Don't answer yet!

Maybe you will be skeptical, but after reading this article, you will find enough reasons to believe it.

First, the problem import

Quote 1 Four people, Tang Priest, the Monkey King, Pig Bajie and Friar Sand, work in different positions in the same company, and now they need to change their positions, asking everyone not to stay in their original positions, so they have different arrangements.

Example 2: Each of the four students wrote a greeting card. First, all of them were put away. Then each student took out a greeting card and asked everyone not to take his own card. Thus, the four greeting cards were distributed in different ways.

Example 3: Put four balls numbered 1, 2, 3 and 4 into four boxes numbered 1, 2, 3 and 4 respectively. It is required to put one ball in each box, and the number of balls cannot be the same as the number of boxes (i.e. 1 no 1, 2 no 2, 3 no 3).

It is not difficult to find that the three examples cited above are all the same type of questions. What is the answer? Let's use enumeration to give you the answer:

Suppose the original order: a, b, c, d.

Pay attention to certain rules when enumerating. If it is regarded as the position of 1, 2, 3 and 4, then in the first step, A can put any position of 2, 3 and 4, in the second step, determine the position of B, and in the third step, determine the positions of C and D:

1 dislocation arrangement: b, a, d, C(A is in position 2, b is in position 1, and the positions of c and d are uniquely determined);

The second dislocation arrangement: D, A, B, C(A is in position 2, B is in position 3, and the positions of C and D are uniquely determined);

The third dislocation arrangement: C, A, D, B(A is in position 2, B is in position 4, and the positions of C and D are unique);

The fourth dislocation arrangement: B, D, A, C(A is in position 3, B is in position 1, and the positions of C and D are unique);

The fifth dislocation arrangement: C, D, A, B(A is in position 3, B is in position 4, and the positions of C and D can be 1, 2);

The sixth dislocation arrangement: D, C, A, B(A is in position 3, B is in position 4, and the positions of C and D can also be 2,1);

The seventh dislocation arrangement: B, C, D, A(A is in position 4, B is in position 1, and the positions of C and D are only determined);

Eighth dislocation arrangement: C, D, B, A(A is in position 4, B is in position 3, and the positions of C and D can be 1, 2);

The ninth dislocation arrangement: D, C, B, A(A is in position 4, B is in position 3, and the positions of C and D can also be 2, 1).

It can be seen that there are nine dislocation arrangements of the four elements. In other words, the answers to the above three cited examples are all nine.

Well, here's the problem: Tu Tu is always wet. I don't want to list them all. My eyes are blurred. What should I do? And what if it's not four elements next time? What if the answer is swollen again?

Please read the following patiently. Let's be clear in advance: the next paragraph needs some math knowledge. If you think your math is not bad, you can read it word for word. If you say no, it doesn't matter, just remember the final conclusion!

Second, theoretical derivation.

In fact, the essence of the three problems involved in the above example is the arrangement of each element that is not in its own numbered position. We call the permutation problem with this restriction the total dislocation permutation problem.

Is a very old mathematical problem, Bernoulli, Euler and other mathematicians have studied it. Although this kind of problem is difficult, we can solve it quickly? Trick? Yes Let Tu Tu Laoshi explain it in detail for everyone:

We record all the dislocations of n elements as Dn.

D 1 = 0 because the1element is not misplaced.

Two elements can be interchanged, that is, there is 1 dislocation arrangement, then D2= 1.

When n? At 3 o'clock, any element ai among n different elements is not arranged in the I position corresponding to its number, but must be arranged in one of the remaining n- 1 positions, so ai has n- 1 permutation.

That is, the first step is to arrange ai, and there are n- 1 species.

Step 2: Arrange the elements corresponding to the positions occupied by ai.

For each arrangement of ai, for example, ai is arranged at J position, and there are two arrangements of elements aj corresponding to J position:

The first case: aj happens to be in I position. At this time, ai is in the J position, aj is in the I position, and the ordering of elements ai and aj has been determined, and there are n-2 elements left, so their ordering problem is transformed into the number of n-2 elements all dislocated, and there should be Dn-2 kinds;

The second case: aj is not in the I position. At this time, ai is still in the J position, and aj is not in the I position, which means that aj also has a position that cannot be ranked. That is to say, except ai, all the other n- 1 elements have a position that can't be arranged, so the problem can be transformed into a total dislocation arrangement of n- 1 elements.

