(1) If you choose one of these books, how many different ways are there?
(2) If you take a math book, a Chinese book and an English book from these books, how many different ways are there?
(3) If you take out two books with different themes from these books, how many different ways are there?
Solution: (1) Because you can get a book from the shelf, you should classify it. Because there are three kinds of books, you can divide them into three categories. So according to addition principle, the number of books you get is: 3+5+6= 14.
(2) Getting 1 math books, Chinese books and English books from the bookshelf needs to be completed in three steps. According to the principle of multiplication, the number of different methods is 3×5×6=90 (species).
(3) If two books with different subjects are taken from the bookshelf, there are three situations (several languages 1 book, several English 1 book and English 1 book), and each situation needs to be completed in two steps. So we should calculate the number of different methods obtained by * * according to the principles of addition and multiplication:
3×5+3×6+5×6=63 (species).
Example 2. Given two sets A = {1, 2,3} and B={a, b, c, d, e}, how many different mappings can be established from a to b?
Analysis: First of all, we must make it clear that "this matter refers to mapping. What is mapping?" That is, for every element in A, there is a unique element in B.. "
Since there are three elements in A, it is necessary to find a home in B, and all three elements should be completed. So we have to go in three steps. When these three steps are completed, a map is established. According to the principle of multiplication, the number of different maps that can be established by * * * is 5×5×5=53 (species).
2. Two formulas of permutation number and combination number
There are two forms of permutation number and combination number formulas. One is the continuous product form, which is mainly used for calculation. The second is the form of factorial, which is mainly used for simplification and proof.
Formal factorial form of continuous product
The equation holds.
Comments: This is a proof problem of permutation number equation, in the form of quotient of factorial, using the nature of factorial: n! (n+ 1)=(n+ 1)！ The deformation process can be simplified.
Example 4. solve an equation
Solution: The original equation can be simplified as:
The solution is x=3.
Comments: When solving the equation given by permutation number and combination number, we should pay attention to the relationship between extracted elements and extracted elements in the definitions of permutation number and combination number, and the important restriction that they are natural numbers before removing symbols.
3. Application of permutation and combination
In the previous math problems of college entrance examination, the permutation and combination problems were mainly applied problems. Generally, certain restrictions will be attached; The contents and scenes of these application problems are varied, and the solutions are still regular. Commonly used methods are: general method and special method.
General methods are: direct method and indirect method.
(1) can be divided into two categories in the direct method. If the problem can be divided into mutually exclusive categories, according to addition principle, classification can be used; If the problem considers order, according to the principle of multiplication, the occupation method can be used.
(2) Indirect method generally solves the problem by eliminating the negative side of the problem.
Special method:
(1) Special element position: give priority to elements or positions with special requirements, and then consider other elements or positions.
(2) Binding method: Some elements must be arranged together, closely combined and glued into a group by "binding method", and arranged separately inside and outside the group.
(3) Interpolation: Some elements must be separated and arranged together by "interpolation", and those that do not need to be separated should be arranged in vacant positions.
(4) Other methods.
Example 5.7 People form a line and find out the number of species in different arrangements that meet the following requirements.
(1) The middle section of platoon A; (2) A is not arranged at both ends; (3) A and B are adjacent;
(4) A is to the left of B (no need to be adjacent); (5) Party A, Party B and Party C;
(6) Party A, Party B and Party C are not adjacent.
Solution: (1) The middle of Row A belongs to a "special position", so resettlement is given priority. There is only one way to stand, and the other six people are randomly arranged, so there are: 1×=720 different arrangements.
(2) Party A's failure to arrange both ends also belongs to the problem of "special position". Priority placement Party A has seeds in any of the five intermediate positions, and the remaining six people can arrange seeds at will, so * * * has = 3,600 different arrangements.
(3) Party A and Party B are adjacent and belong to the "binding mode". A and B are combined into an "element", and the other five people ***6 elements are randomly arranged, and then arranged in group A and B, so * * * has = 1400 different arrangements.
(4) A is on the left side of B. Considering that among all the permutations formed by a row of 7 people, the permutations of "A is on the left side of B" and "A is on the right side of B" are in one-to-one correspondence, and each permutation accounts for half of all permutations without adjacency, so A has =2520 different permutations on the left side of B.
(5) The juxtaposition of A, B and C also belongs to the arrangement in which some elements must be together. By using the "binding method", Party A, Party B and Party C are combined into an "element", and the other four people ***5 "elements" are randomly arranged. Now Party A, Party B and Party C exchange positions, so there are =720 different arrangements.
