High School Mathematics Permutations and Combinations, High School Math One Permutations and Combina

There are three types at the end of row A> 1 and 234 in row A, 3 types at the end of B no row, and the remaining 3 are all arranged 3*2*1

There are 3*3*3*2*1=54 kinds.

2> At the end of row a, the remaining 4 numbers are all arranged.

There are 4*3*2*1=24 types.

So there are 78 kinds in total.

1) B, there are 4*3*2*1=24 types.

2) If c starts and b is not at the end, then there are 2*3*2*1=12 types.

3) If D starts with D and B is not at the end, then there are 2*3*2*1=12 types.

4) If E starts with and B is not at the end, then there are 2*3*2*1=12 types.

There are a total of 60 permutations.

A is in the first place, there are 4*3*2 kinds, B is in the last place, there are 4*3*2 kinds, A is in the first place, B is in the last place, there are 3*2 kinds.

Therefore, there is 5*4*3*2-4*3*2-4*3*2+3*2=78

First find a45 = 5 * 4 * 3 * 2 = 180....All the order in which they are arranged.

A in front: a34 = 4 * 3 * 2 = 24

b is at the end: a34=24

A is in the front and B is in the back, A23=3*2=6

then 180-24-24+6=138

a54-a43-a43+a32=78

Choose 4 of the 5 elements in a random arrangement, subtract the arrangement with a first, subtract the arrangement with b as the end, and add the repeat subtraction to get the arrangement (a is the first and b is the last combination).

You've got to have duplicates.

For example, make boys 1, 2, 3.

Follow your method: one situation: choose No. 1 to go to the rest of the hospital first, then go to the rest of the hospital after 2 and 3, maybe No. 2 and 3 will also be assigned to the rest of the hospital.

Another case: choose No. 2 to go to the rest of the hospital first, and then go to the rest of the hospital after 1 and 3, maybe No. 1 and 3 will also be assigned to the rest of the hospital.

So the two situations are repeated

There are other situations that will be repeated, so you will inevitably count too much Correct algorithm: It is not wrong to divide the girls first, i.e. a32

Then it depends on the situation of the remaining hospital:

There are three types:1If there is only one boy, there is c31*a21*a21=12 kinds of 2If there are two boys, there are c32*c21=6 species.

3.Three boys are left, and there is one.

It adds up to 12 + 6 + 1 = 19 types.

There are a total of a32*19=114 species.

The remaining two boys, each of whom chooses 1 c(3,1)*c(3,1)=9 from 3 hospitals.

It's obviously wrong, and there are a lot of repetitions

If there must be ab bb ba aa, there must be a shared aba to meet the ab ba number is equal.

So first write out without AABB:

abababa or bababab (three times each for ab and ba. 5 A's and 3 B's to be inserted

abababa

Insert 5 A (insert at A, do not change AB, BA) (there are four blanks).

So there is. A 5 series of 4 kinds (insert 5 A's in a row) such as aaaaaabababa

1 4 series 12 species 4 3 (divided into two groups 1a plug and then another 4a) 2 3 series 12 species 4 3

113 series, 12 types, 4 3 2 2 types.

The 122 series has 12 types in the same way.

1112 series 4 types.

11111 series 1 kind.

56 kinds of inserts 3 Bs

There are three of them) one 3 series 3 kinds (insert 3 Bs in a row).

1 2 series 6 types 3 2 (divided into two groups 1b plug and then another 2b) 111 series 1 kind.

10 kinds, so abababa has a total of 56 10=560 kinds.

abababa

Insert 5 A (insert at A, do not change AB and BA) (there are three blanks).

So there is. 0+5 series 3 types.

1 4 series 6 types 3 2 (in two groups 1a plug then another 4a) 2 3 series 6 species 3 2

113 series 2 types 3 2 2

122 series 2 3 2 2

21 kinds of inserts 3 Bs

There are four of them) one 3 series 4 types (insert 3 Bs in a row).

