Lottery with powerballs: from the point of view of the player vs. organizer My text offers the following exercise:

Discuss the probabilities for the following lottery: $$\text{5 out of 35, with two powerballs}$$ In this lottery $5$ "correct" balls are drawn from $35$ and then $2$ more which we call powerballs. What are the probabilities of "$4$ correct and $1$ powerball".

Version 1 (from the point of view of the player)
Putting myself in the player's shoes, I could think on the drawing as something which happened in the past but is kept in secret. So $5$ of the $35$ balls are good, $2$ are so-so, $28$ are bad. Hypergeometric distribution comes to mind, therefore the answer seems to be $$\frac{\binom{5}{4}\binom{2}{1}\binom{28}{0}}{\binom{35}{5}}$$
Yet the answer offered by my text is different: $$\frac{\binom{5}{4}\binom{2}{1}\binom{28}{2}}{\binom{35}{5}\binom{30}{2}}$$
It was clear from the textbook solution that they look at the problem differently, I put the book aside and tried to solve it in the alternative manner.
Version 2 (from the point of view of the organizer)
For the organizer the process of a player choosing some $5$ numbers on the ticket lies indeed in the past. The organizer now picks $5+2$ out of $35$ in such a way that the order is irrelevant within the $2$ groups but not across them. $$\blacksquare\blacksquare\blacksquare\blacksquare\square \ | \ \color{red}{\blacksquare\square}$$ There are $\binom{35}{5}\binom{30}{2}$ ways of picking. Furthermore, there are $\binom{5}{4}$ ways to fill the $4$ vacant places among the $5$ with some numbers chosen by the player (that also determines the $1$ vacant powerball-place). Now we are left with $2$ vacant places: one on the left, one on the right. They are not interchangeable, so there are $30\cdot29$ ways to fill them in. So the answer seems to be
$$\frac{\binom{5}{4}30\cdot29}{\binom{35}{5}\binom{30}{2}}$$
It still differs from the textbook solution yet coincides with my answer in the first version. I'm almost sure that the text is wrong yet with probabilities one never knows, since it's sometimes surprisingly counterintuitive.
Edit. A little description of what physically happens:
A player marks $5$ numbers out of $35$. The organizer has a box with $35$ numbered balls. First he draws $5$ balls from it (those become "correct" numbers) and then he draws $2$ more from the same box (which by the time has $30$ balls). These $2$ balls are then called "powerballs". So what are the chances that among the $5$ numbers marked by the player there are $4$ "correct" ones and $1$ "powerball".
 A: We can also calculate the probability in two steps (to agree however with your result and not with the one in the book). 
The process of the lottery is in reality described as follows. The player marks $5$ balls. In the first step $5$ balls out of $35$ will be drawn and the player wants to match $4$. So the probability for this event is equal to $$\frac{\dbinom{5}{4}\dbinom{30}{1}}{\dbinom{35}{5}}$$ Now there is only $1$ marked ball left and $2$ more will be drawn. The player wants that the remaining marked ball will be drawn, so the probability for this event is equal to $$\frac{\dbinom{1}{1}\dbinom{29}{1}}{\dbinom{30}{2}}$$ By the multiplication rule, you have that the desired probability is equal to $$\begin{align*}\frac{\dbinom{5}{4}\dbinom{30}{1}}{\dbinom{35}{5}}\cdot\frac{\dbinom{1}{1}\dbinom{29}{1}}{\dbinom{30}{2}}&=\frac{\dbinom{5}{4}\dbinom{2}{1}\dbinom{30}{2}}{\dbinom{35}{5}\dbinom{30}{2}}=\frac{\dbinom{5}{4}\dbinom{2}{1}}{\dbinom{35}{5}}\end{align*}$$
or equivalently
$$\begin{align*}\frac{\dbinom{5}{4}\dbinom{30}{1}}{\dbinom{35}{5}}\cdot\frac{\dbinom{1}{1}\dbinom{29}{1}}{\dbinom{30}{2}}&=\frac{\dbinom{5}{4}\cdot30\cdot29}{\dbinom{35}{5}\dbinom{30}{2}}\end{align*}$$
as you already found with both your approaches. Perhaps, there is some different assumption concerning the procees of the lottery made in the book.
