Want this question answered?
If repeated digits are allowed, then the largest 12-digit number is 999,999,999,999 .If repeated digits are not allowed, then the largest 12-digit number is 989,898,989,898 .If the same digit can't be used more than once, then the largest possible number has only10 digits. The number is 9,876,543,210 .
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
5040
The first digit can be 0 through 9, ten possibilities. Having selected the first digit, you have 9 digits to pick from the second digit. Having selected the second digit, you have 8 digits to pick from the third digit. Hence total possibilities = 10 x 9 x 8 = 720
Assuming a 3-digit password can begin with a zero, the answer is 720.
If repeated digits are allowed, then the largest 12-digit number is 999,999,999,999 .If repeated digits are not allowed, then the largest 12-digit number is 989,898,989,898 .If the same digit can't be used more than once, then the largest possible number has only10 digits. The number is 9,876,543,210 .
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 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.
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
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.
If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.