High School Mathematics, Permutations and Combinations

Solution: For the first arrangement: 11123 is arranged in the following ways: (a5,5) a(3,3)=5*4*3*2*1 (3*2*1) =20 kinds of arrangement, where a(5,5) means that the number of ways in which the 5 numbers are arranged without considering the repeated numbers, because there are 3 identical numbers, so it is necessary to divide by a(3,3).

For the second arrangement: the arrangement has a(5,5) [a(2,2) a(2,2)]=5*4*3*2*1 [2*2] = 30 likewise, where a(5,5) means that the number of ways in which the repeated numbers are all arranged without considering the repeated numbers, since there are two kinds of identical numbers, so it is necessary to divide by a(2,2)*a(2,2).

Note: (The digit before the comma is in the lower right corner of A or C, and the digit after the comma is in the upper right corner of A or C. )

11123 c(5,1) x c(4,1) x c(3,3) Look at 3 Choose 1 out of 5 positions, 2 choose 1 in the remaining 4 positions, and 1 in the last 3 positions

Or c(5,3) x c(2,1) x c(1,1) from 1, pick 3 positions from 5 positions for 1, pick 1 for the remaining 2 positions for 2, and last 1 position for 3

11223 c(5,2) x c(3,2) x c(1,1) from 1, choose 2 out of 5, choose 2 for the remaining 3 positions, and give 2 for the last 1 position

c(5,1) x c(4,2) x c(2,2) from 3 .

Question 1. a55/a33 =5×4=20（5！/3！）

a55/(a22*a22)=120/4=30

This type of problem to be rescued is ann (a number of repetitions, factorial of the number of second repetitions......）

Repeat these five numbers and divide a number by a number several times, for example, 11123 first row a5 in the division by a3 11223 first in the full row a5 and then divide by (a2*a2) to get 30 This method is rote memorization.

8. Individual numbers 1-8

Let's say you pick 1-4 in circle A, and the rest are in circle B; And if you select 5-8 in A and 1-4 in B, these two situations are the same, because there is no difference between the two circles of A and B. 4 out of 8 people is c84... But there is no difference between the two circles, so divide by a22

Choose four people to form a circle of C84, there is no difference between the two circles, so except for A22.

For example, if you choose a group of 1, 2, 3, and 4, it is the same as if you chose a group of 5, 6, 7, and 8.

Since x+f(x)+xf(x) is an odd number.

So if x is odd, f(x) is odd or even.

If x is an even number, f(x) is an odd number.

And then that's it in four steps.

The first step is to take the two odd numbers in a, and there is one way to take them.

The second step is to combine the two odd numbers with any element of b, there are a total of 25 ways to take the even number in a, there is 1 way to take it.

The fourth step is to take any odd number in b, and there are 2 ways to take it.

Then use the multiplication principle to get a total of 25*2=50 ways to take it, that is, mapping, and get your answer.

From the analysis of the topic, to make x+f(x)+xf(x) is an odd number, then x and f(x) are either odd numbers, or an odd and an even, can not be even at the same time, when x is an odd number, f(x) odd and even, so it is 5 2 When x is an even number, f(x) must be an odd number, so it is 2 1, according to the map definition, the three elements in a must be taken, so this is a step-by-step, applying the principle of multiplication, you can get 5 2 * 2 1 = 50

1-1)^2011=1-c2011 1 + c2011 2 -c2011 3 + c2011 4 - c2011 5…Grinding dies....-c2011 2011=0.

c2011 1 + c2011 2 -c2011 3 + c2011 4 - c2011 5……Cong Zhi -c2011 2011 = -1

The 5 projects are contracted by 3 teams, and each team contracts at least one, and there are two schemes 1+1+3

1. One team contracts 3 items, and the other two teams each contract 1 item, first 3 teams are lined up, and then 1, 1, 3 are selected: a(3,3)*c(5,1)*c(4,1)*c(3,3) a(2,2), because the number of projects contracted by two teams is the same, so it is necessary to divide by a(2,2), which is a heap problem.

2. One team contracts 1 item, and the other two teams each contract 2 items, first 3 teams are fully platooned, and then select 1, 2, 2: a(3,3)*c(5,1)*c(4,2)*c(2,2) a(2,2), and there are also two teams contracting the same number of projects.

The total number is the sum of these two schemes, which should be 150, choose C