how many 3 digit combinations from 0 to 9?
44314
If what you want are all possible three digit numbers with no repetition of the digits then you have 10 choices for the first digit, you have 9 choices for the 2nd digit, and you have 8 choices for the 3rd digit giving you 10x9x8 = 720 in all.
529550
If we assume your letters range from A to Z and your digits from 0 to 9, you have 26 possibilities per letter and 10 per digit. This gives you 26^4*10^4 which is roughly 4 billion and a half. 8! / (4!
849988
If three different numbers have to be used, then 10 x 9 x 8 = 720 possible combinations.
551730
Berry analyzed passwords from previously released and exposed tables and security breaches, filtering the results to just those that were exactly four digits long [0-9]. There are 10,000 possible combinations that the digits 0-9 can be arranged into to form a four-digit code.
434189
10 10 = 100 two digit combinations. 10 10 10 = 1000 three digit combinations. There are 60 of them. Using the same argument as above there are 60 60 = 3600 two "digit" combinations and 60 60 60 = 216000 three "digit" combinations.
1310477
Or perhaps you meant the number of 8 digit numbers (repetitions permitted) using the digits 0-9. In that case, each digit has 10 choices and there are 8 of them, so the answer is 108 = 100 000 000.
171400
I also know that there are 100 combinations of two digits from 0-9, and 10 ombinations of one digit from 0-9.
316456
There are about 1,000 combinations possible for a 3-digit number/code. There are 10,000 combinations possible for a 4-digit number/code.
876640
Hi Mike. If order mattered, we would say select the first number (10 choices), then the second (9 choices) then the third (8 choices). So there would be 10 x 9 x 8 = 720 possible choices.
22327
There are 1 billion 9 digit numbers (000,000,000 through 999,999,999). There are 45 different combinations of two different numerals (10 x 9 divided by 2). There are 512 (2 to the 9th power) different permutations for any two numbers to be used in a 9 digit number.
1004049
So there are possible numbers of length 4. It's easy. The first digit can be 1,2,3,4,5 or 6 (6 combinations). We choose the second,third and fourth digit the same way we did it with the first (so it's 6*6*6 combinations) and the final answer is 6*6*6*6=1296 combinations.
275484
If we assume your letters range from A to Z and your digits from 0 to 9, you have 26 possibilities per letter and 10 per digit. This gives you 26^4*10^4 which is roughly 4 billion and a half.
405563
How many combinations of three number codes can occur with a standard combination lock? The range of possible numbers is 0-39 which allows 40 different possible choices each turn of the dial. Each combination requires exactly three numbers (i.e. three turns of the dial). E.g. 35-40-5.
431600
We conclude that there are 35−3×25+3=150 combinations. If there were no restrictions, the number of five digit codes that could be formed with the digits 1, 2, 3 would be 35 since there are three ways to choose each of the five digits.
350931
Clearly, from 1 to 100, there are ten numbers with 3 as the unit's digit- 3, 13, 23, 33, 43, 53, 63, 73, 83, 93; and ten numbers with 3 as the ten's digit - 30, 31, 32, 33, 34, 35, 36, 37, 38, 39.
362961
There are 6 = 3x2x1 ways to order 3 digits in a row. Thus the number of combinations of 3 of the 10 digits is 720/6 = 120 combinations.
575111
Explanation: The trick is to realize that a number can not start with a zero! Now, there are 105 ways in which the digits 0-9 can be chosen for the five places of a five digit number. Out of these, 104 start with zero (once we start with 0, there are only 4 slots to fill, where we have 10 choices each).