If a member $\{a,b\}$ of $S$ is picked at random, what is the probability that $a+b$ is even? 
Let $S$ be the set of all unordered pairs of distinct two digit integers. If a member $\{a,b\}$ of $S$ is picked at random, what is the probability that $a+b$ is even?

Solution:
The pairs (unordered are): $$\{10,11\},\{10,12\},\{10,13\}\cdots,\{10,99\}$$
$$\{11,12\},\{11,13\},\{11,14\}\cdots,\{11,99\}$$
$$\cdots$$
$$\{98,99\}$$ which are in total:$89+88+87+\cdots +1$
and the pairs $\{a,b\}$ such that $a+b$ is even are: $$\{10,12\},\{10,14\},\cdots,\{10,98\}$$
$$\{11,13\},\{11,15\},\cdots\{11,99\}$$
$$\cdots$$
$$\{97,99\}$$
Please tell me whether my approach is correct or not and how to proceed further. I would like an easy solution /  shortcut to the above problem. My solution is very time taking I believe.
 A: It's easier to count the number of cases in which $a + b$ is odd. In particular, note that the sum $a + b$ is odd if and only if one of $a, b$ are odd and the other is even. There are $45$ two-digit odd numbers, and there are $45$ two-digit even numbers. Therefore, there are $45 \cdot 45 = 2025$ ways to choose $a, b$ such that $a + b$ is odd. 
Next, note that there are ${90\choose 2} = 4005$ ways to choose $a, b$ without any constraint. By complementation, there are $4005- 2025 = 1980$ unordered pairs $\{a, b\}$ such that $a + b$ is even.
Thus, the required probability is given by
$$\frac{1980}{4005}  = \boxed{\frac{44}{89}}$$
A: It doesn't matter if the pair is ordered or unordered, provided that $a$ and $b$ are distinct. If we choose two numbers in order without replacement, then every pair $\{a,b\}$ is chosen in two ways: once as $(a,b)$ and once as $(b,a)$. The distribution of $a+b$ is unaffected.
No matter which first element $x \in \{10, \dots, 90\}$ we choose, there are $45$ numbers $y$ with the opposite parity to $x$ (so that $x+y$ is odd), but only $44$ numbers $y \ne x$ with the same parity (so that $x+y$ is even), because we can't choose $x$ again. So there's a $\frac{44}{44+45} = \frac{44}{89}$ probability of getting an even total, and a $\frac{45}{44+45} = \frac{45}{89}$ probability of getting an odd total.
