• a3 has a total of 3 ways to choose from.
• a4 has a total of 2 ways to choose.

So the number n you want to find is 1 x 2 x 3 x 4 x 5 = 120 numbers.

Exercise 3

Given a set A consisting of the numbers 0, 1, 2, 3, 4, 5, 6. From this set A, it is possible to establish several natural numbers consisting of 05 digits, each pair of which is different and ensure that the number 5 and number 2 do not stand next to each other.

Prize:

1. Find a 5-digit natural number that is arbitrarily different in pairs:

A natural number consisting of 05 different digits with an arbitrary pair of ones has the form n = a1a2a3a4a5. In there:

• a1 has a total of 6 ways to choose (a1 ≠ 0).
• a2 has a total of 6 ways to choose from.
• a3 has a total of 5 ways to choose from.
• a4 has a total of 4 ways to choose from.
• a5 has a total of 3 ways to choose.

So we have 6 x 6 x 5 x 4 x 3 = 2169 natural numbers.

2. Find a natural number consisting of 05 digits that are different in pairs and have numbers 2 and 5 that cannot be next to each other:

Suppose we have the number 2 with the number 5 as a random digit a. We will find a natural number with 04 digits:

Case 1: a1 = a

• a1 has a total of 5 ways to choose from.
• a2 has a total of 4 ways to choose from.
• a4 has a total of 3 ways to choose from.

So we will have 5 x 4 x 3 = 60 numbers.

Case 2: a1 ≠ a so:

• a1 has a total of 4 ways to choose (Because a1 ≠ 0,2,5).
• Suppose a2 = a then there are 3 positions for number a.
• a3 has a total of 4 ways to choose from.
• a4 has a total of 3 ways to choose from.

So we will have 4 x 3 x 4 x 3 = 204. The numbers 2 and 5 can be interchanged. Therefore, we can deduce: 204 x 2 = 408 numbers.

According to the problem requirements: 2160 – 408 = 1572 ways.

Through the above exercises, you can understand how to apply formulas in mathematics. That is the formula for calculating alignment, combination, and permutation in specific problems. From there, you will do the exercises quickly and absorb knowledge faster.

Combination, alignment, and permutation formulas are basic knowledge in high school. This formula has appeared in a few high school graduation exam questions over the years. Therefore, students must firmly grasp the above formulas for calculating combinations and combinations to facilitate the process of studying for exams and taking important tests.

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