20 18 what are the problem-solving skills of the arrangement and combination of national civil service examinations?

The permutation and combination problem is a common problem in the logical judgment part of judgment and reasoning module in administrative ability test. However, due to the complexity of the known information, it is difficult for many students to solve this problem in a short time. Painting education reminds the majority of candidates preparing for the 20 18 National Civil Service Examination that to solve the problem of permutation and combination, we must carefully examine the questions and make it clear whether they belong to permutation and combination or a mixed question of permutation and combination; At the same time, we should grasp the essential characteristics of the problem, flexibly use the basic principles and formulas for analysis, and pay attention to some strategies and methods.

1. Indirect method

That is, excluding some qualified methods and adopting the strategy of positive difficulty and reverse equivalent conversion to complete the number of methods of something. If the number of methods in one step is uncertain or repeated counting, then we should consider using classification, which is an effective means to solve complex problems, and when there are many positive classifications, we should consider using indirect counting.

Example: Choose 4 out of 6 boys and 5 girls to participate in the competition, which requires at least 1 male and female. How many different ways are there?

a . 240 b . 3 10c . 720d . 1080

Correct answer b

Analysis: If this problem is considered from the front, there are many cases. If the indirect method is adopted, the opposite of at least one man and one woman is to choose only boys or girls respectively, which can be changed to C (1 1 4)-c (6,4)-c (5,4 4) = 3 10/0.

2. Scientific classification

In the problem, there are both the restriction of elements and the arrangement. Generally, the elements are arranged first (that is, the combination).

For complex permutation and combination questions, because there are many situations, it is necessary to scientifically classify different situations in order to solve them in an orderly manner and avoid repetition or omission. At the same time, it is clear that the classification is in line with addition principle, and the addition operation should be done.

Example: A company invites 10 teacher to attend the meeting. If Party A and Party B cannot attend at the same time, there are () different invitation methods.

a . 84b . 98c . 1 12d . 140

Correct answer d

Analysis: According to the requirements; Party A and Party B cannot participate in the following categories at the same time:

A. If A participates and B does not, then choose 5 teachers from the remaining 8 teachers, with C (8 8,5) = 56 kinds;

B.b participates and A does not participate, which is the same as (a).

C. If neither Party A nor Party B participates, then six teachers are selected from the remaining eight teachers, with C (8 8,6) = 28 kinds.

So there are 56+56+28= 140 kinds of * *.

3. Special priority method

Special elements, priority treatment; Special position, priority. For the permutation and combination problem with additional conditions, the method of considering special elements and positions first and then other elements and positions is generally adopted.

Four of the six volunteers were selected to do four different jobs, namely, translation, tour guide, shopping guide and cleaning. If two volunteers, A and B, can't do translation work, then the different option is ().

(A)280 species (B)240 species (C) 180 species (D)96 species.

Correct answer: b

Analysis: Because volunteers A and B can't do translation work, which is a "special" post, one of the remaining four volunteers has four different ways to choose translation work, and then three of the remaining five volunteers are chosen to do three different jobs, namely, tour guide, shopping guide and cleaning, with A (5 5,3) = 6544.

4. Binding method

The so-called binding method is that when solving the problem of requiring several elements to be adjacent, the adjacent elements are considered as a whole first, and then the order between elements in the whole is considered separately. Note: its primary feature is adjacency, and secondly, the binding method is generally applied to the sorting problem of different objects.

Example: 5 boys and 3 girls are lined up, and 3 girls must be lined up together. How many different arrangements are there?

A.240B.320C.450D.480

Correct answer b

Analysis: Using the binding method, three girls are regarded as one element and arranged with five boys. * * There are two kinds of A (6,6 6) = 6x5x3x2, and then three girls are arranged internally. There are six kinds of A (3 3,3) = 6, step by step. Multiplication should be adopted, so the arrangement method * * * includes: A (6).

5. Choose the method of "one", similar to division.

For some elements arranged in a certain order, these elements can be arranged with other elements first, and then the total arrangement number is divided by the total arrangement number of these elements. The "alternative" here means that there are many arrangements that are "similar" to what we want, and we only choose one.

Example: How many ways can five people rank A before B?

A.60B. 120C

Correct answer a

Analysis: Five people have five arrangements! = 120, including two cases: A is in front of B and A is behind B (whether A and B are adjacent or not is not mentioned, so it can be ignored). The topic requires that A comes before B, so the answer is A (5,5) ÷ A (2,2) = 60 kinds.

6. interpolation method

The so-called interpolation method is to solve the problem that some elements are not adjacent, first arrange other elements, and then insert the specified non-adjacent elements into the gap or both ends of the arranged elements.

Note: A. The first feature is non-adjacency, and the second feature is that the sorting problem generally adopts interpolation method.

B. When inserting nonadjacent elements in an arrangement element, it is necessary to pay attention to whether it can be inserted at both ends.

C. The difference between bundling method and interpolation method can be simply recorded as "bundling method for adjacent problems and interpolation method for non-adjacent problems".

For example, if there are five people in line, A, B, C, D and E, and it is required that A and B cannot stand together and A and B cannot stand at both ends, how many queuing methods are there?

A.9B. 12C. 15D.20

Correct answer b

Analysis: Arrange C, D and E first, and then insert A and B into the two spaces formed by C, D and E respectively. Because A and B are not standing at both ends, there are only two spaces to choose from. The total number of methods is a (3,3 3) × A (2 2,2) =12.

7. plug-in method

The so-called plug-in method refers to the problem-solving strategy of inserting less than the required number of groups of 1 blocks between elements when solving the grouping of several identical elements and requiring at least one element in each group.

Note: Its primary feature is that the elements are the same, and secondly, each group contains at least one element, which is generally used for combination problems.

For example, put eight identical balls into three different boxes, and each box should contain at least one ball. How many ways are there in a box?

A.2 1B.24C.28D.45

Correct answer a

Analysis: To solve this problem, you just need to divide the eight balls into three groups, and then put each group in a box in turn. So we just need to divide the eight balls into three groups, so that we can arrange the eight balls in a row, and then insert two boards in the space formed by the eight balls, and then we can divide the eight balls into three groups smoothly. Among them, the ball in front of the first board is placed in the first box, the ball between the first board and the second board is placed in the second box, and the ball behind the second board is placed in the third box. Because there is at least one ball in each box, two plates can't be placed in the same empty space, and plates can't be placed at both ends, so the number of ways to put plates is C (7 7,2) = 21. (Note: There is no difference between boards. )