999.
Think about it 001 002 003.....etc.
It just goes in order up to the 999
plus 000 makes 1,000
To calculate an ordered sequence (a permutation), you take the number of options for the first choice, times the number of options for the second choice, times the number of options for the third choice (etc).
In this case, you have 10 options for each digit (0 thru 9), so: 10 x 10 x10 = 1,000
Oh, what a happy little question! If we have 10 digits to choose from (0 through 9) and we can't repeat any, then for the first digit we have 10 options, for the second digit we have 9 options left, and for the third digit we have 8 options remaining. So, the total number of possible three-digit passwords would be 10 x 9 x 8 = 720. Isn't that just delightful?
If the digits can be used more than once, then 900. If not, then 648.
1 set
5 x 10 x 5 = 250 different numbers, assuming there is no limit to each digits' use.
99997
Oh, what a happy little question! If we have 10 digits to choose from (0 through 9) and we can't repeat any, then for the first digit we have 10 options, for the second digit we have 9 options left, and for the third digit we have 8 options remaining. So, the total number of possible three-digit passwords would be 10 x 9 x 8 = 720. Isn't that just delightful?
The number of 4 digit codes, using different digits, is 10*9*8*7 = 5040. However, computer passwords can, usually, have repeated digits and, if that is allowed, you can have 104 = 10000 codes.
5040
10,000
36
5040 different 4 digit numbers can be formed with the digits 123456789. This is assuming that no digits are repeated with each combination.
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
If the digits can be used more than once, then 900. If not, then 648.
500
24
27 three digit numbers from the digits 3, 5, 7 including repetitions.
1339