This is a factorial problem. The first number can be any of ten digits, the second any of nine (because you can't repeat a digit), the third any of eight and the fourth any of the remaining 7 digits. 10x9x8x7=5040 combinations.
10!/3! = 604800 different combinations.
575757
There are a huge number of combinations of 5 numbers when using the numbers 0 through 10. There are 10 to the 5th power combinations of these numbers.
When using the numbers one through seven, the total number of different combinations depends on how many numbers you want to select. If you're looking for all possible combinations of any length (from one to seven), you can calculate it using the formula for combinations, which is (2^n - 1) (where (n) is the total number of items) to account for all subsets except the empty set. Therefore, for seven numbers, there are (2^7 - 1 = 127) different combinations. If you specify a particular number of selections, the calculation would differ accordingly.
Through the magic of perms and coms the answer is 729
1000
10!/3! = 604800 different combinations.
15
10,000
575757
There are a huge number of combinations of 5 numbers when using the numbers 0 through 10. There are 10 to the 5th power combinations of these numbers.
When using the numbers one through seven, the total number of different combinations depends on how many numbers you want to select. If you're looking for all possible combinations of any length (from one to seven), you can calculate it using the formula for combinations, which is (2^n - 1) (where (n) is the total number of items) to account for all subsets except the empty set. Therefore, for seven numbers, there are (2^7 - 1 = 127) different combinations. If you specify a particular number of selections, the calculation would differ accordingly.
Through the magic of perms and coms the answer is 729
9!/6!, if the six different orders of any 3 digits are considered distinct combinations.
There is 1 combination of all ten numbers, 10 combinations of one number and of nine numbers, 45 combinations of two or eight numbers, 120 combinations of three or seven numbers, 210 combinations of four or six numbers and 252 combinations of five numbers. That is 1023 = 210 - 1 in total.
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.
There are 360 of them.