Permutations and combinations need to be known! Talent comes in

1.There are 3 straight lines, and at most you can determine a plane for every two, so there are 32 = 3 in total.

2.There are 3 planes, and they don't have a common intersection line, which means that every two planes determine an intersection line, and there are c32=3.

3.It's 6 intersections, it's not a mistake, I'll give you 6 points.

A total of 6 points Every 3 non-collinear means that every 3 points can determine a triangle (respectively, vertices), a total of c63=20.

Hehe, in fact, the problem of arrangement and combination is very simple, they have an order, they are arranged, and the arrangement without order is a combination, if you don't understand, he, I'll talk to you in detail.

That c32 or something, you just read it in order, I won't hit that symbol.

I'm in a hurry, I made a mistake, hehe.

Any two can determine 1 plane, so it is c23 (the front is above, and the back is below), a total of 3.

3 Because any two planes intersect have 1 intersection line, but the title says that there is no common, so it is still c23 = 3.

Any 3 points out of 6 can form a triangle, so c36 = 20 kinds.

It's not easy to type my bar.

The idea is represented by Cartesian coordinates.

2. I also think of the Cartesian coordinate system, where there are 3 straight lines and 6 rays.

3. Choose 3 out of 6 simple permutations and combinations, a total of 20 situations.

The answer to the first question is 3 planes.

The answer to the second question is 3 intersections.

These two also know by comparing their fingers.

The third question is 20, and the stupid way is that the 6-sided one counts by himself, and the god man upstairs c32=6, admire.

There are 3*9*8 2=108 combinations of the letters o, q and 0.

There are only 0, the number of combinations without the letters o and q is 3*9=27, only o or q, and the number of combinations without 0 is 2*3*9*8 2=216, so the letters o, q and the number 0 appear at most one in a total of (108+27+216)*4*3*2*1=8424.

2. (1) can form 4*5=20 different two-digit numbers (2) there are 4 numbers larger than 2 in b; There are 3 larger than 4, 2 larger than 6, and 1 larger than 8.

Therefore, it can form a two-digit number with ten digits less than a single digit 4+3+2+1=10.

3 6 = 729, the number of people in each intellectual competition is not limited, everyone can choose 3 items, and the choices of 6 people are independent of each other, so there are 3*3*3*3*3*3 ways to register! You're thinking too complicated!

The answer is 729? How do I calculate it's 720... 6*5*4*3！

There are 6 types of candidates for the first project, 5 types of candidates for the second project (one project per person, so the second project is selected among the remaining 5 people), and 4 candidates for the third project. The three items are indefinite, and the rank is 3! , so 6*5*4*3!

I didn't understand what your formula meant yesterday, but today I understand that the method you considered was correct, but there was a small mistake in the calculation. Your formula should be 3c(6,6)+a(3,3)*[c(6,1)c(6,2)c(6,3)+c(6,1)c(5,4)c(1,1) 2+c(6,1)c(5,5)+c(6,2)c(4,2)c(2,2) 3+c(6,3)c(3,3) 2]; You are miscalculated because you are dividing by a factorial of 2 or 3

In the group, you have a total of 420 less group

Suppose there is only 1 dot on the paper, plus the 4 dots of the quadrilateral, there are 5 in total, and you can cut it into 4 triangles.

After that, add points in turn on Sun Xiao's loss paper, and note the divine will:

Each time a point is added, the point is inside a triangle that was previously cut, and this point divides the triangle into 3 triangles, i.e. this point increases the total number of triangles cut out on the paper by 2.

As a result, the number of triangles cut out increases by 2 for each point of infiltration.

So what is sought is 4+2*(10-1)=22

How to cut? Do the three vertices of the triangle have to be three of the 14 points?

5 different teachers are assigned to 3 schools, 3 of the 5 teachers are selected from 3 schools respectively, there are a (5,3) methods, the remaining 2 arbitrary points, there are 3 to the 2nd power of the division, and the two steps are multiplied to get 540

1.A and B both go to school A, and the remaining three teachers first choose two to B, C, there are A(3,2) methods, and finally the remaining one has 3 methods, multiplying 18 The probability is 18 540 = 1 30

2.The 5 teachers are divided into 3 parts, and at this time, the number is not considered, but only the number of teachers. There are 6 divisions of and a total of 6 divisions.

When a=1, there are 3 ways, expect=3 6=1 2, when a=2, there are two, expectation=2 6=1 3, and when a=3, 1, expectation=1 6

If you have any questions, you can ask me.

All of the above answers are questionable.

is an average sail leak grouping.

For example, (ab) (cd) e f

cd ) ab ) e f is the same.

Answer: Closed c(6,2)c(4,2) a(2,2) The remaining two people do not need to be divided, and they are naturally divided into 2 groups.

15*6 2=45 species.

If you want to work differently from the 4 of the Banquet Thing.

It is 45*a(4,4)=1080, which is not found in this question.

In the first category, the twin sisters of the freshman year are on the first car, and the remaining two on the first car are from different grades, and two of the three grades are selected as c(2,3), and then one more student is selected from the selected grade as c(1,2),c(1,2), so there are 3 2 2 = 12 species.

In the second category, if the twin sisters of the freshman year are not in the first car, then two students from one of the remaining three grades are selected to be in the first car, which is c(1,3), and then one person is selected from the remaining two grades (the same as the first category), then there are a total of 3 2 2 = 12 types.

So there are 24 different ways to get on the bus.