Math permutation problems, help math permutation problems

0 is the mantissa number c43 a33

2 is the mantissa c43 a33-c32 a22, 0 in the first time to remove!

4 is the mantissa of C43 A33-C32 A22, the reason is the same as above.

c43×a33+c43×a33-c32×a22+c43×a33-c32×a22=60

a54-c43 a33-2(c43 a33-c32 a22)=60Both methods are 60, is it wrong??

Think about it yourself, where do you look at it, what did your teacher say?

Upstairs, random answers, this score has 0 no 0 two steps, no 0 is p4 4, there is 0 when 0 can not be in the first place.

To be even, the end should be 0,2,4.

And 0 can't start with it, so there should be.

There are 16 kinds of 0, 2, and 4 endings in 1 dozen heads.

There are also 16 types of 3 dozen heads.

There are 16 types of 2 dozens.

There are 16 types of 4 dozen heads.

0 starts without. So there are a total of 16 + 16 + 16 + 16 = 64 kinds.

I can't count them.

Discussion of different situations: 1 case with zero 0 may be on a 1000 place, 3 and then in addition to the zero bit, choose 3 out of 4 numbers in the other three digits 4 4 3 2 result 3 4 3 2

2. In the case of excluding zeros, 4 numbers are all rowed: 4 3 2 1, odd numbers are single digits, 1 3, choose one 2

Thousands 2 4 1 (3) Choose 1 3 Other 3 2 3 Total 2 3 3 So the even number is 3 4 3 4 3 2 1 2 3 3 ?

I just came back and looked at it, and this question is still quite interesting to many people There are many ideas for this question My answer is 48 Don't say that others are wrong if you don't have the same ideas

When ending with 0, there are 4!Piece.

There is a 4 when it ends with 2 or 4-3!Piece.

So there is a total of 24 + 24-6 + 24-6 = 60

The focus of this question is how to deal with the position of 6, and now use "x" to represent a certain number in , and the arrangement simulation is as follows:

xx8x75x6

xx8x756x

xx8x765x

xx867x5x

x68x7x5x

6x8x7x5x

From the above arrangement, it can be seen that there are four 6 kinds of arrangement, and there are four numbers in each arrangement that need to be filled, which is the whole arrangement process, so there are: p(4,4)*6=144 kinds.

No problem with the title.

There are some problems with the problem itself, but if the ordinal number of 5 in the example is 2, then one of the solutions to the problem is as follows, and the order of 578 must be 8, 7, 5In other cases, if it does not meet the topic, press 6 and put it before 8, between 87, between 75, and after 5.

1 When boy A stands on the left, because the position is fixed, so don't think about it, 2 girls together only have 2 situations, because the girl is next to A, so the position is also fixed, as long as the other 4 boys are all lined up, that is, a 4 4 and the first 2 girls can swap positions between each other, there is a 2 times, so when A is on the left, 2 women are next to each other, the formula is 2*a4 4

In the same way, A is also counted on the right, so as long as the above formula is *22, the middle position of the boy A is actually fixed, in fact, don't think about it, two girls are in one situation, but 2 girls can change positions between them, so there is a 2 times, and then the 2 girls are regarded as a whole (the binding method is used here) and the other 4 boys are all lined up, that is, a5 5, so the final formula should be 2*a5 5

Because I learned math eight years ago, but I'm more interested in permutations and combinations, I don't know if it's the right thing to do, I hope it helps you.

Analysis: Treat 2 girls as a big girl (there are 2 kinds of rows inside), now it has become 6 people, 5 men and 1 woman, first row A, there are 2 kinds, and the remaining 5 are casually arranged, so there are 2 * 5 * 4 * 3 * 2 * 1 = 240, because there are 2 rows inside the big girl, so the correct row = 2 * 240 = 480 kinds, and the row formula = 2 * 2 * 5!

2) It is much easier, because A is in the middle, it is fixed, there is only one arrangement, first fix the A, there are 3 positions on the left, there are 3 positions on the right, because the girls must be adjacent, so the two girls are either on the left or on the right, look at it as a whole, there are 2 kinds of rows on the left, (there are 2 kinds of rows inside), so there are 2 * 2 = 4 rows, because there are also two rows on the left and right, so the girls have 2 * 4 = 8 rows, and the remaining 4 boys are casually arranged, there are 4 * 3 * 2 = 24 rows, So there is a total of 24*8=192 arrangement, and the sorting formula = 2*2*2*4! =192

(1) Treat 2 girls as a whole, and arrange all 5 elements with 4 boys except A, a total of 5 elements, a total of a(5,5) methods;

The 2 girls have their own a(2,2) arrangement;

Male A can be ranked at the left end or the right end, and there are 2 ways.

According to the principle of multiplication, the number of common methods is .

a(5,5) a(2,2) 2=480 species.

2) Citing the results of (1), without considering boy A, the total number of permutations in a group of 6 people consisting of 4 men and 2 women is .

a(5,5) a(2,2)=240 species;

The situation that boy A is in the middle is 1, so the number of ways to put boy A into the group of 6 people that has been arranged is also 240, but this also includes the original 2 girls in the middle of the two adjacent positions, and the addition of boy A makes the 2 girls become non-adjacent, a total of a(2,2) a(4,4)=48 kinds, so the number of eligible arrangement methods is.

240-48=192 species.

Considering that one of the 4 independent lights is in different positions, if it is at the position of 1 or 8, then the remaining adjacent lights have a total of 6-2 = 4 positions, so there are 2*4=8 types; Similarly, bits 2 and 7 have 2 * (5-2) = 6 kinds, 3 and 6 bits have 2 * (4-2) = 4 kinds, 4 and 5 bits have 2 * (3-1) = 2 kinds, there are 20 different positions, and each position has 2 4 = 16 non-signal light combinations, so there are a total of 20 * 16 = 320 kinds of signals.

How many lights are there in a row? The topic is not clear.

Don't tell me how many lights there are.

In this permutation problem, the last step is wrong, and there is no last step.

Because you first choose 4 out of 8 to stand in a row, is your row the first row or the second row? In the first row, the remaining four people are all lined up in the second row. If you have to say that I want to choose first, not to mention the first row or the second row, then you have a duplication, for example, you choose the first row of 1234, and the second row of 5678, then you must have the first row of 5678, and then choose the 1234 row, so isn't it a repetition, right?

There is no need for the third step, in fact, the answer is that all 8 elements are arranged.

Let's assume that five numbers are coquainous, then take two different numbers to form the fraction a b, a total of 4*5=20 combinations.

LGA-LGB = LG(A B), so A B has 20 different combination results, and LGA-LGB has 20 different results.

But the reality is that 1 3 = 3 9 and 3 1 = 9 3, i.e., there are two duplicate cases. Then the final answer is 20-2=18.

So the answer is C.