Lottery with ordered draw - check my proof Let's assume that the order in which the numbers are drawn shall be considered. 
6 numbers will be drawn out of 30 numbers and each number can only be drawn once.

Calculate the probability that your lottery ticket has exactly 3 numbers which also appear on the winning ticket. 

My approach:

Let be $\Omega:=\{(\omega_1, ..., \omega_6) \vert \omega_i \in \{1, ..., 30\}\}$ and $\vert \Omega \vert = { 30! \over 24!}$. 

Let $B:=\{\omega \in \Omega \vert \omega \text{ contains 3 winning numbers}\}$. $\color{red}{\text{How do I rigorously write this down?}}$

Let $B^c$:= complement of $B$

Now I am not sure how to rigorously define $A$ the set of positive outcomes. It should contain three elements of $B$ and three elements of $B^c$ considering all possible permutations ot them.
$\color{red}{A:=(\{\omega_1, ...,\omega_6,)\vert \omega_i \in ~??\} }$. 

Then the cardinality of $A$ should be: $\vert A\vert = {6! \over 3!}\cdot { 24! \over 3!}\cdot {6 \choose 3}$. Hence, $P(A)=\frac{{6! \over 3!}\cdot { 24! \over 3!}\cdot {6 \choose 3}}{30! \over 24!}\approx 0.068$

Is this correct? Any comments are appreciated.
What we are interested in is whether exactly three of the six numbers on the lottery ticket match the winning numbers drawn in the lottery.

If there are exactly three matches, then the lottery ticket must contain exactly three of the six winning numbers and three of the other $24$ numbers.  Since six of the thirty numbers are winning numbers, the probability that this occurs is
$$\Pr(\text{exactly three matches}) = \frac{\dbinom{6}{3}\dbinom{24}{3}}{\dbinom{30}{6}}$$
which agrees with the answer you obtained.

A favorable outcome is any set of six numbers containing exactly three winning and three losing numbers.  If you wish to take the order of selection into account, favorable outcomes consist of ordered sextuples with three winning and three losing numbers.
What we are interested in is whether exactly three of the six numbers on the lottery ticket match the winning numbers drawn in the lottery.

If there are exactly three matches, then the lottery ticket must contain exactly three of the six winning numbers and three of the other $24$ numbers.  Since six of the thirty numbers are winning numbers, the probability that this occurs is
$$\Pr(\text{exactly three matches}) = \frac{\dbinom{6}{3}\dbinom{24}{3}}{\dbinom{30}{6}}$$
which agrees with the answer you obtained.

A favorable outcome is any set of six numbers containing exactly three winning and three losing numbers.  If you wish to take the order of selection into account, favorable outcomes consist of ordered sextuples with three winning and three losing numbers.