Only three: 12, 13 and 23.
Remember that the combinations 12 and 21 are the same.
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
20 There can be two answers to this: I'll assume this is what you meant-> let's say you mean like pairing the numbers 1-5 with the numbers 6-9. 5*4=20
10 Combinations (if order doesn't matter). 3,628,800 Possiblilities (if order matters).
15
6 ways: 931,913,139,193,391,319
10,000
It is: 9C7 = 36
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
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.
Each digit can appear in each of the 4 positions. There are 9 digits, therefore there are 9⁴ = 6561 such combinations.
There are 9C3 = 84 combinations.