575757
10!/3! = 604800 different combinations.
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.
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.
1000
10!/3! = 604800 different combinations.
15
10,000
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.
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.
There are 15C3 = 15*14*13/(3*2*1) = 455 combinations, and I am NOT even going to try listing them.
Including the null combination, there are 2^14 = 16384 combinations.