Combinations - at least and at most There are 20 balls - 6 red, 6 green, 8 purple
We draw five balls and at least one is red, then replace them.  We then draw five balls and at most one is green.  In how many ways can this be done if the balls are considered distinct?
My guess:
$${4+3-1 \choose 3-1} \cdot {? \choose ?}$$
I don't know how to do the second part...at most one is green?
Thanks for your help.

Event A
Number of ways to choose 5 balls = $ _{}^{20}\textrm{C}_5 $
Number of ways to choose 5 balls with no red balls = $ _{}^{14}\textrm{C}_5 $
Hence the number of ways to choose 5 balls including at least one red ball =$ _{}^{20}\textrm{C}_5 - _{}^{14}\textrm{C}_5 $
Event B
Number of ways to choose 5 balls with no green balls = $ _{}^{14}\textrm{C}_5 $
Number of ways to choose 5 balls with exactly one green ball = $ _{}^{14}\textrm{C}_4 \times 6 $ (we multiply here by 6 because we have 6 choices for the green ball we choose.)
Since the above two choice processes are mutually exclusive we have that, 
Hence the number of ways to choose 5 balls including at most one green ball =$ _{}^{14}\textrm{C}_5 + _{}^{14}\textrm{C}_4 \times 6 $

Events A and B are independent.
Therefore, the total number of ways of doing A and B = $ (_{}^{20}\textrm{C}_5 - _{}^{14}\textrm{C}_5) \times (_{}^{14}\textrm{C}_5 + 6_{}^{14}\textrm{C}_4  )$
