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, use 1, 2, 3, ..., 9, and 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)
Sophomore? Permutation and combination exercise (1)
First, multiple-choice questions:
1, put three different balls into four boxes, and the number of different kinds of balls is ().
a . 8 1 b . 64 c . 12d . 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.
64 BC to 60 BC
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 ().
2 160
5. Arrange a program list with five solos and three choruses. If the choral programs can't be ranked first and the choral programs can't be adjacent, then 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.24 B.36 C.46 D.60
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。
Second, fill in the blanks
9 、( 1)(4p 84+2p 85)÷(P86-P95)×0！ =___________
(2) if P2n3= 10Pn3, then N = _ _ _ _ _ _ _ _ _
10. According to the arrangement of four different elements in A.B.C.D, the arrangement of three different elements is _ _ _ _ _ _ _ _ _ _ _ _ _.
1 1, 4 boys and 4 girls in a row, girls don't have to row at both ends, there are _ _ _ _ _ different rows.
12, 3 RMB for 10 cents, 4 RMB for 10 cents 1 RMB, 1 RMB. These RMB can be used to form _ _ _ _ _ _ different currencies.
Third, answer questions.
13, with 0, 1, 2, 3, 4, 5, these six numbers form five digits, and there is no duplicate sign.
(1) What are the following situations?
① Odd numbers, ② divisible by 5, ③ divisible by 15.
④ less than 35 142, and ⑤ less than 50000 and not a multiple of 5.
(2) If these five digits are arranged from small to large, what is the number 100?
14, 7 people in a row, how many different arrangements are there in the following situations?
(1) A takes the lead;
(2) A does not lead or 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 arranged from high to low and from left to right;
(8) A does not take the lead, and B does not rank among them.
15, three numbers are randomly selected from the five numbers of 2, 3, 4, 7 and 9 to form a three-digit number, and there is no duplicate sign.
(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?
Senior two mathematics
Arrangement and combination exercise
Reference answer
First, multiple-choice questions:
1.B
2.B
3.A
4.D
5.C
6.C
7.B
8.A
Second, fill in the blanks
9.( 1)5; (2)8
10.abc、abd、acd、bac、bad、bcd、cab、cad、cbd、dab、dac、dbc
1 1.8640
12.39
Third, answer questions.
13.( 1)①3× =288
②
③
④
⑤
(2) ellipsis.
14.( 1) =720
(2)5 =3600
(3) =720
(4) =960
(5) = 1440
(6) =2520
(7) =840
(8)
15.( 1)
(2)
(3)300×( 100+ 10+ 1)=33300
Example 1. A computer user plans to use no more than 500 yuan's funds to purchase single-chip microcomputer software and boxed disks with unit prices of 60 yuan and 70 yuan respectively. Buy at least 3 softwares and at least 2 boxes of disks as needed, so the different purchase method is ().
(a) Five species (b) Six species (c) Seven species (d) Eight species.
Solution: The number of purchased software is X and the number of disks is Y, depending on the meaning of the problem.
When x = 3, y = 2, 3, 4; When x = 4, y = 2,3; When x = 5, y = 2；; When x = 6, y = 2. There are seven solutions to the inequality group * * * above, so there are seven different ways to buy * * *, so choose C.
Solution 2 According to the meaning of the question, (x, y) is located on the boundary and interior point of a triangle surrounded by three straight lines L 1: x = 3, L2: y = 2, L3: 60x+70y = 500 (coordinates are all integer points), as shown in Figure 7-2- 1
Comment that this is an application problem of counting, the first solution is transformed into finding the number of integer solutions of inequality groups; The second solution is transformed into finding the number of whole points in a specific area on the coordinate plane. In fact, both solutions finally adopt the exhaustive method, which is one of the basic methods to solve the counting problem.
Example 2. In a field with 10 ridges side by side, choose two ridges to plant two crops, A and B, one for each crop. In order to be beneficial to crop growth, it is required that the interval between two crops should be no less than 6 ridges. How many different planting methods are there?
× ○ ○ ○ ○ ○ ○ × ○ ○
× ○ ○ ○ ○ ○ ○ ○ × ○
× ○ ○ ○ ○ ○ ○ ○ ○ ×
○ × ○ ○ ○ ○ ○ ○ × ○
○ × ○ ○ ○ ○ ○ ○ ○ ×
○ ○ × ○ ○ ○ ○ ○ ○ ×
Scheme 1: As shown in the table, X represents the ridge where crops are planted and о represents the ridge where crops are not planted, so there are six different ridge selection methods. Because A and B are two crops, there are 12 different planting methods * *.
The ridge selection method of Solution 2 can be divided into three categories: the first category is 6 ridges apart, and there are three selection methods: 1-8, 2-9, 3-10; The second one is 7 ridges apart, and there are two options: 1-9 and 2- 10. The third type is separated by 8 ridges, and the selection method is only 1- 10, so there is a 6-ridge selection method, and the planting method is 12.
Commented that this is an application problem of counting. The first solution adopts the method of picture frame. Scheme 2 directly applies the principle of addition principle sum multiplication.
If cases 1 and 2 are judged as permutation and combination problems, and formulas containing permutation or combination numbers are listed, the thinking of the problems will be complicated and it will be difficult to draw correct conclusions. Therefore, the counting problem can not be simply attributed to the permutation and combination problem, nor can it be solved only by calculating the number of permutations or combinations.
Example 3.7 People form a line and find out the number of different arrangements that meet the following requirements.
