How many three digit numbers have only one 5?

In addition to the fact that it should be 225, there are 72 with a single digit of 5, 72 with a ten-digit of 5, and 81 with a hundred-digit of 5.

225 pcs. There are 72 with a single digit of 5, 72 with a 10-digit of 5, and 81 with a 100-digit of 5.

The sum of the three numbers is 10

From large to small: 910, 901, 820, 811, 802, 730, 721 ,..

The fifth largest number is 802

There are 54 qualifying three-digit numbers, with the fifth largest being 802

Write the first 6 digits of these three digits from largest to smallest, in order, 910, 901, 820, 811, 802, 730, 721...

So, the 5th largest is 802

There are 15 three-digit numbers where the sum of the numbers is equal to 5, and they are:

Thick and tall posture 131

Iwajue 401

Let three be x, y, z because the three-digit number is a multiple of 5 so z is 0 or 5 because x+y+z=10 is the difference, x+y=10 or x+y=5 when x+y=10, there are 9 when x+y=5, there are 5, so there are 14 in total.

Meet the number sum is 10, 910 901 820 802 730

So the fifth largest number is 730

5 5 produces 6 empty positions, assuming that there are no two 3s put together, there are c(3,6)=20, there are 2 repentant 3 put together, there are a(2,6)=30, 3 manuscripts are all put together, there are c(1,6)=6 so there are 20 + 30 + 6 = 56 different octet digits in the honorifice.

Divided into two categories and 5 only appears once, then the other number 8, that is, 8a (3, 3), but the first digit of the three digits can not be 0, and the first digit is 0 There are two cases, so this kind of last 8a and 5 have a number that appears 2 times, with the interpolation method, that is, 2a (3, 1). The final result is 8a(3,3)-2+2a(3,1)=52

35x, 3x5, 5x3, 53x, x take (0-9) each group has 10 possibilities, so there are 40.

x35 and x53 are three-digit numbers, so these two groups of x can't take 0 x take (1-9), and each group has 9 possibilities, so there are 18.

40 + 18 = 58 in total.

There are four cases of such a three-digit number:

1.consists of , x) arrangement, x is not equal to 0, the total number is 42;

2.It is composed of 4 in total;

3.It is composed of 3 in total;

4.It is composed of 3 in total;

So there are 42 + 4 + 3 + 3 = 52.

There are 15 digits with a sum of 5:

There are 15 three-digit numbers where the sum of the numbers is equal to 5, and they are:

5 = 1 + 1 + 3 = 1 + 0 + 4 = 1 + 2 + 2 + 2 + 3 + 0 = 5 + 0 + 01 + 1 + 3 Yes

1+0+4 has +2+2

2+3+0 has +0+0 has: 500

There are 15 of these three-digit numbers.

From 135 to 935, from 153 to 953, from 305 to 395, from 350 to 359, from 530 to 539, from 503 to 593 and subtract the duplicates, there are 54 in total.

The original problem is to find out how many permutations of length 3 contain .

Any two digits are , and the remaining one digit is ten numbers from 0 to 9. Then there is a total of 3*2*10=60Since the requirement is a three-digit number, then the need to start with 0 needs to be eliminated, there are two cases: 035 and 053, so it can be determined that there are a total of 60-2=58 three-digit numbers that meet the question.

100 of 199-26.

200-299 out of 26.

300-399 has 100.

400-499 has 26.

500-599 has 100 pcs. In common.

: There are 10 in total.

5: There are 9 in total, deducting 355

3. 35: There are 8 in total, deducting 335,035

There are 9 in total, deducting 535

3: There are 9 in total, deducting 533

6. 53: There are 7 in total, and after deducting 353,053,553, there are 10 + 9 + 8 + 9 + 9 + 7 = 52.

The correct solution is as follows: 3*5: There are 10.

5*3: There are 10 of them.

35*: There are 10-1=9.

53*: There are 10-1=9.

35: There are 9-2=7.

53: There are 9-2=7.

52 in total.

It is better to use the permutation combination formula, the high position cannot be 0, first distinguish the permutation containing 0, and then the permutation that does not contain 0 at all. If it contains 0, it can only appear in the ten or single digits, there are 2 ways, and the others can only be 3 and 5 (all 3 or all 5 are wrong), that is, the arrangement with 3 and 5, with the formula a(2, 2) = 2, 2 * 2 = 4, and the effective arrangement with 0 is 4.

If it does not contain 0, and it must contain 3,5, there are two cases, the first is 3,5 is not repeated, that is, 3 occupies a digit, there are 3 methods, 5 occupies a digit, there are 2 methods, the third number is divided from 1-9 3,5 is selected, there are 7 methods, 3 * 2 * 7 = 42 methods. The second is 3,5 repeats. It is to let 3 occupy two digits, there are 3 ways, the remaining digits can only choose 5, let 5 occupy two digits, there are also 3 methods, and the remaining digits can only choose 3, in this second case:

3+3=6 types.

Together, there are: 4 + 42 + 6 = 52 methods. If you use other methods, you can do it.