There are 6 white balls and 9 black balls. Probability of drawing two white, then two black? From A First Course in Probability (9th Edition):

3.5 An urn contains 6 white and 9 black balls. If 4 balls are to be
  randomly selected without replacement, what is the probability that
  the first 2 selected are white and the last 2 black?

This method is straightforward and results in the correct answer (according to the book):
$$\frac{6}{15} \cdot \frac{5}{14} \cdot \frac{9}{13} \cdot \frac{8}{12} = \frac{6}{91} $$
(This is just the multiplication principle and probability of drawing the color of that ball at that time) 
However, I want to understand this in terms of conditional probability. I don't understand why this doesn't work:
$$P(E \mid F) = \frac{P(E \cap F)}{P(F)} ={\frac{{6 \choose{2}}{9 \choose 2}}{{15 \choose{2}}{13 \choose 2}}}÷{\frac{{6 \choose{2}}}{{15 \choose{2}}}} = {\frac{{9 \choose 2}}{{13 \choose 2}}} = \frac{6}{13} \ne \frac{6}{91}$$
$\frac{6}{13}$ is exactly 7 times more than the previous answer. Why does this method fail to work? What mistake have I made? I tried to use the exact same method used in question 3.3, where this resulted in the correct answer. 

Optional – About 3.3
3.3 Use Equation (2.1) to compute in a hand of bridge the conditional 
probability that East has 3 spades given taht North and South have a 
combined total of 8 spades.

Here, we see that:
$$P(E \mid F) = \frac{P(E \cap F)}{P(F)} ={\frac{{13 \choose{8}}{39 \choose 18}{5 \choose 3}{21 \choose 10}}    {{52 \choose{26}}{26 \choose 13}}}÷{\frac{{13 \choose{8}}{39 \choose 18}}{{52 \choose{26}}}} = {\frac{{5 \choose 3}{21 \choose 10}}{{26 \choose 13}}} = \frac{29}{115} \approx 0.339$$
Which is the answer in the back of the book.

Method 1.
$$
\text{Probability}={\frac{{{6}\choose{2}}{{9}\choose{2}}}{{{15}\choose{2}}{{13}\choose{2}}}}=\frac{6}{91}.
$$
$$
\binom{6}{2}\text{ is the quantity of ways to choose 2 any white balls of 6 white balls},
$$
$$
\binom{9}{2}\text{ is the quantity of ways to choose 2 any black balls of 9 black balls},
$$
$$
\binom{15}{2}\text{ is the quantity of ways to choose 2 any white balls of}
$$
$$
\text{ the total amount of balls (15)},
$$
$$
\binom{13}{2}\text{ is the quantity of ways to choose 2 any black balls of}
$$
$$
\text{ the total amount of balls (13), considering that 2 any white balls are chosen already}.
$$

Method 2.
Also this problem can be performed using the more detailed binoms:
$$
\text{Probability}=\frac{\binom{6}{1}\binom{5}{1}\binom{9}{1}\binom{8}{1}}{\binom{15}{1}\binom{14}{1}\binom{13}{1}\binom{12}{1}}=\frac{6}{91}.
$$
