Take any three numbers from 1, 3, 5, 7, 9, and any two numbers from 0, 2, 4, 6, 8 to form a five digit number without repetition______ How many?

Take any three numbers from 1, 3, 5, 7, 9, and any two numbers from 0, 2, 4, 6, 8 to form a five digit number without repetition______ How many?

Take three out of the first five numbers, and take two out of the last five numbers, which are arranged in order, with a total of: c53c52a55. ∵ there may be 0 in the selected five numbers, ∵ to subtract the zero start, ∵ the zero start has a total of: c53c41a44, so the number of five digits without repeated numbers = c53c52a55-c53c41a44 = 11040

It is composed of the numbers 1, 2, 3, 4, 5, 6, 7, 8 and 9. All the possible four digit numbers without repetition are______ ．

There are 9 × 8 × 7 × 6 = 3024 four digits without repetition. When 1 is in thousand digits, there are 8 × 7 × 6 digits, that is, when 1 is in thousand digits, it needs to be added 336 times. When 1 is in hundred digits, ten digits and one digit, there are 8 × 7 × 6 digits, which are added 336 times. The sum of number 1 is 1111 × 336 The sum is 1111 × 336 + 2222 × 336 + +9999×336=336×（1111+2222+… +9999) = 336 × 49995 = 16798320 A: the sum of these four digits is 16798320

How many five digit numbers can be formed by taking any three numbers from 1, 3, 5, 7 and 9 and any two numbers from 2, 4, 6 and 8
I know the answer, but the point is I don't know why?

Take any three numbers from 1, 3, 5, 7, 9, there is no order requirement, it is a combination, so there are C (53) = 5 × 4 ／ 2 = 10 ways to take them. Similarly, take any two numbers from 2, 4, 6, 8, there are C (42) = 4 × 3 ／ 2 = 6 ways to take them. After taking them twice, the five numbers will be arranged completely, and there are a (55) = 5 × 4 × 3 × 2 × 1 = 120 ways to arrange them

How many five digit numbers can be formed by taking any three numbers from 1, 3, 5, 7 and 9 and any two numbers from 2, 4, 6 and 8?

There are C35 methods to take any three numbers from 1, 3, 5, 7 and 9, and C24 methods to take any two numbers from 2, 4, 6 and 8. Then the five numbers are arranged in full, with a total of c35c · 24 · A55 = 7200, so a total of 7200 five digit numbers without repetition can be formed

Permutation and combination: how many three digit numbers without repetition can be formed with the numbers 1, 2, 3 and 0? If the numbers are allowed to repeat, the total number can be formed··
Permutation and combination: how many three digit numbers without repetition can be formed by numbers 1, 2, 3 and 0?
If the number is allowed to repeat, how many different three digit numbers can be formed?

Take 1, 2 and 3 first, and the combination number is 3 × 3 × 3 = 27
Then take 0 and two of the remaining three numbers (there are three methods), 0 cannot be in the first place
Then the number of combinations is (2 × 2 × 1) × 3 = 12, so there are 27 + 12 = 39 combinations when no repetition is allowed
As for the second question, since the first question has calculated all the combinations without repetition, only three digits with repetition are left
There are three combinations when all three digits are the same
The combination of two numbers is 3 × 6 + 3 + 2 × 3 = 27
The meaning of 3 × 6 is to take any one of 1, 2 and 3 numbers as a repeated number. For example, if 1 is taken and then 2 and 3 numbers are taken, there are 6 combinations. When the repeated number is 2, there are 6, so it is 3 × 6
The reason for adding 3 is that there are three combinations when 0 is repeated
The last 2 × 3 is because if you take 0 as the non repeating number, and then take any one of the other three numbers as the repeating number, it will be 2 × 3
Therefore, 39 + 27 + 3 = 69 combinations are allowed

Permutation and combination have the numbers of 0, 1, 2, 3 and 4, which are required to form a 4-digit number without repetition
There are numbers 0, 1, 2, 3 and 4, which are required to form a 4-digit number without repetition. How many such numbers are there?
Please list the formula

4 * 4 * 3 * 2 = 96

How many combinations are there for 3 out of 10 numbers?
1. How many numbers are there in the combination of 3 out of 10 numbers from 0 to 9? The numbers in the combination can be repeated, such as 011 or 111
2. How many numbers are there in the combination of 7 out of 10 numbers from 0 to 9? The numbers in the combination can be repeated, such as 011 or 111

Each person can choose from 0 to 9, and each person can choose from 10
1.10*10*10=1000
2.10^7

How many even four digit numbers ()
A. 6B. 10C. 12D. 24

When the last digit is 0, the first three digits are randomly arranged with A33 = 6. When the last digit is 2, the first digit can only be selected from 1 and 3, and then the middle two digits are arranged with A12 · A22 = 4. According to the principle of classification and counting, there are 6 + 4 = 10 even digits without repetition

[permutation and combination] use the numbers 0, 1, 2, 3, 4 and 5 to form a four digit number without repetition
It can be composed of 300 different four digit numbers, which can be arranged in a sequence from small to large. What is the 85th item?

From small to large
How many are there at the beginning of 1
And then how many of them start with 2
The first one is to find the last three digits, and the fifth one is to choose the third row, that is, 5 * 4 * 3 = 60
2 starts with finding the last 3 for 5 and selecting 3 rows is 5 * 4 * 3 = 60, so item 85 starts with 2
Then look for a hundred
It's item 61 from 2
From 0 to 20, 4 choose 2 rows, 4 * 3 = 12
Since the beginning of the first 100, we have selected 2 rows of 4 * 3 = 12 24 items, and now we have 84 items
The next item is what you want
It's a hundred. The 85th project of 2301 started from 3 is 2301

Take any two digits from the number 1.2.3.4.5.6.7.8.9 to form a double digit without repetition. Is this double digit a multiple of 9?
The answer is one in nine
Isn't 9 a multiple of 9

The multiples of 9 are 18,27,36,45,54,63,72,81, which are completely symmetrical. There are 72 kinds of 9 * 8, so 8 / 72 = 1 / 9