(6) Party A, Party B and Party C are not adjacent to each other and belong to separate arrangements where some elements cannot be together. With the method of "inserting blanks", four people other than Party A, Party B and Party C are arranged in a row, and five "blanks" are formed between left, right and every two people. Then insert a, b and c in the three "spaces", so * * * has it.
= 1440 different arrangements.
Example 6. Use 0, 1, 2, 3, 4 and 5 to form a five-digit non-repeating number, and calculate the following categories of numbers respectively:
(1) odd number; (2) multiples of 5; (3) Numbers greater than 20300; (4) Numbers excluding the number 0 and 1, 2 are not adjacent.
Solution: (1) Odd number: To get a 5-digit odd number, there are three steps. First, considering that the number of digits must be odd, choose a number from 1, 3,5 to arrange the number of digits. In the second step, consider that the first digit cannot be 0, and choose one of the remaining four digits that are not 0 to rank first.
Step 3: Choose 3 numbers from the remaining 4 numbers and rank them in the middle. According to the multiplication principle, * * * has =388 (number).
(2) A multiple of 5: 0 is used for classification.
Category 1: If 0 is a bit, it will have = 120.
The second category: 0 is not a bit, or 5 is a bit, then =96.
Then * * * has such a number: =2 16 (pieces).
(3) Five digits greater than 20300 can be divided into three categories:
Category I: 3xxxxx, 4xxxxx and 5xxxxx.
Category 2: 2 1xxx, 23xxx, 24xxx, 25xxx.
Category III: 203xx, 204xx, 205xx, with one,
Therefore, the five-digit * * greater than 20300 is: =474 (pieces).
(4) The sum of numbers excluding the number 0 1 2 is not adjacent: it is completed in two steps. Step 1, three numbers, 3, 4 and 5, are arranged in a row; In the second step, 1 and 2 are inserted into two positions of four "blanks", so * * * has =72 five digits without the number 0, and 1 and 2 are not adjacent.
Example 7. A straight line separates from a circle. The six points on the straight line are A 1, A2, A3, A4, A5, A6, and the four points on the circle are B 1, B2, B3, B4. How many straight lines can you get at most? How many people at least?
Solution: When the most straight lines are obtained, that is, straight lines without three points can be divided into three categories:
The first category is that the number of straight lines connecting a point on a known straight line and a point on a circle = 24;
The second category is that the number of straight lines taken by any two points on a circle = 6;
The third category is that there are 1 lines, so the largest number of lines is n1=++1= 31(lines).
When the number of lines obtained is the least, that is, the number of overlapping lines is the most, it is more convenient to subtract the number of overlapping words by exclusion method, and the overlapping lines are the lines connected by two points on the circle. Excluding duplicates is the minimum number of lines: N2 = n1-2 = 31-12 =19 (lines).
Senior two math permutation and combination exercises.
Li gang
Arrangement exercise
1, put three different balls into four boxes, and the number of different kinds of balls is ().
a、8 1 B、64 C、 12 D、 14
2, n∈N and n
A, B, C, D,
3. Four numbers (1, 2, 3, 4) can be used to form the number () of natural numbers whose numbers are not repeated.
a、64 B、60 C、24 D、256
4. Three different movie tickets are distributed to 10 people, and each person has at most one ticket, so the number of different kinds of tickets is ().
a、2 160 B、 120 C、240 D、720
5. Arrange a program with 5 solos and 3 choruses. If the chorus program can't be ranked first, and
Chorus programs cannot be adjacent, so the number of different arrangements is ()
A, B, C, D,
6, 5 people in a row, of which at least one of Party A and Party B is at both ends. The number of rows is ()
A, B, C, D,
7. Use the numbers 1, 2, 3, 4 and 5 to form five digits, which are not complex, and the even number less than 50000 is ().
A, B, C, D, 60 years old
8. A class committee will be divided into five people, who are the vice monitor, study committee member, labor committee member and sports committee member.
Among them, A can't be a monitor, B can't be a study committee member, and the number of different division schemes is ().
A, B,
C, D,
Answer:
1-8 BBADCCBA
Fill in the blanks
1 、( 1 )( 4p 84+2p 85)÷(P86-P95)×0！ =___________
(2) if P2n3= 10Pn3, then N = _ _ _ _ _ _ _ _ _
2. From the arrangement of four different elements A, B, C and D, the arrangement of three different elements is as follows.
__________________________________________________________________
3, 4 boys, 4 girls in a row, girls don't row at both ends, there are _ _ _ _ different arrangements.
4. There are three dimes, 1 dimes and four dimes 1 yuan, which can be composed of these dimes.
_ _ _ _ _ _ Different currencies.
Second, answer the question.
