0 1 2 3 4 5 choose 3 numbers... You can t use it once. How many can be odd?

Note: It can be found using simple multiplication rules.

There are 3 kinds of single digits, 4 kinds of 100 digits when the 10 digit is 0, and 3 kinds of 100 digits when the 10 digit is non-0 (the 100 digit cannot be equal to 0).

That is: 3c1*(4c1+4c1*3c1)=48

3c1 times 4c1 times 4a2 = 3 times 4 times 6 = 72

If you don't go to high school 2, you don't need to study this kind of problem......

021 031 041 051 013 023 043 053 015 025 035 045 0 The first 12.

103 123 143 153 105 125 135 145 1 First 8.

201 231 241 251 203 213 243 253 205 215 235 245 2 12.

301 321 341 351 305 315 325 345 3 First 8.

401 421 431 451 403 413 423 453 405 415 425 435 4 12

501 521 531 541 503 513 523 543 5.

So a total of 12 + 8 + 12 + 8 + 12 + 8 = 60.

(1) Permutation and combination problem: 5x5x4x3x2=600 five-digit odd number, that is, the last number is 135, the first term is not zero, and the same is according to the previous method, that is, the last item is 024, and the first term is not zero, and also according to the previous method, I hope it will help you,

1 2 3 4 5 can form an unrepeating 4-digit number:

a54=5×4×3×2=120

Even numbers mean that the single digit is an even number, that is, 2 or 4 is in the single digit, and the other 3 digits are selected in the remaining 4, with ranking.

c21×a43=2×4×3×2=48

Can make 5 4 3 2 = 120 distinct four digits.

The even number of digits can only be 2 or 4, so there are 4 3 2 1 = 24 that meet the conditions.

The number of 12345 numbers that can form distinct numbers is 5x4x3x2 120

The even number is 2x4x3x2 48

With five numbers, there are 5 choices for the first place, 4 choices for the second place, 3 choices for the third place, and 2 choices for the fourth place, for a total of 5 4 3 2 = 120 choices.

Among them, the even numbers are: the fourth place has 2 kinds of choice 4 or 2, the third place has 4 kinds, the second digit has 3 kinds, the first place has 2 kinds, and the even number that makes up the even number is: 2 4 3 2 = 48 kinds.

A total of 6 + 4 + 2 = 12 (pretending to infiltrate a Wu call) cavity is lacking.

The number divisible by 25 must be a multiple of 2*3*5=30, so the integer ends with 0, and the sum of the vertical collisions of the three-digit segment is a multiple of 3.

So the hundred digits and the ten digits can be a combination of 1 and 2 or a combination of 1 and 5, so move 2*2 = 4 kinds, respectively: 120, 210, 150 and 510

5 4 + 4 4 = 36 (pcs).

Side, ode to the manuscript

Wild filial piety

You can arrange 3 digits in single digit and on the top of the list.

The 100 digits can be called to arrange 4 numbers.

There can be 4 numbers in the 10th place.

According to the principle of multiplication, 3*4*4 = 48 punches.

Because the 10,000-digit number cannot be 0, it can only be selected from the five numbers of 1 5, and there are 5 methods; Then the other four digits can choose four of the remaining five numbers to permutate, yes.

a（5，4）=5×4×3×2=120

According to the principle of multiplication, it is possible to form a five-digit number 5 120 = 600 without repeating numbers.

An odd number of them:

There are 3 ways to take a single digit from 1, 3, 5; There are 4 ways to choose a non-zero number other than a number for 10,000 digits; The remaining three positions are arranged by selecting 3 of the remaining 4 numbers, and there is.

a（4，3）=4×3×2=24

According to the principle of multiplication.

3 4 24 = 288 odd numbers, 600-288 = 312 even numbers.

If it's not clear, keep asking!

1) Permutation and combination problem: 5x5x4x3x2=600 five digits.

Five odd digits, i.e. the last number is 135, and the first term is not zero.

Also follow the previous method.

Five even digits, i.e. the last item is 024, and the first item is not zero.

Also follow the previous method.

Hope it helps,

24 species; The single digit of the even number is an even number, so the source friend is classified first, divided into two categories, the first Ming Qing class is a digit of 2, the hundred and ten digits are arranged from the other 4 numbers, A4 (2) = 43 = 12;

The second type is a single digit of 4, and the hundred and ten digits are also arranged from two of the other four digits, a4(2)=43=12

Then the principle of addition is 12 + 12 = 24 species.

1-5 take any three numbers, these three numbers are even, how many ways to choose them?

2 4 3 = 24 (species).

Five consecutive even numbers are set to 2n, 2n n n n+8, according to the condition: 2n+4-(2n+2n+8) 4=18, solve: n=16, and then find the result of each formula from the equal difference series:

2n+2n+8）/2*5=10*16+20=180