Comprehensive application of permutations and combinations of Grade 5 Olympiad problems!! For more

Hello, let's start with the first type of situation: who wins the first two games. There are two types of cases.

Category 2: Hit the third game.

There are two situations in the first game, that is, 2, if A wins the second sentence in the first game, there is only one situation, so multiply by 1, and anyone can win in the third game, so multiply by 2, that is, 2*1*2=4

In the end, the sum of the above two types of cases is 2+4=6

6 possibilities. AA, A, B, B

There are 14 types in total.

A wins: A A, A B A, A B A A, A B B A B A, B B A A, B A, B A, B A, B A B A.

B wins: B B, B B B, B A B B, B A B B, B A B B, B

A A A B B

B B B

6 possibilities. (1) A wins the first two games; (2) B wins the first two games; (3) A wins the first game, B wins 2 games in a row; (4) B wins the first game, and A wins 2 games in a row; (5) A wins the first game, B wins the second game, and A wins the third game; (6) B wins the first game, A wins the second game, and B wins the third game.

Looking for a topic? There's a competition math network that's good.

Question 1: Because of the <, so d Question 2: Column equation a+b=,a-2=b+2+, from the second equation we can get a=b+2+2+, and then substitute the first equation, b+, we can get b=21,a=

Question 3: The answer is 4005. The process of solving the problem is as follows:

1. If the decimal point is added between the ten digits and the single digit, then let the four digits be x, then, the solution equation can get x=, not that the integer does not meet the requirements, so the decimal point is not added between the single digit and the ten digits.

2. If it is added between the ten and the hundred, the equation can be obtained in the same way, and the equation can be solved to get 4005, which meets the requirements. 3. If the decimal point is added between the hundred and the thousand, the equation can be obtained in the same way, and the equation can be solved to obtain x=, not an integer that does not meet the requirements.

4. If the decimal point is added before the thousand, the equation can be obtained, and the equation can be solved to get x=, not an integer that does not meet the requirements.

Question 4: Answer: Number A = , Number B = The process of solving the problem is as follows:

According to the problem, it can be seen that the number A is less than the number B, then you can list the equation: number B - number A =, 10 * number A = number B, substituting the second equation into the first equation, then 10 * number A - number A = , solve the equation to get number A = number A =, then number B =

Additional question: The answer is $ per bag of milk powder. Problem Analysis:

Because Dongdong buys 3 bags less than Fangfang, the remaining money must be greater than or equal to yuan, and the money worse than Dongdong to buy 3 bags is yuan, and because they bring the same amount of money, so the money left after Dongdong buys 5 bags of milk powder should be yuan, so the money for 3 bags of milk powder should be yuan, and each bag of milk powder should be yuan, Dongdong and Fangfang both brought yuan.

1,, b because, from this it can be seen that d 0

And because, visible shrinkage 10 times, so.

2. Basket: kilograms

Basket: kilograms

The digits become 6 digits, and the decimal point is reduced to the original, i.e. it is only a multiple of the original number.

Equation: 4, from "the decimal point of the number A is moved to the right by one place, then it is equal to the number B", it can be seen that the number B is 10 times the number A.

Number A: Number B:

Question 1: So; c c, so e is 1 or 6, so I have to choose b;

Question 2: After: B: (;

Original: B: 23-2=21kg

A: Question 3: 4 digits become 6 digits, and the decimal point is reduced to 1 100.

Ten digits and everyone: 05 (because 95), then take 05 to count thousands and hundreds (note that 5 has been borrowed 1): 40

This four-digit number is 4005

Question 4: A x, then B 10x, 10x-x=, x=, 10x=

1 You can try this, if it is a, bring it into the letter, try slowly, the answer is b2, you can know from the question, A is 2*2+ kg more than B, and then use a formula, ((

3 solutions; Let the original number be x, now it is 1 100x

x-1/100x=

x=4005

4. Solution: Let B be x, then A is (x=

1、b。Problem solving ideas: judging from the end, c*e is equal to c, then e can only be 1, and a and d are excluded first; D and e are different numbers, so it can't be c, and the rest of the answer is b.

2. a kilogram; B - 21 kg. Solution ideas:"Basket A also has more kilograms than basket B", then the second basket adds another kilogram, and it weighs as much as A, the total weight is kilograms, and each basket weighs 49 2 = kilograms, and the second basket should now weigh kilograms, and then A = kilograms, and B = 23-2 = 21 kilograms.

is a two-digit decimal place, that is to say, the decimal point is added between the hundred and ten places, that is, it is reduced by 100 times, assuming that the original four digits are 1, the current number is that the current number is smaller than the original four digits; with the corresponding .

4、。Similar to question 3, moving one place to the right is to expand by 10 times, and the number A is equal to the number B after expanding by 10 times, indicating that the number B is 10 times the number of A, and the unit "1" is 1, and the number B is 10, the difference is 9, and 9 is used.

A = kilograms, B = 23-2 = 21 kilograms.

4.A: B:

Speed increased by 1 9

The current speed is 10 9 of the original speed, and the time is 9 10, and the original plan will take hours).

When returning to 280 km, the speed is increased by 1 6.

The speed at 280 km is 7 6 of the original speed, and the time is 6 7.

After 280 km, the original time required is 5 3 (1 6 7) 35 3 (hours).

Distance: 280 (15 35 3) 15 84 15 1260 (km).

A: The distance between Beijing and Shanghai is 1,260 kilometers.

The speed of the vehicle is V, and the distance is S

s/ ……s/v-(5/3)=280/v-(s-280)/[1+(1/6)]v ……

Simplification yields s=15v, and substituting it yields v=84 => s=1260 km.

Let the original speed be V, the time required to travel at the original speed is t, and the distance is S Beijing to Shanghai: s=10 9 V* (t-3 2) Shanghai to Beijing: s=280+7 6 V*(t-5 3-280 V) If you keep driving at the original speed, there is s=vt

The three equations can be solved.

Pay attention to the hidden conditions that are not explicitly given in the mining question).

Column equation solving: Let the planned speed be v and the distance between the two places be s.

s/v-3/2=s/[（1+1/9）v]

280 V+(S-280) [(1+1 6)V]=s V-5 3 gives s as 1260 km.

Set up a class of x people, a second class of y people, 1-1 3-1 4)x+(1-1 4-1 3)y=30;

1/3x+1/4y=11/10*(1/4x+1/3y);

The solution is x=48 and y=24

There were 48 people in the original class.