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. _-_
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
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.
450.
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.
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.
100122 or, if negative numbers are permitted, -998877
There are 900 three-digit numbers.
Using the digits of 1345678, there are 210 three digit numbers in which no digit is repeated.
For a 3 digit number, the left most or the most significant digit cannot be zero. So it can be 1,2,3,4,5,6,7,8 or 9 which is 9 possibilities. The middle number can be 0,1,2,3,4,5,6,7,8,9 which is 10 possibilties but one of the digits has been chosen already as the first digit, so the possibilities are only 9. The right most number can be 0,1,2,3,4,5,6,7,8,9 which is 10 possibilities but two of the digits have been already used by the left most and the middle digits. That leaves only 8 possibilities. So the total number of three digit numbers that have three distinct digits is 9 x 9 x 8 = 81 x 8 = 648 possibilities
27 three digit numbers from the digits 3, 5, 7 including repetitions.
757
24 three digit numbers if repetition of digits is not allowed. 4P3 = 24.If repetition of digits is allowed then we have:For 3 repetitions, 4 three digit numbers.For 2 repetitions, 36 three digit numbers.So we have a total of 64 three digit numbers if repetition of digits is allowed.
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
21
There are only two smaller 3-digit numbers and both of them have repeated digits.
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.
It is possible to create a 3-digit number, without repeated digits so the probability is 1.