A: There are 7 schemes, and the number of methods for each scheme is as follows:
Scheme 1: 20? 20? 60? 100 ?
First, choose two to store 20, and arrange the remaining two different numbers.
C(4，2)* P(2，2) = 12？
Option 2: 20? 40? 60? 80
Make a row
P(4，4) = 24
Option 3: 20? 40? 40? 100
Choose two first, save 40, and arrange the remaining 20 different numbers?
C(4，2) *P(2，2) = 12
Option 4: 20? 60? 60? 60
Choose three first and save 60, and the rest will naturally save 20.
C(4，3) = 4
Option 5: 20? 80? 80? 20?
First choose two to save 20, and the remaining two will naturally save 80.
C(4，2) = 6
Option 6: 40? 40? 40? 80 ?
Choose three first and save 40, and the rest will naturally save 80.
C(4，3) = 4
Option 7: 40? 40? 60? 60?
First choose two to save 60, and the remaining two will naturally save 40.
C(4，2) = 6
Total number =12+24+12+4+6+4+6 = 68 species.
2. It is known that A is a set composed of all factors of 54, and let% be an integer on A..
In addition to relationships,
(1) drawing posets
(2) Determine the length of the longest chain in A and write all the longest chains in A in dictionary order.
(3) How many disjoint anti-chains can the elements in A be divided into at least and written completely.
These reverse chains
A: the set of A = {1, 2, 3, 6, 9, 18, 27, 54}.
? 1)COVER(|)={( 1，2)，( 1，3)，(2，6)，(3，6)，(3，9)，(6， 18)，(9， 18)，(9，27)，( 18，54)，(27，54)}
? 2) There are four totally ordered subsets with the most elements:? The longest chain length is 5.
L 1={54，27，9，3， 1}
L 1={54， 18，9，3， 1}
L 1={54， 18，6，3， 1}
l 1 = { 54 18，6，2 1 }
? 3) It can be divided into at least three disjoint chains: {2,3}, {6,9}, {18,27}.
3. Find the number of integer solutions of the equation t 1+t2+t3+t4=20, where T 1 ≥ 3, T2 ≥ 1, T3 ≥ 0 and T4 ≥ 5.
A:
Because t 1 ≥ 3, t2 ≥ 1, t3 ≥ 0, t4 ≥ 5, t 1 is taken as 3, t2 is 1, t3 is 0, t4 is 5, and the sum of the values is 9, that is to say, the difference of 20-9 is allocated to T650.
At this time: t1'+T2'+T3'+T4' =11.
So there is c (1 1+4- 1,1) = c (14, 1 1).
4 Let S = {∞ 2, ∞ 4, ∞ 5, ∞ 7, ∞ 9} be a given heavy set, where 2, 4, 5, 7, 9.
There are five different elements in S, and each element can have an infinite number in the set. Set hn
It means that n elements are taken from S (can be taken repeatedly), and 2 and 4 are required to appear even several times.
Number of permutations, hn?
Solution: It is known that two elements 2 and 4 only appear even times, and three elements 5, 7 and 9 appear any times. According to the modified generating function equation, there are:
Let g (x g (x) = (1+x 2/22/2! + x^4/4！ +....+ x^n/n！ )^2 * ( 1 + x + x^2/2！ + x^3/3！ +...+x^n/n！ )^3
? = (e^x + e^(-x))/2 * e^3x
? = 1/4(e^x + 2* e^3x + e^5x)
? = 1/4 (∑ x^n/n！ + 2* ∑ (3x)^n/n！ + ∑ (5x)^n/n！ )
? = 1/4∑( 1+2 * 3^n+5^n)x^n/n！
Hn = 1/4( 1 + 2 * 3^n + 5^n)
5. Four students are interviewed in English and German at the same time, and each subject only requires an interview 1 person. The two subjects are not in the same order. How many orders are there?
Solution: This problem can be understood as that four students go to the English interview in any order, and they can't go to the German interview at the same time.
That is, it turns out that a classmate can't be in the same position (misplacing problem).
So, the solution of this problem is 4! D_{4} = 4！ *4! *( 1- 1/ 1! + 2/2! -3/3! +4/4! ) = 24*9 = 2 16。
6. A restaurant has three kinds of desserts, and there are infinite varieties. How does Xiao Wang choose four kinds of desserts?
Solution: t 1+t2+t3=4.
? C (3+4- 1, 4) = c (6,4) =15 species.
7. In the expansion of (2x 1-3x2+x3) 6, find the coefficient of x 1 3 * x2 * x3 2.
Solution:
(2x 1-3 x2 +x3)^6 =(2x 1-3 x2+x3)*(2x 1-3 x2+x3)*(2x 1-3 x2+x3)*(2x 1-3 x2+x3)*(2x 1-3 x2+x3)*(2x 1-3 x2+x3)
The expansion is equivalent to the product of six polynomials. Each item is regarded as the relationship between addition principle and others.
