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Number of possible outcomes for 2 numbers given that one number is greater than the other

Answer 1

+6 A:

If you choose:

1: 2 to 10 = 9 possibilities
2: 3 to 10 = 8 possibilities
...
9: 10 = 1 possibility

So

9 + 8 + ... + 2 + 1 = 45

This is called an arithmetic progression the sum is equal to:

f(n) = n * (n+1) / 2 = 9 * 10 / 2 = 45

cletus 2010-01-17 04:06:04

For example if the first number I choose is 3 , shouldn't the number of possible options left for the second number be 2? (i.e the numbers 1 and 2). From what you have written, it would be 7.

blackgeneral 2010-01-18 17:45:01

@blackgeneral: I've taken the approach of choosing a number and then all numbers higher than it. You could do it the other way: choose a number and then all numbers below it (which is what you do with 3 yields 1 and 2 instead of 4 through 10). The result will be the same.

cletus 2010-01-18 22:28:48

Answer 2

+10 A:

45.

Imagine your grid of 10x10 pairs. There's a diagonal line from 1,1 to 10,10 where A=B, so those aren't A>B. The line divides the other cases into two halves.. one with A>B and one with B < A. Each of these partitions are equal size. There's 90 values left (10 taken by the diagonal line) so there are 45 pairs where A > B.

SPWorley 2010-01-17 04:06:25

The grid is a useful analogy for 2 variables and I certainly think about Backgammon and similar things in those terms but it quickly falls down when you get to 3 or more. My approach is a more robust and general way of thinking about combinatorial problems.

cletus 2010-01-18 13:35:54

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Number of possible outcomes for 2 numbers given that one number is greater than the other

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