There are 2 possible digits for the first digit (3 or 4), leaving 3 possible digits for the second digit (5 and 6 and whichever was not chosen for the first), leaving 2 possible digits for the third. Thus there are 2 × 3 × 2 = 12 possible 3 digit numbers.
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.
There are two possible digits for the first and last digit, and two possible digits for the centre digit, making 2 × 2 = 4 possible 3 digit palindromes from the set {1, 2}, namely the set {111, 121, 212, 222}.
what is the greets possible 9 digit number that uses each of the digits 1-3 times
If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.
5040
Assuming a 3-digit password can begin with a zero, the answer is 720.
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?
There are 2 possible digits for the first digit (3 or 4), leaving 3 possible digits for the second digit (5 and 6 and whichever was not chosen for the first), leaving 2 possible digits for the third. Thus there are 2 × 3 × 2 = 12 possible 3 digit numbers.
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.
There are 30,240 different 5-digit numbers. Math: 10*9*8*7*6 1st digit has 10 possible choices (0-9) 2nd digit has 9 possible choices (one of the digits was used in the 1st digit) 3rd digit has 8 possible choices 4th digit has 7 possible choices 5th digit has 6 possible choices
There are two possible digits for the first and last digit, and two possible digits for the centre digit, making 2 × 2 = 4 possible 3 digit palindromes from the set {1, 2}, namely the set {111, 121, 212, 222}.
There are seven possible digits for the first digit and 6 digits for the second (minus one digit for the digit used as the first digit) and 5 options for the last digit (minus one again for the second digit) and then you just multiply them all together to get a total possible combination of 210 numbers that are possible.
what is the greets possible 9 digit number that uses each of the digits 1-3 times
If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.
Assuming that 2356 is a different number to 2365, then: 1st digit can be one of four digits (2356) For each of these 4 first digits, there are 3 of those digits, plus the zero, meaning 4 possible digits for the 2nd digit For each of those first two digits, there is a choice of 3 digits for the 3rd digit For each of those first 3 digits, there is a choice of 2 digits for the 4tj digit. Thus there are 4 x 4 x 3 x 2 = 96 different possible 4 digit numbers that do not stat with 0 FM the digits 02356.
The answer depends on which digit is 5 and what the other digits are!