1 2 series 12 types 4 3 (in two groups 1b plug then another 2b) 111 series 4 types.

20 kinds, so there are 21 20=420 bababs.

So the two are 420+560=980 in total

How many of the 15-letter sequences arranged with the letters A and B meet the following conditions?

Two consecutive letters aa appear five times, ab, ba, bb three times each Solution: Starting from the first term, let the paragraphs composed of a be A1, A2, A3 and ,...... in turnThe paragraphs composed of B are B1, B2, B3, ,......

The permutations that meet the conditions can only be such combinations.

a1 b1 a2 b2 a3 b3 a4 or b1 a1 b2 a2 b3 a3 b4 Each segment contains at least 1 a (or b), which occupies 7 bits, and 8 bits are required;

How to divide the remaining 5 A's and 3 B's so that there are 5 Aa and 3 BBs If a paragraph is composed of (5 + 1) A's, then only 5 Aa can be formed in this paragraph;

The remaining 5 A's and 3 B's allocation methods can only be like this.

Yes, put a has c(8,5) 56 (species), put b has c(5,3) 10 (species), a total of 56 10 560 (species).

Yes, put a has c(7,5) 21 (species), put b has c(6,3) 20 (species), a total of 21 20 420 (species).

Therefore, the total number of permutations that meet the conditions is 560 + 420 980 (species) This question is too difficult.

ab, ba, bb three times each, so at least: -abbabbabba - there are 10 letters here, and there are 5 letters left, to make it aa appear five times, and then what, can't think of it. Let me give you a hint and see if you can do it.

1 out of 4 c 7 4

2 Choose one female c 4 1 Then choose four out of eight c 8 4 Multiplication principle 3 Divide into three cases 0 Female 5 male, 1 female 4 male, 2 female 3 male. Principle of Addition 4 First select the representative c 5 3 times c 4 2. Then sort A 5 5 .

Principles of multiplication.

1.Boy A (mandatory) and girl A (mandatory) are proposed, and the remaining seven are randomly selected four, C 7 4, 35 species.

2.List c41*c54+c42*c53+c43*c52+c44*c51The answer is 125

3.Consider the case where the girl is zero (boys only), c55, when one girl, c54*c41, when two girls, c53*c42, add, the answer is 81, 4Three boys C 5 3 and two girls C 4 2 were selected first, sorting, A 5 5, multiplication principle, 7200 species.

1.There are 35 kinds.

2.There are 125 kinds.

3.There are 81 species.

4.There are 7,200 species.

Give me extra points.

Are you saying that the first position cannot be 1

The first two are not 12

The top 3 is not 123

The top 4 is not 1234

That's what it means.

All 5 numbers are arranged in a(5,5)=120 types.

Subtract the above 4 cases.

120-a（4，4）-a（2，2）a（3，3）-a（3，3）a（2，2）-a（4，4）

There are 48 types that meet the conditions.

Hello, I am Teacher Qingxin, what is your topic? I'll see if I can help?

Questions. <>

Why is the case of a girl repeating an A22?

Please wait a minute, the parsing process will take a little time.

If you still don't understand something, please say it, let's work together until you understand it thoroughly, and it will be easy to start the questions in the future.

Questions. C22 can't do it.

Is C22 a combination, or is it possible to choose A or B repeatedly

Questions. I don't really understand, but it doesn't seem like you need to sort two of the two girls in Korean.

Because Korean must have a girl, C12 chose one, multiplied by A33, is it possible that the girl who chose Korean is also in A33? Shouldn't it be a22 to subtract the chance of repetition?

Questions. Wouldn't there be no situation where both girls chose were subtracted? And the Russian and Japanese ones don't have to be cut together, didn't they happen at the same time?

Russian and Japanese do not need to be cut out, because there are no special requirements for Russian and Japanese, as long as there is one person in each of the two languages.