4! = 4*3*2*1 = 24 of them.
two
Starting with three different letters, six two-letter combinations can be made, if the order of the two letters is important. Only three combinations can be made if the order of the two letters is not important. Example: ABC AB AC BA BC CA CB - six variations But if (for your purposes) BA is the same as AB, Then there are only three: AB AC BC
4435236.. I think
32C3 = 4960
-8
two
10
2600
Starting with three different letters, six two-letter combinations can be made, if the order of the two letters is important. Only three combinations can be made if the order of the two letters is not important. Example: ABC AB AC BA BC CA CB - six variations But if (for your purposes) BA is the same as AB, Then there are only three: AB AC BC
4435236.. I think
16 x 15 / 2 ie 120
Two . . . . . 38 and 83.
There are many different combinations. One example is two and 31.
Simple enough to solve. The answer is a power of two. Assuming you have two possible digits, say for example, 3 and 4, then you simply have to multiply it by how many numbers you want to get the total number of combinations. Each number can be 3 or 4 in this case, and you have 5 numbers. That's two to the fifth. Five combinations of any two numbers. 2x2x2x2x2. The answer is 32 combinations.
32C3 = 4960
-8
There are 100 combinations of two numbers ranging from 0 to 9. This is because each of the two numbers can independently be any digit from 0 to 9, resulting in (10 \times 10 = 100) combinations. Each combination includes pairs like (0,0), (0,1), ..., (9,9).