Choose any three of the numbers 0134 to form a few even numbers with no repeats?

It must be even, so one bit must have three possibilities and six. You can use targeting methods. When one digit is 2 and a hundred digits are 1, four numbers can be formed; The hundredth digit is 3, which can form four numbers; The hundredth digit is 4, which can form four numbers; The hundredth digit is 5, which can form 4 numbers; The hundredth digit is 6, which can form 4 numbers; It was concluded that when the number of bits is 2, 20 three-digit even numbers can be formed by analogy, when the bits are 4, it can also form 20 three-digit even numbers, and when the bits are 6, it can form twenty three-digit even numbers, so there is such an even number.

The requirement is even, so there must be three possibilities for a single bit and 6. You can use targeting methods. When one digit is 2 and the hundred digit is 1, four numbers can be formed; The hundredth digit is 3, which can form 4 numbers; The hundredth digit is 4, which can form 4 numbers; The hundredth digit is 5, which can form 4 numbers; The hundredth digit is 6, which can form 4 numbers; The conclusion is that when the number of bits is 2, it can be analogous to form 20 three-digit even numbers, and when the number of bits is 4, it can also form 20 three-digit even numbers; When the bit is 6, it can form 20 three-digit even numbers, so there are 20 + 20 + 20 = 60 such even numbers.

The number of pairs is required, so a bit must have three possibilities and 6. When one digit is 2 and a hundred digits are 1, you can create four numbers; One hundredth is 3, which can form 4 numbers; One hundredth is 4, which can form 4 numbers; One hundredth is 5, which can form 4 numbers; One hundredth is 6, which can form 4 numbers; We found that when the number of bits is 2, 20 three-digit even numbers can be created. Similarly, when the bit is 4, it can also create a 20-bit triple even number, and when the bit is 6 bits, it can create a 20-bit triple even number.

So, there are 20-20-60 such even numbers.

The requirements are even, so each one has three possibilities: 2, 4, 6. The positional method can be used when two:

100 is one, which can be four; a hundred to three, which can be made up of four; A hundred is four, which can be four; One hundred is five, it can be four, and one hundred is six, it can be four. This means that 20 even numbers of three digits are produced in binary numbers. It can also be said that this number is 4:

00 can be an even number of 20 three-digit numbers, 6 digits, and 20 three-digit numbers, so there are 20 + 20 + 20 = 60 such even numbers.

It must be an equal number, so a single bit must have three options and 6. When one digit is 2 and a hundred digits are 1, four numbers can be formed; 100 is 3 and can form 4 numbers; 100 is 4 and can form 4 numbers; 100 is 5 and can form 4 numbers; 100 is 6 and can form 4 numbers; It was concluded that when the number of digits is, a three-digit even number can be simulated to form, when the number of digits is 4, it can also form 20 three-digit even numbers, and when the number of digits is 6, it can form twenty three-digit even numbers, so there are 20 + 20 + 20 = 60 such even numbers.

The requirement is even, so there must be three possibilities for bits. The positioning method can be used, when the bit is 2:100 is 1, it can form 4 numbers; 100 is 3 and can be made up of 4 numbers; A hundred is four, which can make up four numbers; 100 is 5 and can form 4 numbers, and 100 is 6 and can form 4 numbers; So, when the bit is 2, it can form 20 three-digit even numbers.

By analogy, if this bit is 4, it can also be made up of 20 three-digit even numbers, and if this bit is 6, it can be made up of 20 three-digit even numbers, so there are 20+20+20=60 of these even numbers.

There are 12 even numbersSolution: As long as the single digit number is even, the three-digit number is even.

The analysis steps are as follows:

The single digit of an even number can only be 2 and 4, a total of 2 possible choices.

There are 3 possible options for even tens (because an even number has been selected for the single digit, and the remaining 3 digits are filled in the tens digits).

There are 2 possible options for even hundreds, because 2 digits have been selected for the 10th digit, and the remaining 2 digits are filled in the hundreds).

Therefore, the final count is an even number = 2x3x2 = 12 (counts).

There are 4 3 2 = 24 three-digit numbers that make up unrepeating numbers, of which there are 2 3 2 = 12 even numbers.

There are 12 types in total.

Step 1: Determine the single digit, since it is an even number, so the single digit can only be 2 or 4.

The second step is to determine the other digits, and on the other digits, you can choose two of the remaining three numbers, and there are 3 2 = 6 choices.

So there are 2 6 = 12 species.

If the last digit is 2, pick two of the other three and sort it a32, and then suppose the last digit is 4, and then choose two of the other three digits and sort it a32, so the result is 2a32=12. Hope.

The three-digit even numbers are: 4 3 2 2 = 12 kinds.

Permutations and combinations, choose one of the first three positions, choose one of the two for the second place, and choose one of the three places for one of the two.

The last digit makes one of 2.

There are 156 four-digit even numbers that make up 012345 distinct numbers.

1. Even numbers when the single digit is 0, there are 5 ways to choose the ten digits, 4 ways to choose the hundreds, and 3 ways to choose the thousands.

Piece; 2. When the single digit is 2 or 4, there are 4 winning methods for thousands of digits, 4 re-selection methods for hundreds, and 3 winning methods for tens of digits.

2 4 4 3 = 96, and there are 96 even numbers when the single digit is 2 or 4;

The total can be composed.

60 + 96 = 156 even numbers.

The whole three-digit number composed of 01234 is: c(4,1)*c(4,1)*c(3,1)=4*4*3=48.

Note that 0 cannot be done in hundreds.

The odd number of all the three suspicious Hu bits composed of 01234 is: c(2,1)*c(3,1)*c(3,1)=2*3*3=18.

Therefore, 01234 is composed of 3-digit even numbers that are not repeated by Li: 48-18 = 30.

Think of it this way, the first stupid of the three positive grades has three choices (the first position cannot be 0), the second place still has three (because it is not repeated), and the third place can only have two. The positive solution is 3*3*2=18, and if it is repeatable, it should be 3*4*4=48

There are a total of 4 even numbers (0, 2, 4, 6), and only 2 or 4 or 6 can be selected when starting with 0, and there are 3 kinds of hail and dry choices when the source of Tongdong starts with 2 or round staring 4 or 6, so a total of 3 4 = 12 kinds of three-digit even numbers without repeating numbers can be formed.

First of all, this number is an even number, then the last digit is 0 or 2, when the last digit is 0, the hundred digit is the Ming Sleepy Cong, there are 3 cases, and the ten-foot orange bit can only be repeated because there are only 2 cases of the cherry blossoms, so it is 3x2 = 6 kinds.

When the last digit is 2, the hundred digits are 2 cases (0 cannot be the first place), and the 10 digits are also 2 cases, so it is 2x2 = 4 kinds.

That adds up to 10 cases.

Among the numbers that can be formed by these four numbers, there are even numbers of 120, 132, 210, 310, 102, 302, and 312, a total of 7.

Even numbers, the last digit may be 0,2,4

When the last digit of Danzhou is 0, there are 4 types of first rubber, 3 types of second place, and 2 types of third place.

So oil 4 * 3 * 2 = 24 kinds.

When the last digit is 2 and 4, the first Liang Chixiao has 3 kinds, the second place has 3 kinds, and the third place has 2 kinds.

So oil 2 * 3 * 3 * 2 = 36 kinds.

It adds up to 24 + 36 = 60 species.