That is to say, there are (Dn- 1+Dn-2) second-step arrangements aj.

According to the principle of multiplication, we can get two-step multiplication: dn = (n-1) (dn-1+dn-2) (n? 3)。

In other words, we get a recurrence formula of the total dislocation arrangement number. For this formula, only when we know the values of the first term 1 D 1 and the second term D2, can we deduce the values of all the following terms.

For example: D 1=0, D2= 1, D3 = 2 (D2+d1) = 2 (1+0) = 2, D4 = 3 (D3+D2) = 3 (2+/kloc).

Remember the conclusion:

D 1=0，D2= 1，D3=2，D4=9，D5=44，D6 = 265 Dn =(n- 1)(Dn- 1+Dn-2)(n？ 3)。

I can't remember. What if it's swollen? Look at the wet Tu Tu.

Third, Tu Tu shorthand.

There is no dislocation arrangement when there are 1 elements, d1= 0;

The dislocation arrangement of the two elements is 1 species, D2= 1, and shorthand: D2 = 2d1+1;

There are two kinds of dislocation arrangement of three elements, D3=2, shorthand: D3 = 3d2-1;

There are 9 kinds of dislocation arrangements of the four elements, D4=9, abbreviated as D4 = 4d3+1;

There are 44 kinds of dislocations in five elements, D5=44, shorthand: D5 = 5d4-1;

Six elements have 265 dislocation arrangements, D6=265, abbreviated as D6 = 6D5+1;

Dislocation is arranged with Dn kinds of n elements, shorthand: Dn=nDn- 1+.

What should I do if I still can't remember?

I want to tell you a very good news. The four elements and five elements are the most frequently tested in the civil service examination. You only need to remember two important figures? 9? And then what? 44? You can handle it! Do you suddenly feel good?

Let's follow Mr. Tu Tu to kill a gun through several questions:

Fourth, the examination room seconds kill

Example 1 (Beijing, 20 14) Four different cars are parked in four adjacent parking spaces. Now, all the cars leave and stop at these four parking spaces again, and all the cars are required not to stop at the original parking spaces. How many different parking methods are there? ( )

A.9 B. 12

C. 14

Answer a

Analyze the problem of total dislocation arrangement. Remember the numbers: D4=9, D5=44, Dn=nDn- 1+, so four cars have D4=9 parking modes. Therefore, the answer to this question chooses option a.

Example 2(20 1 1 Zhejiang) Four chefs each cook a special dish at dinner. Now everyone is required to taste a dish, but they can't taste their own cooking. How many different ways are there to ask * *? ( )

A.6 species B. 9 species

C. 12 species and D. 15 species.

Answer b

Analyze the problem of total dislocation arrangement. Remember these numbers: D4=9, D5=44, Dn=nDn- 1+. It can be seen that the total dislocation arrangement number corresponding to the four elements is D4=9. Therefore, the answer to this question is option B.

Example 3(20 15 Sichuan Luzhou institution) Four computers A, B, C and D are arranged in a row, counting from left to right. If A is not in the first position, B is not in the second position, C is not in the third position and D is not in the fourth position, then there are () different arrangements.

A.9 B. 10

C. 1 1 D. 12

Answer a

Analyze the problem of total dislocation arrangement. Remember these numbers: D4=9, D5=44, Dn=nDn- 1+. It can be seen that the total dislocation arrangement number corresponding to the four elements is D4=9. Therefore, the answer to this question chooses option a.

Example 4 (Shandong, 20 15) A company transferred 1 staff from five subordinate departments to communicate with other departments. If each department can only accept one person, how many different personnel arrangements are there? ( )

A. 120

C.44 D. 24

Answer c

Analyze the problem of total dislocation arrangement. Remember these numbers: D4=9, D5=44, Dn=nDn- 1+. It can be seen that the total dislocation arrangement number corresponding to the five elements is D5=44. So, the answer to this question is C.

To sum up, for the problem of total dislocation arrangement, numbers? 9? And then what? 44? Better match! Everyone must remember!