Combinatorics Problem: Finding Groups of 3 Numbers with Average Condition

[Mr Davis 97]

Homework Statement

In how many ways can we pick a group of 3 different numbers from the group $1,2,3,...,500$ such that one number is the average of the other two? (The order in which we pick the numbers does not matter.)

Homework Equations

The Attempt at a Solution

I start by noting that in order to get a number in the group $1,2,3,...,500$ from the average of two other numbers from the group, those two numbers must be odd, since their sum must be even. There are 250 odd numbers from 1 to 500. Since the numbers have to be different, we have $250 \cdot 249$ ways to find two odd numbers from the list. However we are given that order does not matter, so we must divide by 2 to get $\displaystyle \frac{250 \cdot 249}{2} = 31125$

However, the correct answer is apparently 62250, which is my answer times 2. My question is, since we are given that order does not matter, don't we have to divide by 2? Since x + y is not different than y + x? Where does my logic go wrong when I think that we should divide by 2?

 

[Fightfish]

I start by noting that in order to get a number in the group $1,2,3,...,500$ from the average of two other numbers from the group, those two numbers must be odd, since their sum must be even.

What is the sum of two even numbers?

 

[Mr Davis 97]

What is the sum of two even numbers?

Ohhh... Don't know how I missed that. So the 31125 is correct for the odd numbers, but then we must add this to 31125, which is the number of even number pairs, which gets us to 62250.