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Take any two of the five numbers 1, 3, 5, 7, 9, and any two from the four numbers 0, 2, 4, 6 Take any two of the five numbers 1, 3, 5, 7, 9, and any two from the four numbers 0, 2, 4, 6 ask 1. How many four digits can be formed without repeating numbers? 2. How many even four digit numbers without repeating numbers can be formed?
You can ask if you don't understand
Take any two of the five numbers 1, 3, 5, 7, 9, and any two from the four numbers 0, 2, 4, 6 Take any two of the five numbers 1, 3, 5, 7, 9, and any two from the four numbers 0, 2, 4, 6 ask 1. How many four digits can be formed without repeating numbers? 2. How many even four digit numbers without repeating numbers can be formed?
One
This question is divided into two parts
When 0: C (5,2) * C (3,2) * P (4,4) = 720
When 0: C (3,1) * C (5,2) * 3 * P (3,3) = 540
So there are 1260 species
Two
If you don't get 0: if you want an even number, it's half of the above part = 360
When 0 is taken and 0 is in bits, C (3,1) C (5,2) P (3,3) = 180
When 0 is taken and 2,4,6 are in bits, C (3,1) C (5,2) C (2,1) P (2,2) = 120
So there are 300 species
How many possibilities are there for four digit permutation?
There are only nine possibilities for the number in the thousand, and there are ten possibilities for the number of other digits. The possible permutation and combination is 9 * 10 * 10 * 10 = 9000
There are several ways to arrange and combine six numbers
If the six numbers are different from each other, there are 720 permutations with a (6,6) = 720
But there are three groups of two identical, so we need to divide by a (2,2) a (2,2) a (2,2) = 8
So in the end, there are 720 △ 8 = 90 permutations
[hope it can help you]
What are the total numbers of 0-9 permutations per six digits?
If six numbers can be repeated
P9,1*P10,1*P10,1*P10,1*P10,1*P10,1=900000
If six numbers can't be repeated
P9,1*P9,5=136080
Remember to give the score
Permutation and combination problem: 123456 this is the original number I gave. There can be 6 × 5 × 4 × 3 × 2 × 1 = 720 permutations and combinations Now I need to remove the combination of Arabic numerals repeated on the same digit as the original number 123456. How many permutations and combinations are left For example: one hundred and twenty-three thousand four hundred and fifty-six one hundred and thirty-two thousand six hundred and forty-five 1 repeats on the same digit Or: one hundred and twenty-three thousand four hundred and fifty-six three hundred and twenty-one thousand six hundred and fifty-four 2 and 5 repeat on the same digit, which are excluded The permutation and combination of numbers can have duplicate numbers in digits, only can't duplicate the original number in digits Please give the answer. It's better to have a formula If you can't understand, I can also come up with a problem. A bookcase full of books, among which 6 books can be exchanged. How many ways do you exchange them? Note: each book must move its position, not one or several books are not allowed to move!
Recurrence formula: an = (n-1) (an-1 + an-1)
A1=0 A2=1 A3=2 A4=9
A5=44
A6 = 53 * 5 = 165 ------ answer
A six digit combination of 0-9 numbers? It can be repeated
It depends on whether your number can be repeated. If it can be: 9c1 * 10C1 * 10C1 * 10C1 * 10C1 = 900000; if not, it is: 9c1 * 9c1 * 8c1 * 7c1 * 6c1 * 5c1 = 136080
All permutations and combinations of 4 out of 6 numbers There are six numbers from 1 to 6, and four numbers are taken at a time. There are several combinations. Please help to list all the combinations
15 combinations:
one thousand two hundred and thirty-four
one thousand two hundred and thirty-five
one thousand two hundred and thirty-six
one thousand two hundred and forty-five
one thousand two hundred and forty-six
one thousand two hundred and fifty-six
one thousand three hundred and forty-five
one thousand three hundred and forty-six
one thousand three hundred and fifty-six
one thousand four hundred and fifty-six
two thousand three hundred and forty-five
two thousand three hundred and forty-six
two thousand three hundred and fifty-six
two thousand four hundred and fifty-six
three thousand four hundred and fifty-six
(n!)/(r!)((n-r)!)
This program is used to calculate the combination, n = 6, r = 4
The six numbers from 0 to 5 make up an even four digit number without repeating numbers, and this even number The six numbers from 0 to 5 can be used to form four digit even numbers without repeating numbers, and the ten digits and hundred digits of this even number are odd numbers! How many of them satisfy the condition?
Forty-two
First of all, there are 1353 odd numbers selected from the six kinds of numbers
Then put them on the tens and hundreds, that is, a (3.2) = 6 kinds
Then arrange the thousand. The thousand can't be 0, so it may be 2, 4 or the odd number left. That is, choose one from three and choose three from three
Then arrange the thousand and the ones
There are two cases,
One kind: the remaining odd number is in the thousand, one bit can be 0.24, all right. There are three kinds
There are two kinds: 2 or 4 in thousand bits, each bit can only be 0 or the remaining 2 (4), a total of 2 * 2 = 4 kinds
The results were as follows: 6 * (3 + 4) = 42 species
Use 0, 1, 2, 3, 4 to make up five even numbers without repeating numbers______ One
When the digit number is 0, the total number of such five digits is: a44 = 24,
When the digit number is 2 or 4, the total number of such five digits is 2 × c31a33 = 36,
Therefore, there are 24 + 36 = 60 five even numbers
So the answer is 60
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