How many 023585 numbers can make up 4 digit odd numbers that are not repeated?

The question you are talking about belongs to permutations and combinations and simple number theory.

My opinion is as follows——, if you form an odd number with 4 digits that are not repeated, then the single digit can only be one of them; A thousand can only be one of .

1. The thousand digits are 3 or 5, and the single digit can only be 5 or 3, then, the 4-digit odd number that can be composed:

p(2,2)*p(2,3)= 12 (pcs);

2. The thousand digits are not 3 or 5, but can only be 2 or 8, then, you can form a 4-digit odd number:

c(1,2)*c(1,2)*p(2,3)=24 (pcs)In summary, the total number of pcs is 12+24 = 36 pcs.

The above answers are for your reference.

This is a permutation and combination of questions.

4-digit odd numbers with no repeating numbers:

The unit digit of the odd number is the odd number, so there are 2 ways to choose the unit digit (choose 3 or 5); The thousand digit (i.e. the first digit) cannot be 0, so the thousand digit can only be selected from the remaining non-zero numbers, and there are 3 ways to choose (remove the odd number chosen by 0 and the single digit, and there are 3 left in the 5 numbers); The 10 and 100 digits are arranged in 2 of the last 3 numbers, and there are 6 types. Therefore, using the principle of step-by-step multiplication counting, a total of 2 3 6 = 36 non-repeating numbers can be composed of 4-digit odd numbers.

It can be composed of so many odd numbers.

Solution, there are 1 from 5 and 3 in single digits

I choose 1 of the dry position from 3 numbers.

then n=c2(1)c3(1)a3(2).

2 3x3x2=36 (pcs).

0, 1, 2, 3, 4, 5 can be composed of 288 types of five-digit odd numbers without repeating numbers.

1. The guarantee is odd, and the last digit must be one of 1, 3, and 5, that is, c31;

2, the first digit cannot be 0, then the first digit is one of the remaining four numbers, that is, c41;

3, at this time, the first and last digits are determined, then you also need to choose three numbers, that is, c43;

4. Since the first and last digits are determined, the other three digits can be positionally uncertain, using the principle of stepwise multiplication, i.e., c31c41c43a33 288.

To sum up: the last digit must be an odd number (1,3,5), the first digit cannot be 0, and the middle three digits c43 (at this time, these three digits contain both 0 and no 0), so there is no need to divide 0 and no 0 to arrange.

1. Single-digit arrangement: 3 types;

2. The arrangement of the first digit: you use one, plus 0 can not be in the first place, then the first number of ways to arrange are: 4 kinds;

3. The remaining digits are arranged casually, including: a(3,4)=24 kinds;

Then: 3 4 a(4,4) = 288 species.

There are 3 choices for the single digit, 4 choices for the 10,000 digits, 4 choices for the 1000 digits, 3 choices for the 100 digits, and 2 choices for the 10 digits.

Multiplication principle: 3 4 4 3 2 = 288 five-digit odd numbers without repeating numbers.

Single digits: 1, 3, 5 - 3 types.

Classification: Choose 0 ten hundred thousand three bits can choose to put 0 3 kinds.

The other three digits choose three out of four numbers to sort a(4,3)=24 - a total of 3*24=72 kinds of this kind.

0a(4,4)=24 is not selected

Therefore, the number of species in these four digits is 72 + 24 = 96.

Multiply by the number of species in the single digit and get 96*3=288

First choose two of the two numbers of 1 and 3 as the last number, there are two ways to choose one from the remaining 3 (divided by 0) numbers as the first number, there are three ways to the last 3 numbers, there are a (3, 3) methods.

2*3*a(3,3)=36

All numbers are arranged in a full order of a66=720... Half of them are even, half are odd. The odd number is 360.

Then count the odd number of 0 in the first place. c31 * a44=72.

I'm still working on the newspaper, hehe288

18 pcs. Explanation: Because of the 4-digit number, 0 cannot be in the first place. The order under the model of the remnants of the world is as follows.

This problem is mathematically permutations and combinations, and it is easier to solve it after learning this knowledge. Under normal circumstances, it should be 5*4*3, according to the requirements of the question, 0 cannot be the first digit of the number, so the number of three-digit numbers should be (5-1)*4*3=48.

The four numbers 02358 can be composed.

4 4 3 = 48 three digits without repeats.

Hope it helps.

Upstairs is correct, a total of 48 different three-digit numbers can be formed, including 023, 025, 028 and so on, such as three-digit numbers starting with 0.

If you don't count the three digits starting with 0, you can make up a total of 36 types.

2 starts with 2x3x2=12.

There are 3x2=6 at the beginning of 3.

A total of 12 + 6 = 18.

The single digit is 0, the hundred digit has c(5,1), the ten digit has c(4, buried 1), the total c(5,1)c(4,1)=20 species;

The single digit is 2, the hundred digit has c(4,1), and the ten digit has c(4,1), a total of c(4,1)c(4,1)=16 species;

The single digit is 4, the hundred digit has c(4,1), and the ten digit has c(4,1) species, a total of c(4,1)c(4,1)=16 species;

Total: 20 + 16 + 16 = 52 species.

Summary: Like this kind of special circumstances can be calculated in the special situation, the rest according to the position of the combination can be, the back of 2, 4, can be combined in the single position to choose 2, 4 method buried liquid combustion c (2, 1), the hundred can not be 0, has been selected to think that there are 4 digits of the combination of the number c (4, 1), there are 4 digits left in the bend of the void ten digits to choose a number of c (4, 1), the total number is c (2, 1) c (4, 1) c (4, 1) = 32, plus the special case that the single digit is 0.

