1000there aren't even 1000 three-digitnumbers...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. _-_
For there to be palindromes, each digit must be replicated. Therefore there are at most three distinct digits.If there are 3 pairs of different digits, then there are 6 palindromes. If there can be more duplicate digits, then there are 27 palindromes.
There are 900 three-digit numbers.
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
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.
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.
27 three digit numbers from the digits 3, 5, 7 including repetitions.
There are 320 such numbers.