Simple enough to solve.
The answer is a power of two.
Assuming you have two possible digits, say for example, 3 and 4, then you simply have to multiply it by how many numbers you want to get the total number of combinations.
Each number can be 3 or 4 in this case, and you have 5 numbers. That's two to the fifth. Five combinations of any two numbers. 2x2x2x2x2.
The answer is 32 combinations.
Only one.
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
There is only one possible combination of a 13 digit number created from 13 digits. In a combination, the order of the digits does not matter so that 123 is the same as 132 or 312 etc. If there are 13 different digits (characters) there is 1 combination of 13 digits 13 combinations of 1 or of 12 digits 78 combinations of 2 or of 11 digits and so on There are 213 - 1 = 8191 in all. If the characters are not all different it is necessary to have more information.
If the same 7 digits are used for all the combinations then n! = 7! = 7*6*5*4*3*2*1 = 5040 combinations There are 9,999,999-1,000,000+1=9,000,000 7-digit numbers.
It depends upon whether the numbers can be used more than once. If the numbers can be used more than once, then there are 1,000 possible combinations; if not, then there are 720 possible combinations. ========== Assuming you are talking about integers you can calculate it this way: you can have any one of 9 digits for the first digit (zero is excluded because that would make it only a 2 digit number) You can have any one of 10 digits for the second and any one of 10 digits for the third digit. That gives you 9x10x10 = 900 different combinations for 3 digit numbers (not 1000). If you can include both negative and positive numbers as different numbers you get twice as many (2x900=18000). If you can count decimal numbers as 3 digit numbers (i.e. not restricted to integers) you can have 900 integers, 900 with the form xx.x, 1000 with the form x.xx (if zero is permitted to be the first digit and count as one of the 3 digits) or 900 (if a leading zero is NOT counted as one of the 3 digits). If a leading zero is NOT counted as one of the 3 digits, you could also have 1000 numbers of the form 0.xxx or just .xxx
There are different numbers of combinations for groups of different sizes out of 9: 1 combination of 9 digits 9 combinations of 1 digit and of 8 digits 36 combinations of 2 digits and of 7 digits 84 combinations of 3 digits and of 6 digits 126 combinations of 4 digits and of 5 digits 255 combinations in all.
about 1,0000000000000
Oh, dude, you're hitting me with some math vibes here. So, if you have 6 digits to choose from to make a 4-digit combination, you can calculate that by using the formula for permutations: 6P4, which equals 360. So, like, you can make 360 different 4-digit combinations from those 6 digits. Math is wild, man.
Assuming the digits cannot be repeated, there are 7 combinations with 1 digit, 21 combinations with 2 digits, 35 combinations with 3 digits, 35 combinations with 4 digits, 21 combinations with 5 digits, 7 combinations with 6 digits and 1 combinations with 7 digits. That makes a total of 2^7 - 1 = 127: too many for me to list. If digits can be repeated, there are infinitely many combinations.
45 In combinations, the order of the digits does not matter so that 12 and 21 are considered the same.
Only one.
To find the number of three-digit combinations, we consider the digits from 000 to 999. Each digit can range from 0 to 9, giving us 10 options for each of the three digits. Therefore, the total number of three-digit combinations is (10 \times 10 \times 10 = 1,000).
There are 167960 9 digits combinations between numbers 1 and 20.
If the digits can repeat, then there are 256 possible combinations. If they can't repeat, then there are 24 possibilities.
The order of the digits in a combination does not matter. So 123 is the same as 132 or 312 etc. There are 10 combinations using just one of the digits (3 times). There are 90 combinations using 2 digits (1 once and 1 twice). There are 120 combinations using three different digit. 220 in all.
There are infinite combinations that can make 3879
If no digit can be repeated then there are 5 combinations, abcd, abce, abde, acde and bcde. If you regard abdc as different from abcd then each of the 5 basic sets could be arranged 24 ways and the total would be 120 combinations.