(1) The middle section of platoon A;
(2) A is not arranged at both ends;
(3) Party A is adjacent to Party B;
(4) A is to the left of B (not necessarily adjacent);
(5) Party A, Party B and Party C are not adjacent.
Solution: (1) In the middle of row A, the other six people are randomly arranged, so * * * has = 720 different arrangements.
(2) If A is arranged at the left or right end, there are different arrangements, then A is not arranged at both ends * * * There are = 3600 different arrangements.
(3) Method 1: First, randomly arrange A and 5 people (***6 people) except B, and then arrange B to the left or right (adjacent) of A, then * * * is there? = 1440 different arrangements.
Method 2: First, combine Party A and Party B into one "element", and the other five * * * six "elements" are randomly arranged, and then Party A and Party B exchange positions, so * * * is there? = 1440 different arrangements.
(4) In the seed arrangement formed by a row of seven people, the arrangement of "A left, B right" and "A right, B left" is one-to-one correspondence (the position of others remains unchanged), so there are = 2520 different solutions to the different arrangement of A on the left side of B. 。
(5) First, four people except Party A, Party B and Party C are arranged in a row to form five "vacancies" between the left and right and every two people, and then Party A, Party B and Party C are inserted into three "vacancies", each of which has 1 person, so * * has = 1 4,40 different arrangements.
This is a group of queuing application problems, which is a typical arrangement problem. Additional restrictions are usually positioning and restriction, adjacency and nonadjacency, left and right, front and back, etc.
Example 4. Use six numbers (0, 1, 2, 3, 4, 5) to form five numbers without duplicate numbers, and calculate the numbers in the following categories respectively:
(1)5;
(2) Numbers greater than 20300;
(3) Numbers excluding the number 0 and 1, 2 are not adjacent.
Solution: The multiple of (1)5 can be divided into two categories: the number of the unit position is 0 or 5,
The unit number is 0, with five digits;
The single digit is 5, and there are 4 in the five digits;
So there is a multiple of 5 * * of +4 = 2 16.
(2) Five digits greater than 20300 can be divided into three categories:
Category I: 3 ×××× years × months × days × months × days × months × days × months × days × days × months × days × months × days × months × days × days × months × days × days × months × days × days × days × months × days × days × days × days × months × days × days × days × days × months × days × days × days × months × days × days × days
Category II: 2 1×××××××× Category II: Category II: 21×××××× Category III: Category IV: Category IV: Category V: Category IV: Category V.
Category III: 203×××××, 204××××, 205××××, with 3 bands.
So there are 3+4+3 = 474 five-digit numbers * * greater than 20300.
(3) Non-adjacent numbers excluding the numbers 0 and 1 2 can be divided into two steps. The first step is to arrange the three numbers 3, 4 and 5 in a row; Step 2: Insert 1 2 into two of the four "blanks" formed in the first step, so * * * has = 72.
Comment on this problem is a set of multi-digit arrangement problems and a typical arrangement problem. Common additional conditions include multiple relations, size relations, adjacent relations and so on. It should be noted that there will be no element duplication in the queuing problem, and the arrangement problem must stipulate that the number that is not repeated is an arrangement problem.
Example 5 There are *** 10 points at the vertex and midpoint of each side of a tetrahedron, and four points that are not * * * are selected in different ways ().
(A) 150 species (b) 147 species (c) 144 species (d) 14 1 species.
The situation of quadric non-* * surface is more complicated than that of quadric * * * surface, so the indirect method can be used. Take four points without restriction, and then subtract these four points * * *.
There are three types of four-point * * * planes (as shown in Figure 7-2-3).
The first category: there are four ways to take one side of a tetrahedron;
The second category: three points on one side of a tetrahedron and the midpoint of the opposite side, such as the plane ABE in the figure, have six methods;
The third category: passing through the midpoint of four sides of a tetrahedron, the plane is parallel to the other two sides, as shown in the figure, EFGM has three planes.
Therefore, the four different ways of taking * * * are -(4+6+3) = 14 1 (species).
So choose d
Comments on lines, surfaces, geometric figures and other figures composed of points are typical combination questions. Common additional conditions are * * * line and non-* * line, point * * surface and non-* * surface, line * * * surface and non-* * surface, etc.
Example 6 (1) has five balls numbered 1, 2,3,4,5 and five boxes numbered 1, 2,3,4,5. Now put these five balls into these five boxes, and ask each box to put a ball, and there are exactly two balls with the same number as the box.
(2) Put four different balls into four boxes numbered 1, 2, 3 and 4, and there is just an empty box.
There is one kind.
Solution (1) Step 1: Throw two balls with the same box number, and there are throwing methods; Step 2: Throw the other three balls. Take the throwing method of the first step as 1, the second ball into the box 1 and box 2 as examples. Since the other three balls cannot be thrown with the same ball number and box number, there are two throwing methods, as shown in the block diagram.
④
⑤
③
⑤
③
④
3 4 5 3 4 5
To sum up, there are * * * 20 delivery methods that meet the meaning of the question.
(2) Step 1: Take out two small balls (planting method) to synthesize an "element" and synthesize three "elements" with the other two balls; Step 2: put three elements into three of the four boxes, and put one element in each box to form an empty box (planting method). So what are the placement methods that meet the meaning of the question? = 144 species.
Comment This is a set of comprehensive counting questions. It should be noted that if it is determined in the question (1) that the remaining three balls in the second step can be put into the remaining three boxes at will, list? Formula, you will make mistakes.