Sample Space of a urn problem involving removing two balls of the same colour

This problem comes from David Stirzaker Elementary Probability and it goes:

An urn contains n heliotrope (H) and n tangerine (T) balls. Two balls are removed from the urn together, at random. Find P(HH) when the sample space is taken to be all unordered pairs of distinguishable balls

The book provides this as the answer:

$ A= \text{{(i, j): 1 ≤ i < j ≤ 2n}}$ and so $|A| = n(2n − 1)$. Likewise, $|{H H}| = \frac{1}{2}n(n − 1)$. Hence, $P(H H) = (n − 1)/(2(2n − 1))$ (A is the sample space).

However I don't get this answer, mainly why $A$ has size $n(2n-1)$ instead of something like $2n(2n-1)$ could someone explain?

Solution 1:

The number of elements in the sample space is the number of ways of selecting $2$ of the $2n$ balls in the urn, which is $$\binom{2n}{2} = \frac{(2n)!}{2!(2n - 2)!} = \frac{2n(2n - 1)(2n - 2)!}{2 \cdot 1 \cdot (2n - 2)!} = \frac{2n(2n - 1)}{2} = n(2n - 1)$$

Similarly, the number of favorable cases is the number of ways of selecting two of the $n$ heliotrope balls, which is $$\binom{n}{2} = \frac{n!}{2!(n - 2)!} = \frac{n(n - 1)(n - 2)!}{2 \cdot 1 \cdot (n - 2)!} = \frac{n(n - 1)}{2}$$

The probability of obtaining two heliotripe balls is therefore $$\Pr(HH) = \frac{\dbinom{n}{2}}{\dbinom{2n}{2}} = \frac{\frac{n(n - 1)}{2}}{n(2n - 1)} = \frac{n(n - 1)}{2n(2n - 1)} = \frac{n - 1}{2(2n - 1)}$$