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Add "+" or "- before the numbers 2, 3, 4, 5, 6, 7, 8, 9 to make the sum of them be - 10. How can you think of several ways?
First of all, choose two numbers so that their sum is - 10, such as - 2 and - 8, - 3 and - 7, - 4 and - 6, and then select the appropriate "+" and "-" signs so that the sum of the remaining numbers is 0. The following answers are for reference: 1, + 2 + 3-4 + 5-6 + 7-8-92, - 2-3-4-5-6-7 + 8 + 93, + 2-3 + 4 + 5 + 6-7-8-94
Add "+" or "-" before the numbers 2, 3, 4, 5, 6, 7, so that their sum is 5. Can you think of several ways?
2+3-4+5+6-7=5
2+3+4-5-6+7=5
Fill in the nine numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 into the box to make the equation true. Can you think of several different ways to fill in? □□□=1 2×□□□=1 3×□□□
According to the stem analysis, we can get the following conclusions
327=1
2×654=1
3×981；
Or 273 = 1
2×546=1
3819．
Fill in the nine numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 into the box to make the equation true. Can you think of several different ways to fill in? □□□=1 2×□□□=1 3×□□□
According to the stem analysis, we can get the following conclusions
327=1
2×654=1
3×981；
Or 273 = 1
2×546=1
3819．
For permutation and combination, take any three of the five numbers 2, 3, 4, 7, 9 to form a three digit number without repeating numbers. Question: what is the sum of these three digits? I can't think of it!
Take three numbers out of five numbers to make up the three digits of unrepeated numbers. A is a hundred. B is ten. C is a single digit. Out of the 60 digits of C (3) (5) = 60, 60 / 5 = 12. Similarly, the other numbers are also 12. There are 60 / 5-12 three digits of 3 taken from B. the sum of these three digits is 12
Permutation and combination: use 0,1,2,3,4,5 to make up 6 digits without repeating numbers. How many can be divisible by 25?
The last two are 25 or 50
If 25
Then the first four are arranged as 4! = 24
But the first place can't be 0
The first zero is 3! = 6
So there are 24-6 = 18
If 50
Then the top four are 4! = 24
So 18 + 24 = 42
Permutation and combination: 1,2,3,4,5,6 several numbers make up six digits without repetition Six digits without repetition are composed of 1,2,3,4,5,6. Even and odd numbers are open, and 1,2 must be adjacent,
One and two make up a group, five groups in total
Three of them are odd books, and two are even numbers
In other words, there are two situations in the five positions: odd Book Beginning and even number beginning
At the beginning of the strange book, it is 3 × 2 × 2 × 1 × 1 = 12
Even starts with 2 × 3 × 1 × 2 × 1 = 12
12 + 12 = 24
Permutation and combination topic: take any three numbers from 1, 3, 5, 7, 8, and then take two numbers from 2, 4, 6, 8, and ask how many 5 digits can be combined without repeating numbers Is there an indirect method
1. The first group takes 8. Then take two from 1,3,5,7. C4 takes 2 = 6; the second group takes 2 from 2,4,6 (ensure no repetition, cannot take 8). C3 takes 2 = 3
2. In the first group, 8.c4 is taken as 3 = 4; in the second group, 8.c3 is taken as 1 = 3
3. In the first group, if 8 is not taken, C4 is taken as 3 = 4; in the second group, if 8 is not taken, C3 is taken as 2 = 3
In the first three cases, the numbers taken in each of the first three cases are not the same, and any five different arrays are divided into five digits, 5 * 4 * 3 * 2 = 120
I'm sorry, I didn't think of an indirect way
Permutation and combination: use 0,1,2,3,4,5 to form a number without repeating numbers, and how many six digits can be divided by 5 How many numbers larger than 240135 can be made?
1. The last digit is 0 a (5,5) = 120
2. The last digit is 5, and the first digit cannot be 0 a (4,1) * a (4,4) = 96
120+96=216
It can form 216 six digits divisible by 5
The problem of permutation and combination: how many different permutations and combinations (repetition allowed) of the ten numbers 0-9? Please write down the detailed formula and ideas I'm poor at math. I don't know whether it's a permutation or a combination There are ten numbers of 0-9: 0,1,2,3,4,5,6,7,8,9 How many combinations can these ten numbers make? Note: the combination of one or several numbers is also considered as a combination. It is not necessary to put ten numbers together each time, and the selected numbers can be repeated For example, a single 0 is a combination; for example, 00 is a combination; for example, 000 is also a combination; for example, 0000 is also a combination Note: different orders can not be considered as the same combination, such as 01 and 10, which are two combinations, and 001101100010 are four combinations Here are some examples of combinations: 100001, 448000841599999999 I don't know if you can understand the question. If the description is not clear, I will add that I haven't learned mathematics and can only describe the problem in this way. Please tell me whether this is an arrangement or combination problem? Please write out the detailed solution formula and ideas,
I see. This is arrangement
The difference between arrangement and combination lies in whether there is order or not
How many different permutations and combinations of the ten numbers 0-9 are allowed to repeat?
How many places do you want to rank? If not, the answer can be long
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