Winning this specific lottery consists of choosing 5 numbers from a list of 50 numbers $\{1,2,3,...,50\}$ and then choosing 2 numbers from a list of 9 $\{1,2,...,9\}$. You win if all your numbers are chosen by the host on tv.
The correct approach is to say that there are a total of ${50\choose 5}\cdot{9\choose2}$ possibilities of choosing the numbers and just one winning combination $\implies$ the probability of winning is $\frac{1}{{50\choose 5}\cdot{9\choose2}}=\frac{1}{76275360}$
But here's a different approach: it doesn't matter if the nubers chosen on tv are chosen before or after you choose your combination, so let's assume the winning numbers are already chosen: $\{x_1,...,x_5,y_1,y_5\}\ \mid\ x_i\in[50], \ y_j\in[9]$
The probability of choosing one right number from the [50] list as your first number is $\frac{5}{50}$, the probability that the second number you choose is right is $\frac{4}{49}$ etc and we get $$\Bbb P[\text{winner}]=\frac{5!}{50\cdot49\cdot48\cdot47\cdot46}\cdot\frac{2\cdot1}{9\cdot8}$$ which is a different probability.
EDIT:
They are actually the same probability 