No repeating!

Let's list them out (the "-" sign used below is inclusive of both integers)

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From 1 to 99

0-9 has 9 (0 is not positive)

10-19 has 9 (not 11)

... Each section of ten numbers has 9 possibilities, which gives us 9*9=81 possibilities between 1 and 99 inclusive.

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From 100 to 999

100-109 has 9 (not 101)

110-119 has 0 (all have duplicates of "1")

120-129 has 8 (not 121, 122)

130-139 has 8 (not 131, 133)

... So each range of a hundred numbers will have 9+9*8=73 possibilities. We have 9 ranges of hundreds (100, 200, 300, ... 800, 900).

Thus the total number of possibilities is 81 + 9*73 = 738

Let us divide this into 3 cases: 1 digit, 2 digit, and 3 digit numbers. The number of ways for 3 digits is 9x9x8=648, while for 2 digit is 9x9=81, and 1 digit is just 9, giving a total of 738

There are 3 types of integers that are less than 1000.

Integers with 1 digit

Integers with 2 digits

Integers with 3 digits

The answer to this question = Number of 1-digit integers without repeating digits + Number of 2-digit integers without repeating digits + Number of 3-digit integers without repeating digits

The number of 1-digit integers without repeating digits is obviously equal to 9.

The number of 2-digit integers without repeating digits is equal to 9*9 = 81

This is because there are 9 possible digits that could occupy the "tens" place of the number. They are {1, 2, 3, 4, 5, 6, 7, 8, 9}. Having selected 1 of these digits, we are still left with 9 numbers to choose from for the units digit. This is because of the fact that "0" cannot occupy the "tens" place.

Following the same logic, the number of 3-digit integers without repeating digits is equal to 9 9 8 = 648.

Hence, the answer is 9+81+648 = 738.

1-digit = 9 integers

2-digits = 9x8+9x1 = 81 integers 3-digit= 9x8x7+9x1x8+9x8x1= 648 integers

9+81+648 = 738

Number of integers from 1 through 999 with no repeated edges =

Number of integers from 1 with no repeated digits + number of digits from 10 through 99 with no repeated digits + Number of integers from 100 through 999 with no repeated digits =

$9 + 9 \times 9 + 9 \times 9 \times 8 = 9 + 81 + 648 = 738$

Thus the answer is 738

How many number is $=$ Number of integers from 1 with no repeated digits $+$ number of digits from 10 through 99 with no repeated digits $+$ Number of integers from 100 through 999 (1000 is not included cause less than ) with no repeated digits $=$

9 + 9 . 9 + 9 . 9 . 8 = 9 + 81 + 648 = $738$ .

Calculate all solutions for three-digit numbers: there are 9 (all digits except 0) possibilities for the first digit, 9 (all digits except the one already used) possibilities for the second digit, and 8 (all digits except the two already used) possibilities.

Calculate all the solutions for two digits and one digit by following the first two and then one steps above, respectively.

9 9 8 + 9*9 + 9 = 738.

Números menores que 1000 , inteiros positivos --> 1,2,3 ... 999 Números de 3 algarismos distintos --> 9(1 ao 9).9(0 ao 8).8(0 ao 7) =648 Números de 2 algarismos distintos --> 9(1 ao 9) .9(0 ao 8) =81 Números de 1 dígito --> 1 ,2,3 --> 9 Números Somando : 648+81+9=738

Number of integers from 1 through 999 with no repeated edges = Number of integers from 1 with no repeated digits + number of digits from 10 through 99 with no repeated digits + Number of integers from 100 through 999 with no repeated digits. = 9 + 9 . 9 + 9 . 9 . 8 = 9 + 81 + 648 = 738

Number of integers less than 1000 with no repeated edges =

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Number of integers from 1 with no repeated digits + number of digits from 10 through 99 with no repeated digits + Number of integers from 100 through 999 with no repeated digits.

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= 9 + 9 . 9 + 9 . 9 . 8 = 9 + 81 + 648 = 738