999,999,999 if you don't include 000000000, otherwise 1,000,000,000.
If you don't allow repeated digits then 9 factorial, ie 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 = 362,880.
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The number of combinations for 9 digits can be calculated using the formula for permutations of n objects taken r at a time, which is n! / (n-r)!. In this case, with 9 digits, there are 9! / (9-9)! = 9! / 0! = 9! = 362,880 possible combinations. This is because each digit can be arranged in 9 different positions, and as each digit is unique, the total number of combinations is the product of these possibilities for each digit.
Oh, dude, you're really testing my math skills here. So, to figure out the number of combinations for 9 digits, you'd use the formula 10^9, which equals 1 billion. That's like a lot of possible combinations, man. So, good luck cracking that code!
There are only 10 combinations. In each combination one of the 10 digits is left out.
9
10000
There are 10 digits, 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Let's say that a three digits number has a form of xyz. There are only 9 possibilities for x (0 cannot be in the first place), 10 possibilities for y, and 10 possibilities for z. So there are 9 x 10 x 10 = 900 combinations. However, there are only 9 digits from 1 to 9. So we have 9 possibilities for x, y, and z. Hence, there are 9 x 9 x 9 = 729 combinations.
To calculate the number of possible combinations of the digits 1, 3, 7, and 9, we can use the formula for permutations of a set of objects, which is n! / (n-r)!. In this case, there are 4 digits and we want to find all possible 4-digit combinations, so n=4 and r=4. Therefore, the number of possible combinations is 4! / (4-4)! = 4! / 0! = 4 x 3 x 2 x 1 = 24. So, there are 24 possible combinations using the digits 1, 3, 7, and 9.