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.
Number of 7 digit combinations out of the 10 one-digit numbers = 120.
56 combinations. :)
This question needs clarificatioh. There are 4 one digit number combinations, 16 two digit combinations, ... 4 raised to the n power for n digit combinations.
The answer will depend on how many digits there are in each of the 30 numbers. If the 30 numbers are all 6-digit numbers then the answer is NONE! If the 30 numbers are the first 30 counting numbers then there are 126 combinations of five 1-digit numbers, 1764 combinations of three 1-digit numbers and one 2-digit number, and 1710 combinations of one 1-digit number and two 2-digit numbers. That makes a total of 3600 5-digit combinations.
It is: 9C7 = 36
Number of 7 digit combinations out of the 10 one-digit numbers = 120.
15
10,000
56 combinations. :)
10!/3! = 604800 different combinations.
Through the magic of perms and coms the answer is 729
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.
9!/6!, if the six different orders of any 3 digits are considered distinct combinations.
There are 360 of them.
This question needs clarificatioh. There are 4 one digit number combinations, 16 two digit combinations, ... 4 raised to the n power for n digit combinations.
9
Each digit can appear in each of the 4 positions. There are 9 digits, therefore there are 9⁴ = 6561 such combinations.