How many three-digit numbers without repeating numbers can be formed with numbers, and how many even numbers can be formed?

Solution: Rank single digit: Take any 1 of the three numbers have: a(3,1)Rank hundreds: Take 1 of the remaining 5 have: a(5,1) (note: it contains 0 as the head).

Middle number: Take 1 of the remaining 4 have: a(4,1) but in a(4,1) in the middle number, because the hundred digit cannot be 0, it appears twice.

Therefore, the number of unqualified species should be subtracted.

Therefore, it is obtained: a(3,1) a(5,1) a(4,1)—2 a(4,1)3 5 4—2 4=52 (pieces).

Answer: There are a total of 52 non-repeating and even three-digit Hui Chunfeng.

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No duplicate numbers:

0 for a good talk: 5x4 = 20.

2, 4 choose one to do a single digit, key Sun La 0 can not be the first place: 2x4x4 = 32.

Total: 52 slips.

There are even three-digit numbers with repeated numbers without beats, and when the single digit is 0, there are c(5,1) c(4,1) 1=20;

When the single digit is not 0, there are c(2,1) c(4,1) c(4,1) c(4,1)=2 4 4=32;

That is, the seller can be grouped, and there are no repeated numbers of three-digit numbers, and the even number of divination is 20 + 32 = 52.

With numbers 012345 a total of 5 5 4 = 100 triple digits without repetition holes, of which 5 4 + 2 4 4 = 52 even numbers.

In total, it can form a three-digit number 5 5 4 = 100, of which the even number is 1 5 4 + 2 4 4 = 52.

0,2,3,4,5,1 is a total of 6 numbers.

Three-digit digits: hundreds, tens, singles; Even: The single digit must be 0,2,4

Hundred: Take from the five numbers of 2, 3, 4, 5, and 1, a total of 5 differences.

Ten digits: Take the five numbers of 100 digits 2, 3, 4, 5, 1, and remove the one that repeats the hundreds, a total of 4.

Single digit: Remove 1 repeat from the 10 digits and 4 numbers, and take 3 for a total of 3.

1. Different three-digit numbers: 5 4 3 = 60 pcs, whereEven:60 2 5 = 24.

Now, add 0 to 10 digits, single digits: 10 digits (1, 100 digits, 1 row spike digits do not move): 1, 3, 5 can add 2 0 differences (e.g. only 1 0 0s can be added (204 or 402 1 each), a total of 3 2 + 2 1 = 8; Unit digits (1, 100, 10 do not move):

The first 5 4 hundred digits and ten digits can be added by 0, a total of 5 4 = 20.

So,2. Different three-digit even numbers: 24 + 8 + 20 = 52. Specifically:

102，104；120，124；130，132，134；140，142；150,152,154 (12).

204；210，214；230，234；240；250,254 (8).

302，304；310，312，314；320，324；340，342；350, 352, 354 (12).

402；410，412；420；430，432；450,452 (8).

502，504；510，512，514；520，524；530，532，534；540,542 (12).

Total: 12 3 + 8 2 = 36 + 16 = 52.

With the number silver rolling hidden 023451 can form a total of 5p(5,2)=5 20=100 without repeating the number of three strikers, where the even number of d is p(5,2)+4 4 2=20+32=52.

A total of 100 three-digit numbers without repeating numbers can be formed 012345 numbers, of which 50 are even.

The hundred digit can't be 0, so there are 5 numbers to choose from, ten digits to choose one from 5 numbers, and a single digit to choose one from guessing the 4 numbers of Fan Kai Sui sedan Chang, namely:

c(5，1)×c(5，1)×c(4，1)=5×5×4=100。The single digits are 0, 2, and 4 are even, and there are 50 digits.

Determine the 100 first, 0 can not be selected, so there are 5 ways to choose a shirt. In determining the ten positions, there are also 5 selection methods, and finally the first position is determined, and there are 4 re-selection methods. So it is possible to form a three-digit number without the number of the collapse of the tomb:

5 5 4 100.

The following trap is the step to solve the problem of Wang file.

Topic. With the four numbers 0, 1, 2, and 3, how many three-digit numbers can be formed without repeating numbers?

Ordinary Student Thoughts:

Note: When using the enumeration method, it is necessary to be organized and enumerated by category.

0 cannot be used as the highest digit.

When 1 is the highest digit, there are 6 three-digit numbers of 102, 120, 103, 130, 123, 132;

When 2 is the highest digit, there are 6 three-digit numbers of 201, 210, 203, 230, 213, 231;

When 3 is the highest digit, there are 6 three-digit numbers of 301, 310, 302, 320, 312, and 321.

Therefore, a total of 6 3 = 18 (pieces) can be formed into three digits.

Latecomer Strategy:

Same method as above. Grinding socks.

Answer: Answer: It can form 18 three-digit numbers without repeating numbers.

Is the legend of the god of the sword good?

When the mantissa is 1, there are 4*3=12.

When the mantissa is 3, there are 4*3=12.

At 5 o'clock, the clan is slow and fierce and there are also 12 Zhaoqiao.

So there are 36 of them.

1. Single-digit arrangement: 3 types;

2. The arrangement of the first digit: you use one, plus 0 can not be in the first place, then the first number of ways to arrange: 4 kinds;

3. The remaining digits are arranged casually, including: a(3,4)=24 kinds; Key Ant Scatter.

Then: 3 4 a(4,4) = 288 species.