A Problem dealing with a Deck of Cards and the Hypergeometric Distribution

Question

Problem:
If $13$ cards are to be chosen at random(without replacement), determine
the probability that (a) $6$ will be picture cards, (b) none will be
picture cards.
Answer: (a)
The probability that a card drawn from a deck of cards will be a
picture card is $\dfrac{3}{13}$. It would tempting to
model this as a binomial distribution but that would be wrong because
it is done without replacement. Instead we use the hypergeometric distribution.
The general form of the hypergeometric distribution is:
$$P(X = x) = \dfrac{ {{b}\choose{x}} { {r} \choose {n-x} } }
{ {{b+r} \choose {n}} } $$
In this case, we have:
\begin{align*}
x &= 6 \\
n &= 13 \\
b &= 4(3) = 12 \\
r &= 52 – 12 = 40 \\
p &= \dfrac{12}{52} = \frac{3}{13} \\
\end{align*}
Let $p_a$ be the probability we seek.
\begin{align*}
p_a &= \dfrac{ {{12}\choose{6}} { {40} \choose {13-6} } }
{ {{12+40} \choose {13}} } \\
%
p_a &= \dfrac{ {{12}\choose{6}} { {40} \choose {7} } }
{ {{52} \choose {13}} } \\
\end{align*}
However, the book gets:
$$ \dfrac{ {{13}\choose{6}} { {39} \choose {7} } } { {{52} \choose {13}} } $$
Where did I go wrong?

Answer 1

Your answer is correct, and the book is wrong.