Sophomore permutations and combinations, sophomore math permutations

Hehe, I'll go online on my phone to answer your questions!

Here's how I do it:

1. Find out all the possible numbers first: the base number cannot be 1 (this is the first thing to consider for logarithms), so there are 8 possibilities for the base number (2-9) as long as the true number is greater than 0, so there are 9 possibilities (1-9), so the logarithm that can be formed is 8 9 = 72 kinds.

2. The next thing to do is to find out the same result:

1) When the true number is 1, the logarithm is 0 regardless of the value of the base, so there are 8 zeros, and 2) When the true number and the base number are the same, the logarithmic result is 1, so there are 8 1s

3) When the base number is 2, the true number is 4 or the base number is 3, and the true number is 9, the value of the logarithm is 2, so there are 2 2s

4) The true number and the base number in 2) above are reversed, and we can see that there are 2 1s 2

Because the question is what are the different logarithms of the total composition, simply:

It's been a long time since I've done a math problem, and I don't know if it's the right thing to do? When I was in school, I was the best at math, and I scored 143 in math in the college entrance examination!

Hehe, sloppy, this time it should be right, right?

Take 2 first, i.e. 9, 8, 2=36

Choose another one to be the bottom and one to be true, and multiply by 2 for 72

2 children are seen as a pit, but there is an order multiplied by 2

There are 21 choices at the beginning and 1 choice at the end.

In the middle three positions, full row A33

c21 a33 2 multiply =24

2×3a3×2a2=2×3×2×1×2×1=24

Explanation: The first row of Zhang Dad or Wang Dad, that is, 2 kinds, two children together as a piece of "child", Zhang's mother, Wang Ma, and the child are in the middle of the team, in any order, that is, 3A3 = 3 2 1, and then consider the two children, you can be in front of Xiao Zhang, you can be in front of Xiao Wang, 2 kinds, such as the best is the answer.

The first division: there is a group of 3 people, the other two groups of 1 person each, a total of C35 A33 = 60 different ways to participate, the second division: there is a group of 1 person, the other two groups of 2 people, a total of C25 C23 A33 = 90 different ways to participate.

There are 60 + 90 = 150 ways to participate.

Then calculate the participation method of A and B to participate in the same project, and regard A and B as 1 person, a total of 4 people, and the possible participation methods are: C24 A33 = 36 The number of arrangement methods that meet the above requirements and A and B do not participate in the same project is 150 36 = 114

Yang Hui triangle.

1 line 1 term c(0,0)=1;

2 rows 1 term c(1,0) = 1,2 terms c(1,1) = 1;

Line 35 14 items C(34,13) = 927983760 and 15 items C(34,14) = 1391975640;

927983760 1391975640=2 3 The ratio of the 14th and 15th numbers from left to right in row 35 of the Yang Hui triangle is 2:3.

The permutations and combinations have been lost for three or four years.,My answer is also 192.,I don't know if I'm right about this idea.,Post it.,How did the landlord say 384?,I hope the landlord can post the solution idea.。

Take 2,4 as a whole, such as x, this x has two possible combinations, now there are only five number combinations: 0,1,x,3,5, use them to form a five-digit number first look at the first digit of the five digits, which can be: 1, x, 3, 5, that is, p41 4 The second number can be:

One of the four (minus the first digit) in 0,1,x,3,5 is also p41=4

The third is p31=3

The fourth is p21 2

The fifth is p11=1

So it adds up to p21*p41*p41*p31*p21*p11=192

If there are 25 boys who choose three to serve as class presidents, school committees, and sports committees, this is the arrangement of 3 out of 25.

25 24 23 = 13,800 different options.

Number of female students???

People are sorted by 1, 2, 3, 4, 5, 6, i.e. 1 is A, 2 is B...

1 goes to A, first arrange workshop B, 2, 3 can't go to workshop B, then there are 3 people left to choose 2 to go to B, that is, C32, so that there are 3 people left, choose one to go to workshop A C31, and the remaining two will go to workshop C, so a total of C32 * C31 = 9

A has to go to workshop A, so don't worry about it.

If B goes to workshop A, then C can only go to workshop C. There are 3 people left, choose one of them to go to workshop C, and the remaining two can only go to workshop B. There are 3 ways to do this.

If C goes to workshop A, then B can only go to workshop C, and there are also 3 ways.

If B and C both go to workshop C, then there are 3 people left, choose one of them to go to workshop A, and the remaining two can only go to workshop B, there are 3 ways.

There are a total of 3+3+3=9 methods.

B to A, then C can only go to C, and then choose one person to C, there are three options, and the remaining two go to B in the same way: C to A, then B can only go to C, and then choose one person to C, there are also three options, and the remaining two go to B

In addition to B, the other 3 people of C will choose one to A, B and C will go to C, and the remaining two will go to B3+3+3=9

If there are 3 people seated, it will be seen as one person in the middle

1. If there is only one seat between him and the left and right people, then there must be more than 2 vacant seats on the other side of one person, so it is impossible;

2. If there are 2 seats between him and the left and right people, then there are 3 groups of seats to choose from: (1,4,7)(2,5,8)(3,6,9). Each group of seats has a p(3,3) way to sit, so there are a total of 3 p(3,3)=18 types.

3. If he has one seat on the left and two seats on the right, then there are also 3 groups of seats to choose from: (2,4,7)(3,5,8)(4,6,9). So there are also 18 ways to sit.

4. If there are two on his left and one seat on his right, it is also 3 groups: (2,5,7) (3,6,8) (4,7,9). There are also 18 ways to sit.

So, there are a total of 18 3 = 54 ways to sit.

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4[7^669+c(669,1)*7^668+..c(669,668)*7+1]

4[7^669+c(669,1)*7^668+..c(669,668)*7]+4

4[7^669+c(669,1)*7^668+..c(669,668)*7] is a multiple of 7.

2 2009 divided by 7 and 4

Today is Wednesday, and the second day of the 2009 calendar of the Tambu Calendar is Sunday.