Hi Carl.

Here's how to do this with pencil and paper.

1. How many choices do you have for the first number? 50.
   
   
2. How many choices do you have for the second number? 49 (because you removed one of them in step 1).
   
   
3. Thus, there are 50 x 49 = 2450 possible "permutations". For permutations order is important, this means, for example, that (40, 20) is a different permutation from (20, 40).
   
   
4. You don't want permutation, you want combinations where order is not important. [You don't want to include (40,20) and (20, 40).] So that means you need to know how many different permutations there are for each combination. If there are two numbers, there are two permutations per combination.
   
   
5. Divide the possible permutations by number of permutations per combination: 2450 / 2 = 1225.

Here's another example: How many combinations are there for picking three numbers (no duplicates) from 1 to 20?

1. First choice: 20 possibilities.
   
   
2. Second choice: 19 possibilities.
   
   
3. Third choice: 18 possibilities.
   
   
4. Number of permutations: 20 x 19 x 18 = 6480.
   
   
5. There are 3 numbers. So the number of permutations (arrangements) of three items equals 1 x 2 x 3 = 6. Thus, there are six permutations per combination.
   
   
6. The number of combinations therefore is 6480 / 6 = 1080.

I didn't use any algebra, but if you want to step up to that, you will find easy formulas using factorials that give you quick answers. You can search for "combinations" in our Quick Search for more info on that.

Hope this helps,
Stephen La Rocque.