5,040
Assuming that the combination uses ALL single digits from 0 to 9, then for the first digit you will be able to use all 10 numbers, for the second digit you will be able to use 9, for the third digit 8 and for the last digit 7, giving a total number of combinations of 10 x 9 x 8 x 7 = 5,040 without the same number being used more than once in each combination.
There are twelve possible solutions using the rule you stated.
A 3-digit safe code can have combinations ranging from 000 to 999. This gives a total of 1,000 possible combinations, as each digit can be any number from 0 to 9. Therefore, the total number of combinations is 10 (choices for the first digit) × 10 (choices for the second digit) × 10 (choices for the third digit), which equals 1,000.
The number of four-digit combinations is 10,000 .Stick a '3' before each of them, and you have all the possible 5-digit combinations that start with 3.There are 10,000 of them. They run from 30,000 to 39,999 .
Assuming that repeated numbers are allowed, the number of possible combinations is given by 40 * 40 * 40 = 64000.If repeated numbers are not allowed, the number of possible combinations is given by 40 * 39 * 38 = 59280.
987654321
There are twelve possible solutions using the rule you stated.
A 3-digit safe code can have combinations ranging from 000 to 999. This gives a total of 1,000 possible combinations, as each digit can be any number from 0 to 9. Therefore, the total number of combinations is 10 (choices for the first digit) × 10 (choices for the second digit) × 10 (choices for the third digit), which equals 1,000.
The number of four-digit combinations is 10,000 .Stick a '3' before each of them, and you have all the possible 5-digit combinations that start with 3.There are 10,000 of them. They run from 30,000 to 39,999 .
Assuming that repeated numbers are allowed, the number of possible combinations is given by 40 * 40 * 40 = 64000.If repeated numbers are not allowed, the number of possible combinations is given by 40 * 39 * 38 = 59280.
The number of possible 3 digit combinations you can make out of 1-9 with outrepeated digits is:9C3 = 9!/(3!(9-3)!) = 84
987654321
10234567
There are 1,000,000 possible 6-digit combinations. This is calculated by considering that each digit can range from 0 to 9 (10 possible values), and since there are 6 digits, the total combinations are (10^6 = 1,000,000).
There are 210 4 digit combinations and 5040 different 4 digit codes.
This question needs clarificatioh. There are 4 one digit number combinations, 16 two digit combinations, ... 4 raised to the n power for n digit combinations.
A 4-digit number can range from 0000 to 9999, which includes all combinations of four digits. Since each digit can be any number from 0 to 9 (10 options), the total number of combinations is calculated as (10^4). Therefore, there are 10,000 different combinations for a 4-digit number.
16