? Since x 1 is a cubic power, three of the six polynomials choose 2x 1, so there is C (6 6,3) * 2 3.
? Since x2 is the power of 1, -3x2 is selected from the remaining three polynomials, so there is C(3, 1)*(-3).
? Since x3 is a power of 2, it is enough to choose x3 from the remaining two polynomials, so there is c (2,2).
So the coefficient x13 * x2 * x3 2 = c (6,3) * 2 3 * c (3,1) * (-3) * c (2,2) =-1440.
Solution 2:
Since the sum of the powers of x 1+x2+x3 is 6, a six-term polynomial can be understood as a finite collinearity consisting of three x 1, one x2 and two x3.
? 6! /(3! * 2! * 1! ) * 2^3 * (-3) * 1^2 = - 1440
8. If 1/( 1-2x) 2 = ∑ AK * x k, then ak =
Solution:
According to Taylor formula
f(x) = f(0) + f 1(0)/ 1！ *x + f2(0)/2！ *x^2 + ....
The second expansion:? f 1(0)=(-2)( 1-2x)^(-3)*(-2)？ = (-2)*(-2) f 1(0)/ 1！ = (- 1)^ 1 *( 1+ 1)*(-2)^ 1
The third expansion: F2 (0) = (-2) (-3) (1-2x) (-4) * (-2) 2 = (-2) (-3) * (-2) 2? f2(0)/2！ = (- 1)^2 *(2+ 1) * (-2)^2
So the k term is expanded as fk(0)/k! =(- 1)^k *(k+ 1)*(-2)^k
Method 2:
Because the formula (1-x) (-n) =? ∑ C(n+k- 1,k)*x^k
Substitute n=2? X = 2x to get the coefficient: c (k+ 1, k) * 2 k.
9. How many different ways are there to put four different balls in three different boxes so that there will be no empty boxes?
? Solution:
? Take two balls C(4, 2) first, and then line up the rest. So the answer is c (4,2) * p (3,3) = 36.
10. Five liberal arts students, five science students, which rows are in the middle?
? Solution:
Arrange five liberal arts students first, and five more! Plant, and then insert the science students into the gap between the liberal arts students, so there are five! This kind of plug-in can be located on the left or on the right.
So there is always 2 * * 5! * 5! Species arrangement.
1 1 Find the number of positive integer solutions of equation x 1+x2+x3+x4= 10.
Solution:
Because of the positive integer solution, x1> is needed; 0，x2 & gt0，x3 & gt0，x4 & gt0
The original formula can be understood as: (x1+1)+(x2+1)+(x3+1)+(x4+1) = 6.
So the number of positive integer solutions is C(6+4- 1, 6) = c (9,3) = 84.
12.？ Expand the function f (x) = (1+x+x 2+x 3+ and find the coefficient of x14 ...) 2 * (x 2+x 3+x 4+...) 3.
Solution:
? This situation can be understood as five different colors of balls (such as black, white, red, green and blue), and each ball is unlimited. Now it is required to take out 14 balls, including at least 2 red balls, 2 blue balls and 2 green balls.
? Then you can distribute 2 red balls, 2 basketball balls, 2 green balls, 14 balls, and take the remaining 8 balls from the balls of 5 colors.
? Let's assume that there are 1 white balls, x2 black balls, x3 green balls, x4 red balls and x5 blue balls. The method is:
? x 1+x2+x3+x4+x5 = 8
? Then c (8+5- 1, 8) = c (12, 4) = 495.
13. There are two red balls, 1 white ball and 1 yellow ball. How many different combinations are there?
Solution: 1) Generation function method:
f(x)=( 1+x+x^2)( 1+x)( 1+x)= 1+3x+4x^2+3x^3+x^4
? So there is a combination number =1+3+4+3+1=12.
2) Case discussion:
Red: 0. 1.2
White: 0. 1
Yellow: 0. 1
Select exit method: 0,0,0.
Select the combination of 1 ball: (0, 0, 1)(0, 1, 0), (1, 0).
Choose the combination of two balls: (0, 1, 1) (1, 0, 1, (1, 1, 0) (2,0,0).
Choose a combination of three balls: (1, 1, 1) (2, 1, 0), (2, 0, 1).
Choose a combination of four balls: (2, 1, 1)
Total * * * combination scheme =1+3+4+3+1=12.
14. Put four different balls in three different boxes, so there are no empty boxes. How many different ways are there?
Solution: It is completed in two steps. The first step is to divide four different balls into three groups, that is, to select two balls as a group. There is a C (4,2) method.
? Step 2, put the balls divided into three groups into three different boxes and fill them up with P (3,3).
So the number of methods is: c (4,2) * p (3,3) = 6 * 6 = 36 (species)
15. Find the number of positive integer solutions of equation x+y+z+k = 10.
Solution: (x+1)+(y+1)+(z+1)+(k+1) = 6.
C(6+4- 1，6) = C(9，6) = C(9，3)