All five bottles are labeled, what is the possibility that all of them are mislabeled?

There are 44 possibilities for all of them to be mislabeled.

Just memorize: 1 element is not all misaligned.

There is 1 type of 2 elements with a total misalignment arrangement, 2 types of 3 elements with a total misalignment arrangement, 9 types of 4 elements with a total misalignment arrangement, and 44 types with 5 elements.

Methods for studying misalignment problems——Enumeration.

For less common permutations, you can use the enumeration method.

When n=1, the full arrangement.

There is only one, not a staggered arrangement, d1 = 0.

When n=2, there are two kinds of full arrangement, i.e., and , the latter is staggered, and d2 = 1.

When n=3, there are six kinds of full arrangement, i.e., , where only and are staggered, and d3=2. In the same way, we can know that d4=9.

The smallest number of misalignments are: d1 = 0, d2 = 1, d3 = 2, d4 = 9, d5 = 44, d6 = 265, d7 = 1854.

All mislabeled: 44 types.

Just memorize: 1 element is not fully misaligned, 2 elements are 1 full misalignment, 3 elements are 2 are fully misaligned, 4 elements are 9 are fully misaligned, and 5 elements are 44 are fully misaligned.

For specific derivations, please refer to:

Fully dislocated arrangement encyclopedia.

2 out of 5 are paired, which is a total of 5c2=5!/2!/(5-2)!= 10 scenarios.

The remaining 3 bottles were all mislabeled. ABC three bottles, if A sticks B, then it can only be B sticks C, C sticks A, one situation. If the bending beam is A pasted with C, it is the same, there is only one case. There are two types of wrong stickers.

Based on the above, there are a total of 10 * 2 = 20 kinds of travel conditions.

To break it down: it's right to pick two out of 5:

c(2,5) = 10 species.

Among the remaining three, there will be any row: all Tong Wuqing right, one blind right, all wrong three game grip situation.

It is known that it is all wrong on the three, so the first and third bottles, the optional labels are: 2 kinds, for the 2nd 3rd, the optional is 1 kind, and the third is also 1 kind.

So, c(2,5)*2*1*1 = 20

Answer] :d paste three wrong according to the question, and paste two on the right mask. First of all, select 3 of the five bottles as C35 10 kinds, these three bottles are mislabeled, and these three bottles are mislabeled in two cases, such as:

A is pasted as B, B is pasted as C, and C is pasted as A; In the second case, A is pasted as C, B is pasted as A, and C is pasted as Hongzhen A. So there are 10 2 20 cases of mislabeling three out of five bottles. So the answer is d.

20 kinds. Event.

1. If there are three of them that are wrong, then two of them must be correct, that is, c(2,5)=10 events. Second, the other three are all mislabeled, you can assume that the number of the three bottles is 1,2,3, and the corresponding three Yuyuan labels are also 1,2,3, and there are only two possibilities if you want to paste all of them wrong.

The first type is 1 (2), 2 (3), 3 (1).

The second type is 1 (3), Qing infiltration 2 (1), 3 (2).

The inside of the parentheses indicates the label, and the outside indicates the bottle shouting base.

Event 1 and Event 2 are obviously independent, and the rule of multiplication is used, that is, 10*2=20

20 2 pairs out of 5, there are 5 * 4 (2 * 1) = 10 possibilities The remaining 3 are mislabeled, there are A mislabeled as B, B mislabeled as C, C mislabeled as A or A mislabeled as C, C mislabeled as B, B mislabeled as A 2 possibilities These 2 independent conditions should be multiplied, so there are a total of 10*2 = 20 kinds [[Unclear, ask again; Satisfied, please adopt! Good luck opening !!

1. First choose three bottles to paste the wrong ones, which is equivalent to choosing 3 out of 5, and there are 10 ways to choose them;

2. If the three bottles are wrongly labeled, use the elimination method to do it: three bottles are labeled with three labels, a total of 3! = 6 stickers; There are 3 cases where only one bottle is right, 1 case where there are 3 bottles, and it is impossible to have only 2 bottles right.

The remaining 2 cases are the case where 0 bottles are pasted correctly;

3. So the possible situation is 10 2 = 20 kinds.

Label bottles.

1 2 3 4 5 a b c d e take 1 2 3 when d e pair.

There are these two 2 3 1

Take 1 2 4 when c e pair.

There are these two 2 4 1

Thus, choosing 3 out of 5 numbers has 123;124；125；134；135；234；245；235；145；345 These 10 ways to arrange, and then each of them has two ways to paste the wrong way, that is, 10 * 2 = 20 kinds.

All mislabeled.

There are 4 types of five sequential movement dislocations.

Two of them are mislabeled with each other [1 medium], and the other three are mislabeled with each other [2 kinds], c(2 5)*1 2=10 2=20 kinds.

There are 4 + 20 = 24 types.

First of all, 3 of the five bottles were selected as C35 10 kinds, these three bottles were mislabeled, and the three bottles were mislabeled in 2*1*1 cases. So there are 10 2 20 cases of mislabeling three out of five bottles.