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?

-
You’ve not given enough information. How many balls of each color are in the set from which you’re drawing? – Brian M. Scott Jul 11 '12 at 5:12
thank you, can't believe I totally left it out! – pauliwago Jul 11 '12 at 5:14

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 = $(_{}^{20}\textrm{C}_5 - _{}^{14}\textrm{C}_5) \times (_{}^{14}\textrm{C}_5 + 6_{}^{14}\textrm{C}_4 )$

-
Thank you! This was very clear. – pauliwago Jul 11 '12 at 5:54

There are $14$ non-red balls, so there are $\binom{14}5$ ways to draw $5$ non-red balls. There are $\binom{20}5$ ways to draw $5$ balls regardless of color, so there are $\binom{20}5-\binom{14}5$ ways to draw $5$ balls including at least one red ball.

Now we’ll calculate the number of ways to draw $5$ balls, of which at most one is green. The easiest way to calculate this is to split it into two cases: no green balls, and exactly one green ball. Since there are $6$ green balls, there are $14$ non-green balls, and the number of ways of drawing $5$ of them is $\binom{14}5$, just as when we counted the ways of drawing $5$ non-red balls. There are $\binom61=6$ ways to draw one green ball and $\binom{14}4$ ways to draw $4$ non-green balls, so there are $6\binom{14}4$ ways to draw $5$ balls, exactly one of which is green. Adding the two cases, we see that there are $\binom{14}5+6\binom{14}4$ ways to get at least one green ball amongst the $5$.

The final answer, as you realized, is the product of these two numbers,

$$\left(\binom{20}5-\binom{14}5\right)\left(\binom{14}5+6\binom{14}4\right)\;.$$

I make that $(15504-2002)(2002+6006)=108,124,016$.

-