PRMO 2013 question 5 red blue green balls

5. There are n−1n - 1 red balls, nn green balls, and n+1n + 1 blue balls in a bag. The number of ways of choosing two balls from the bag that have different colors is 299. What is the value of nn?
[PRE-RMO–2013]

Partially correct approach:
Choose 1 red and 1 non-red ball = (n-1)(n+n+1) = (n-1)(2n+1)
Choose 1 green and 1 non-green ball = (n)(n-1+n+1) = (n)(2n)
Choose 1 blue and 1 non-blue ball = (n+1)(n+n-1) = (n+1)(2n-1)

Adding all we get 6n^2 - 2 = 299
But wait, LHS is even and RHS is odd. Something is wrong here.
Oh, gotcha! We have done double counting. We have counted each combination twice.
For e.g., in the first line we counted red + green and red + blue combinations.
In the second line, green + red and green + blue.
In the third line, blue + red and blue + green.

We were supposed to count each one only once.
So, divide LHS by 2.
3n^2 - 1 = 299 => n^2 = 100 => n = 10

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