Rolling a die twice in either order with not mutual exclusive events I am reading the following problem:

If a single die is rolled twice, find the probability of rolling an
odd number and a number greater than $4$ in any either order

My solution:
Probability of rolling an odd number = $$\frac{3}{6}$$
Probability of rolling a number greater than $4$ = $$\frac{2}{6}$$
Since the events are not mutually exclusive ($5$ is counted twice) the probabilty of throwing an odd number followed by a number greater than $4$:
$$\frac{3}{6}\cdot \frac{2}{6} - \frac{1}{36} = \frac{5}{36}$$
The probablity of throwing a number greater than $4$ followed by an odd number is: $$\frac{2}{6}\cdot \frac{3}{6} - \frac{1}{36} = \frac{5}{36}$$
Therefore the my final solution is $$\frac{5}{36} + \frac{5}{36} = \frac{10}{36}$$
But it is wrong, as the solution states that it should be $$\frac{11}{36}$$
What am I doing wrong here?

The outcome potentially counted twice is $(5,5)$, not $5$.
The number of pairs $(a,b)$ where $a$ is odd and $b>4$ is $3\times2=6$. This includes $(5,5)$.
The number of pairs $(a,b)$ where $a>4$ and $b$ is odd is $2\times3=6$. This includes $(5,5)$.
The sum of these is $6+6=12$, but $(5,5)$ was counted twice; there are actually only $11$ pairs in the union of these two sets.
