The numbers 0, 1, 2, 3, and 4 can make up a five digit number without repeating numbers

If the five digits are not repeated, there are: 4 4 3 2 1 = 96.

If the five digits can be repeated, there are: 4 5 5 5 5 = 2500.

1. Divisible by 4, the mantissa can only be these six kinds, and these three mantissa arrange the remaining three numbers, each with 6.

When making mantissa numbers, 0 cannot be the first position, so there are 4 of each.

So there are 30 of them.

2. There are three types larger than 21034, the first one.

The first place is 2, the single digit is 4, there are these 3, the first place is 2, and the single digit is 0 is larger than 21034, there are 6; 9 in total.

There are 6 in the first position of 3 and 0 in the single digit, and in the same way, there are 6 in the 4 digit and 6 in the 2 digit; A total of 18.

There are 6 in the first place and 6 in the single digit 0, and in the same way, there are 6 in the single digit 2, for a total of 12.

The final result is 9 18 12 = 39.

3. According to the meaning of the title, it should be even odd and even, the first digit can choose 2, the second digit can choose 2, the third digit can choose 2 except the first digit, the fourth digit can choose 1, and the last one is left at the end, and the result is 2 * 2 * 2 * 1 * 1 = 8.

The number of unduplicated digits consisting of the numbers 0, 1, 2, 3, 4 is"The number of 5" permutations: 5x4x3x2x1=120 (pcs).

The number that starts with 0 is a four-digit number, and the rest is a five-digit number. The number of 4-digit arranges starting with 0 is "4 takes 4": 4x3x2x1=24 (pcs).

120--24=96 (pcs).

Therefore, the five-digit number without repeating digits should be 96 (pcs).

The numbers 0, 1, 2, 3, and 4 can be made up of 96 five-digit numbers without repeating numbers.

p(5,5)-p(4,4)=120-24=96.

First arrange by 5 numbers, a total of p(5,5)=5 4 3 2 1=120.

Subtract the p(4,4) at the top of the 0 row.

You can make up five digits without repeating numbers: 5! -4！= 96 pcs.

Let's assume that 0 can be put in the first place, there are 5x4x3x2x1 = 120 kinds and 0 can no longer be the first place, remove the possibility of 0 in the first place, a total of 4x3x2x1 = 24 kinds 120-24 = 96 kinds.

0 cannot be the first digit, so the number of five digits without repeating digits is.

c(4,1)*p(4,4)=96

Let's assume that 0 can be put in the first place, there are a total of 5x4x3x2x1 = 120 kinds and 0 can not be the first place, remove the sail type possibility of 0 in the first place, and guess the coarse 4x3x2x1 = 24 kinds 120-24 = 96 kinds.

600 pcs. This five-digit number can be expressed as:

The first digit can have 5 situations (0 cannot be taken) to call the buried liquid fiber;

The second digit can have 5 cases (0-5 six numbers, the first digit takes 1), the third digit can have 4 cases (the first two digits take 2), the fourth digit and limb position can have 3 cases (the first three digits take 3), and the fifth digit can have 2 cases (the first four digits take 4), so 5*5*4*3*2=600.

96 (pcs) Answer: Makes up 96 five-digit numbers without repeating numbers

There are 2 4 = 16.

They are 10, 12, 13, 14

If you don't understand, please ask about this state, if it helps, please Cai Sen Fengna, thank you for closing Sun Xie!

= 96 pcs.

A: With 0, 1, 2, 3, 4, these five numbers can make up 96 four-digit numbers without repeating numbers

The arrangement of 3 out of 5 is missing.

is a53=5!/(5-3)!=60

100 Hornab can be 0

And the hundred digit is 0, there is a42=4!/(4-2)!=12 so there are 60-12 = 48.

Because it is a three-digit number, the hundred digit cannot be 0, so there are 5 options for the hundred digit number: 1 2 3 4 5

Because it is a three-digit number without repeating, the ten-digit number cannot be the same as the hundred-digit, but the ten-digit number can be 0, so the ten-digit Qi Annihilation key also has 5 options.

Then there are only 4 single-digit options.

So at the end there is 5*5*4=100 three-digit highs.

Step 1: Set the hundredth digit: a(4,1) (you can't use 0 in the first book).

Step 2: Set ten digits and single digits: a(4,2).

a(4,1)*a(4,2)=4*12=48 to form 48 non-repeated trimatches.