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.
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 .
987654321
10234567
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.
There are twelve possible solutions using the rule you stated.
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 .
987654321
10234567
There are 210 4 digit combinations and 5040 different 4 digit codes.
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.
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.
The number of possible 3 digit combinations you can make out of 1-9 with outrepeated digits is:9C3 = 9!/(3!(9-3)!) = 84
16
The short answer is 1000. This is very easy to visualise: Simply consider each number in the combination to be a digit in a decimal number. We then end up with a three-digit number. Such a three-digit number ranges in value from 000 to 999, or 1000 unique combinations.
There are 36 possible characters (26 letters + 10 numbers) that can be used in each position of the 11-digit combination. Therefore, the total number of possible combinations is 36^11, which is approximately 7.52 x 10^17. This means there are over 750 quadrillion possible 11-digit combinations of letters A-Z and numbers 0-9 when combined.
Oh, dude, let me break it down for you. So, to find the number of 5-digit combinations from 1 to 60, you just do 60 minus 1 plus 1, which gives you 60. So, there are like 60 different 5-digit number combinations you can make from that range. Easy peasy, lemon squeezy!