Probability of taking out coins by order 
The question goes as following:
  There is a jar with 65 coins. 30 are green, 15 are yellow and 20 are red.
  When taking out 20 coins from the jar, one by one, what is the probability that the first three coins and last three coins selected will be green?

My logic was that if I only look at the first 3 and last 3 selected, there are $30 \cdot 29\cdot 28\cdot 27\cdot 26\cdot 25$ options for them to be green, and a total of $65\cdot 64\cdot 63\cdot 62\cdot 61\cdot 60$ options to choose the first and last 3.
So the probability should be $ (30 \cdot 29\cdot 28\cdot 27\cdot 26\cdot 25)\div (65\cdot 64\cdot 63\cdot 62\cdot 61\cdot 60) $
Would appreciate any comments regarding if this answer makes sense or not.

Let's see for $7$ balls:
$$P(\color{green}{G_1G_2G_3}(\color{green}G \text{ or } G')_4\color{green}{G_5G_6G_7})=\\
P(\color{green}{G_1G_2G_3G_4G_5G_6G_7})+P(\color{green}{G_1G_2G_3}G'_4\color{green}{G_5G_6G_7})=\\
\color{green}{\frac{30}{65}\cdot \frac{29}{64}\cdot \frac{28}{63}}\cdot \color{green}{\frac{27}{62}\cdot \frac{26}{61}\cdot \frac{25}{60}\cdot \frac{24}{59}}+
\color{green}{\frac{30}{65}\cdot \frac{29}{64}\cdot \frac{28}{63}}\cdot \frac{35}{62}\cdot  \color{green}{\frac{27}{61}\cdot \frac{26}{60}\cdot \frac{25}{59}}=\\
\color{green}{\frac{30}{65}\cdot \frac{29}{64}\cdot \frac{28}{63}}\cdot \color{green}{\frac{27}{62}\cdot \frac{26}{61}\cdot \frac{25}{60}}\cdot \left(\frac{\color{green}{24}+35}{59}\right)=\\
\color{green}{\frac{30}{65}\cdot \frac{29}{64}\cdot \frac{28}{63}}\cdot \color{green}{\frac{27}{62}\cdot \frac{26}{61}\cdot \frac{25}{60}}$$
Similarly:
$$P(\color{green}{G_1G_2G_3}(\color{green}G \text{ or } G')_4...(\color{green}G \text{ or }G')_{17}\color{green}{G_{18}G_{19}G_{20}})=\\
\color{green}{\frac{30}{65}\cdot \frac{29}{64}\cdot \frac{28}{63}}\cdot \color{green}{\frac{27}{62}\cdot \frac{26}{61}\cdot \frac{25}{60}}.$$
