To find how many integers from 1 to 1000 have none of their digits repeated, we can analyze the problem by considering the number of digits in the integers. For 1-digit numbers (1-9), there are 9 options. For 2-digit numbers (10-99), the first digit can be chosen in 9 ways (1-9), and the second digit in 9 ways (0-9 excluding the first digit), giving 9 × 9 = 81 combinations. For 3-digit numbers (100-999), the first digit can be chosen in 9 ways, the second in 9 ways, and the third in 8 ways, resulting in 9 × 9 × 8 = 648 combinations. Adding these, we find the total: 9 + 81 + 648 = 738 integers with non-repeating digits from 1 to 1000.
There is only 1, the number 54.
952 of them.
1, 2, 4, 5, 8, and 10, so only 6 digits.
An infinite amount
192 ways
648
There are 120 of them.
How many
52
There are 125 of them.
There are 45 integers between 11 and 999999 which consist of only one digit being repeated. There are 831430 integers that contain at least one repeated digit.
9*9*8*7 = 4536
10*9*8=720
There are 870 such numbers.
There is only 1, the number 54.
952 of them.
1, 2, 4, 5, 8, and 10, so only 6 digits.