Probability and Conditional Probability Could someone please provide a worked solution to this question?
A box contains 5 amber, 7 blue and 9 green balls. Six of the balls are removed from the box at random and without replacement. 
Find the probability that
(a) three out of the six balls are blue;
(b) four of the balls are blue and two are green;
(c) the second ball to be selected is amber, given that the final ball to be selected is blue. 
I was wondering if there is a combinatorial argument for these types of problems,  because multiplying fractions here seems a little time consuming and messy. 
My answers:
(a) $\frac{7}{21}\times\frac{6}{20}\times\frac{5}{19}\times\frac{14}{18}\times\frac{13}{17}\times\frac{12}{16}\times\frac{6!}{(3!)^2}$
(b) $\frac{7}{21}\times\frac{6}{20}\times\frac{5}{19}\times\frac{4}{18}\times\frac{9}{17}\times\frac{8}{16}\times\frac{6!}{(4!)(2!)}$
Any help very much appreciated, as always. 
 A: (A) The probability that when selecting $6$ from $21$ balls, you select $3$ from $7$ blue and $3$ from $14$ non-blue is: $$\dfrac{\binom 73\binom {14}3}{\binom {21}6} \tag{$= \dfrac{\frac{7\cdot 6\cdot 5}{3\cdot 2\cdot 1}\cdot \frac{14\cdot 13\cdot 12}{3\cdot 2\cdot 1}}{\frac{21\cdot 20\cdot 19\cdot 18\cdot 17\cdot 16}{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}}$}$$
Which matches your answer, in more compact form.
The count of Blue balls so selected follows a Hypergeometric Distribution.
(B) The probability that when selecting $6$ from $21$ balls, you select $4$ from $7$ blue, $2$ from $9$ green (, and $0$ from $5$ amber) is: $$\dfrac{\binom 74\binom {9}2\color{silver}{\binom 5 0}}{\binom {21}6}$$
Which again matches your answer.
(C) The probability that when given that $1$ of the blue balls will be reserved for the sixth draw, the second ball selected will be one from the $5$ of the $20$ unreserved balls that are amber, is just:

 $$\mathsf P(C_2=a\mid C_{6}=b) = \frac{5}{20}$$

No need to overcomplicate it.
A: Well, for such combinatorical questions there are typically multiple different "counting schemes" to get to the same answer. Here is the thought process I would have to arrive at those. Hope it helps:
a) There are $\binom{5+7+9}{6}$ ways to choose the $6$ balls. There are $\binom{7}{3}$ ways to choose $3$ blue balls and $\binom{5+9}{3}$ ways to choose $3$ non-blue balls. So the Answer is
$\begin{align}\frac{\binom{5+9}{3}\binom{7}{3}}{\binom{5+7+9}{6}}\end{align}$
b) very similar: $\binom{7}{4}$ ways to choose $4$ blue balls. $\binom{9}{2}$ ways to choose $2$ green balls. So the answer is
$\begin{align}\frac{\binom{7}{4}\binom{9}{2}}{\binom{5+7+9}{6}}\end{align}$
Generally: It is always about the "binomial coefficient" $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ which is the number of ways to select $k$ objects (without replacement) out of a total of $n$ objects.
c) Here, order is important, so you can't use binomial coefficients in the same way. The best way here is to multiply fractions for each step like this:
$P(\text{last is blue}) = \frac{7}{5+7+9}$.
$P(\text{second is amber and last is blue}) = \frac{5}{5+7+9} \cdot\frac{7}{4+7+9}$
$P(\text{second is amber}|\text{last is blue}) = \frac{P(\text{second is amber and last is blue})}{P(\text{last is blue})} = \frac{5\cdot7\cdot21}{7\cdot21\cdot20}$
