High School Permutation Questions, High School Math Permutation Questions

6 people in a row, if the position is indicated by 123456, then A, B and C can only be in 135 or 246, otherwise it will not meet the topic. Because they stand in a row, there is a difference in order, so it is a matter of arrangement. So what to do now is to have several ways to arrange A, B and C in these three fixed positions, namely:

p(3,3)=6 species; If there are three remaining people, which also correspond to three empty spaces, then the arrangement method is p(3,3)=6. Therefore, when A, B and C are in the 135 position, the number of species arranged is 6 * 6 = 36 species, and when A B and C are in 246, the number of species arranged is 6 * 6 = 36 species, so the total number of permutations is 36 + 36 = 72 species, I don't know if you can understand this expression, I hope you can like it!

a33*a43=144 ranks A, B and C first, so it is A33=6. Then the other three people plug in the four blanks, there is a sequential interpolation station, so it is a43=24. Finally, 24*6=144

The numbers 0, 1, 2, 3, 4, 5 are common to four digits without repeating numbers.

When the mantissa number is 0, 2, 4, it is an even number, and the mantissa number is 0.

5*4*3=60 species.

There are 4*4*3=48 kinds with a mantissa of 2.

There are 4 * 4 * 3 = 48 kinds with a mantissa number.

So there are 156 kinds in total.

Among them, the mantissa 0 and the hole orange 5 can be divisible by the early group 5.

There is a mantissa of 0.

Seed. There is a mantissa of 5.

Seed. So there are 108 kinds of hands.

(1) There are 1+2+2 2+2 2+2 (n-1=n 2-1 numbers in the first n rows of the number table.

Then the first number in line i, is 2i-1

So a ij = 2 (i-1) + j-1

Because 2 10< 2010< 2 11, and a ij = 2010

So i=11

Then there is 2 10 + J-1 = 2010

The solution is j=987

2）∵an=a11+a22+a33+…+ann

1+2+2^2+……2^(n-1)]+1+2+……n-1）]

2^n-1+n(n-1)/2

So a n-n n=2 n-(n 2+3n+2) 2

When n 4, it is easy to know a nn 2+n

When n=4, the inequality is obviously true, and suppose that when n=k(k 4), the conjecture holds, i.e., 2 k>(k 2+3k+2) 2

When n=k+1, 2 (k+1)=2*(k 2+3k+2) 2=k 2+3k+2

Because k 2+3k+2-[(k+1) 2+3(k+1)+2] 2=(k+2)(k-1)>0

Then 2 (k+1)> k+1) 2+3(k+1)+2] 2, i.e., when n=k+1, the conjecture is also correct

By , when n 4, is established

When n 4, an n 2+n

In summary, when n=1,2,3, an n2+n; When n 4, an n 2+n

1. The last position is 0: there can be 5 choices for the first place, 4 kinds for the second place, and 3 kinds for the second place, a total of 5 4 3 kinds.

Last non-0: 2 last position, 4 first position, 4 second position, 3 second position, a total of 2 4 4 3 types.

A total of 5 4 3 + 2 4 4 3 = 156 species.

2. The first place is 2 or more: 4 types in the first place, 5 types in the second place, 4 types in the second position, and 3 types in the last position, a total of 4 5 4 3 kinds.

The first position is 1, and the second position is 4 or more: 2 types of secondary position, 4 types of second position, 3 types of last position, a total of 2 4 3 types.

The first position is 1, the second position is 3, and the second position is 4 or more: 2 types of second position, 3 types of last position, a total of 2 3 types.

A total of 240 + 24 + 6 = 270 species.

a46-a35-a35

The subtracted a35 is the first stick of A, and then the A35 that is subtracted, and B runs the fourth bar, in the middle, A runs the first bar, contains the fourth bar of B, and in the same way, B runs the fourth bar, contains the first stick of A run, that is, A runs the first bat, and B runs the fourth bat, which is lost twice.

So + once.

First of all, choose 4 out of 6 people, minus the first in the first row of A, and subtract the 4th in the B row, so that the first in the first row of A and the fourth in the fourth row are added.

Because the subtracted a35 contains a24, minus two a35s, so you subtract an extra a24

Only then do I add an a24 at the end

This a24 is when the first baton of A and the fourth baton of B exist at the same time.

There are 3 options for the recreation committee.

There are 4 options for study committee members.

There are 3 options for the Sports Commissioner.

So there are 3*4*3=36 species.

Understand what the job is first.

Schedule work. Who is special?

A and B, A work.

Special preferred. By special classification:

A and B are arranged: c42 * c21 * a33 = 72

Arrange one of A and B: C21 * C43 * C31 * A33 = 144 Do not arrange A and B: C44 * A44 = 24

So there are 240 schemes.

This one is relatively simple, A can have four people to choose from, B has five people, because six people have five left, C has four, and D has three. That is, 4*5*4*3=240 people.

If there is no A and B, then A44 (4*3*2*1).

If there is one person A and B, then C43*3*A33 *2 (A or B) If A and B have both, then C42*A22*A22

Add the above three cases.

Hello: A mapping is a mapping that defines a domain to a value range, which can be one-to-one or many-to-one.

That is, for every x in the defined domain, there is a unique y corresponding to it in the value field, but the reverse is not required.

1) A has five choices for each element in the definition, there are five elements, so there are 5 mappings to the fifth power, and there are 3125 mappings for chain leakage.

2) One-to-one mapping from A to A.

Arrange all the elements within the definition domain.

There are 5 options for the first element, the remaining 4 for the second element, 3 for the third element, 2 for the fourth element, and 1 for the fifth.

So the number of mappings is 5*4*3*2*1=120.

3) There happens to be an elemental preimage.

We subtract the number of mappings in (1) from the number that happens to have all the elements as the same element, which is the number we are looking for.

All elements correspond to one element, and there are 5 types.

So there are 3125-5=3120 mappings.

The shed is rotten. Thank you.