5. Use the six numbers of 0, 1, 2, 3, 4 and 5 to form a five-digit number, and there are no duplicate numbers.
(1) What are the following situations?
① odd number
② divisible by 5.
③ divisible by 15.
④ less than 35 142
⑤ Less than 50000 and not a multiple of 5.
6. If these five digits are arranged from small to large, what is the number 100?
1 × × × ×
1 0 × × ×
1 2 × × ×
1 3 × × ×
1 4 × × ×
1 5 0 2 ×
1 5 0 3 2
1 5 0 3 4
7. How many different ways are there for 7 people in a row under the following circumstances?
(1) add the card head
(2) A does not occupy the head, nor does it occupy the tail.
(3) Party A, Party B and Party C must be together.
(4) There are only two people on both sides.
(5) Party A, Party B and Party C are not adjacent.
(6) A is to the left of B (not necessarily adjacent)
(7) Party A, Party B and Party C are in the order from high to low and from left to right.
(8) Party A does not take the lead, and Party B is not in the middle.
8. Choose three numbers from the five numbers 2, 3, 4, 7 and 9 to form a three-digit number, and there are no duplicate numbers.
(1) How many such three digits are there?
(2) What is the sum of the digits of all three numbers?
(3) What is the sum of these three digits?
Answer:
One,
1、( 1)5
(2)8
Second,
abc、abd、acd、bac、bad、bcd、cab、cad、cbd、dab、dac、dbc
3、8640
4、39
5、
①3× =288
②
③
④
⑤
6、
= 120 〉 100
=24
=24
=24
=24
=2
7、( 1) =720
(2)5 =3600
(3) =720
(4) =960
(5) = 1440
(6) =2520
(7) =840
(8)
8、( 1)
(2)
(3)300×( 100+ 10+ 1)=33300
Arrangement and combination exercise
1, if, then the value of n is ()
a、6 B、7 C、8 D、9
There are 30 boys and 20 girls in one class. Now we should select five people from them to form a propaganda group, including boys and girls.
The selection method of no less than 2 students is ()
A, B,
C, D,
3. There are 10 points in the space, five of which are on the same plane, and the rest have no * * * plane, so 10 points can be determined.
The number of coplanar planes is ()
a、206 B、205 C、 1 1 1 D、 1 10
Six different books are distributed to Party A, Party B and Party C, with two books each. The number of different kinds of books is ().
A, B, C, D,
5, by five 1, two 2 arranged into a series containing seven items, then the number of different series is ().
a、2 1 B、25 C、32 D、42
6. Let P 1, P2…, P20 be the points corresponding to the 20 complex roots of the equation z20= 1 on the complex plane, and take these points as the tops.
The number of points in a right triangle is ()
a、360 B、 180 C、90 D、45
7, if, then the value range of k is ()
a 、[5， 1 1] B 、[4， 1 1] C 、[4， 12] D、4， 15]
8. There are four different red balls and six different white balls in the pocket. Take out four balls at a time and take out a ball of thread.
Points, take out a white ball and mark 1 point, so the total score is not less than 5 points.
A, B,
C, D,
Answer:
1、B 2、D 3、C 4、A 5、A 6、B
7、B 8、C
1, calculation: (1) = _ _ _ _
(2) =_______
2. Put seven identical balls into 10 different boxes. If there are no more than 1 balls in each box, there will be _ _ _ _ _ _ _ _.
Different versions.
3.∠AOB has five points on the edge OA and six points on the edge OB, plus the point O *** 12, and this 12 is the top.
There are _ _ _ _ _ triangular points.
4. Take any four numbers from the number 1, 2, 3, …, 9 to make their sum odd, then * * * has _ _ _ _.
Different methods.
5. Known
6.( 1) How many triangular pyramids are there with the vertex of the cube as the vertex?
(2) How many four pyramids are there with the vertex of the cube as the vertex?
(3) How many pyramids are there with the vertex of the cube as the vertex?
7. Set A has 7 elements, set B has 10 elements, set A∩B has 4 elements, and set C satisfies
(1)C has three elements; (2)C A∪B； (3)C∩B≠φ, C∩A≠φ, and find one of such sets C.
Count.
8. From 1, 2, 3, ... 30, take three unequal numbers at a time so that their sum is a multiple of 3.
* * * How many different ways are there?
Answer:
1、490
2、3 1
3、 165
4、60
5. Solution:
6. Solution: (1)
(2)
(3)58+48= 106
7. Solution: There are elements 7+ 10-4= 13 in A ∪ B.
8. Solution: Divide these 30 numbers into three categories according to the remainder after division by 3:
A={3，6，9，…，30}
B={ 1，4，7，…，28}
C={2，5，8，…，29}
(1)