0
1000
there aren't even 1000 three-digit
numbers...
There are the digits 1 through 9 for the first digit. Then, we have 0 through 9 for the second digit - excluding the first digit. For the third digit, we have 0 through 9, excluding the two previous digits for a total of 9*9*8 =
81*8 =
648 three digit numbers with distinct digits.
EDIT- ... to be a little more specific, if your talking about a 3-digit password or unlock code its would be 1000. _-_
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How many three digits are there in which all the digit are distinct?
There are the digits 1 through 9 for the first digit. Then, we have 0 through 9 for the second digit - excluding the first digit. For the third digit, we have 0 through 9 excluding the two previous digits
How many even 3-digit numbers can be formed from the digits 1 2 3 and 5 if each digit can be used only once?
The final digit must be a 2 to form an even number. The first digit may be any of three remaining digits (1, 3, 5) while the second digit may be any of the two remaining digits. All together, that makes 3*2 = 6 distinct even numbers.
How many three digit numbers are there such whose digits sum is odd?
450.
How many three even digit numbers by using 12457 if each digit is used once?
None, because you do not have three even digits. None, because you do not have three even digits. None, because you do not have three even digits. None, because you do not have three even digits.
How many odd numbers with 4 distinct digits can be formed using the digits in 234567?
The last digit must be odd: three possibilities. The other digits can be anything else, as long as they're different: five times four times three possibilities. Thus, the answer is 3 x 5 x 4 x 3 = 180.
