Oh, dude, let me grab my calculator for this super thrilling math question. So, you've got 9 numbers and you're trying to figure out how many ways you can arrange them. That's like 362,880 combinations, but who's counting, right? Just a casual number of possibilities for your 9-digit number extravaganza.
To calculate the number of 9-digit combinations that can be made from the numbers 1-9 without repetition, we use the permutation formula. Since there are 9 numbers and we are choosing 9 without repetition, the formula is 9P9 = 9! / (9-9)! = 9! / 0! = 9! = 362,880. Therefore, there are 362,880 9-digit combinations that can be made from the numbers 1-9.
If there are no restrictions on duplicated numbers or other patterns of numbers then there are 10 ways of selecting the first digit and also 10 ways of selecting the second digit. The number of combinations is therefore 10 x 10 = 100.
The answer is 10C4 = 10!/[4!*6!] = 210
Each digit can appear in each of the 4 positions. There are 9 digits, therefore there are 9⁴ = 6561 such combinations.
Through the magic of perms and coms the answer is 729
Using the formula n!/r!(n-r)! where n is the number of possible numbers and r is the number of numbers chosen, there are 13983816 combinations of six numbers between 1 and 49 inclusive.
It would be from 1111111 to 9999999, and because 9999999 - 1111111 = 8888888, the answer is 8888888.
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
10 Combinations (if order doesn't matter). 3,628,800 Possiblilities (if order matters).
It is: 9C7 = 36
10,000
15
6 ways: 931,913,139,193,391,319
If there are no restrictions on duplicated numbers or other patterns of numbers then there are 10 ways of selecting the first digit and also 10 ways of selecting the second digit. The number of combinations is therefore 10 x 10 = 100.
To calculate the number of different 4-digit combinations that can be made using numbers 0 through 9, we use the concept of permutations. Since repetition is allowed, we use the formula for permutations with repetition, which is n^r, where n is the number of options for each digit (10 in this case) and r is the number of digits (4 in this case). Therefore, the number of different 4-digit combinations that can be made using numbers 0 through 9 is 10^4, which equals 10,000 combinations.
The answer is 10C4 = 10!/[4!*6!] = 210
Only three: 12, 13 and 23. Remember that the combinations 12 and 21 are the same.
Each digit can appear in each of the 4 positions. There are 9 digits, therefore there are 9⁴ = 6561 such combinations.