Imagine that even though you listen to just the first $10$ songs, the iPod has actually generated a random ordering of the entire set of $100$ songs. We’re going to count the ways to distribute the $10$ Beatles songs in this list so that the first Beatles song is the fifth song overall.
In order for this to happen, one Beatles song must go into the fifth slot in the order, and the other $9$ must go into the last $95$ slots. We don’t care which Beatles song is which $-$ they could be $10$ copies of the same song, for all we care $-$ so all that matters is the positions, not which Beatles song goes into which of the $10$ positions chosen for Beatles songs. We have no choice about one Beatles song: it must go into the fifth slot overall. The other $9$ must go into $9$ of the last $95$ slots, and there are $\binom{95}9$ ways to pick those $9$ slots. Thus, there are altogether $1\cdot\binom{95}9=\binom{95}9$ ways to pick $10$ slots for the Beatles songs so that the first Beatles song is the fifth song in the whole list.
There are $\binom{100}{10}$ ways to pick $10$ positions in this ordering, so there are $\binom{100}{10}$ different sets of $10$ positions in which the $10$ Beatles songs could occur. $\binom{95}9$ of these $\binom{100}{10}$ do what we want: they put the first Beatles song in the fifth position overall. The $\binom{100}{10}$ possibilities for the positions of the $10$ Beatles songs are all equally likely, and $\binom{95}9$ yield the desired outcome, so the probability of that outcome is
$$\frac{\binom{95}9}{\binom{100}{10}}\;.$$
Addendum added to the end of my answer, to respond to the comment of Dmitrii I.
The confusion is actually between (for example) the quoted odds of
To a Mathematician, the first syntax above is the most rational, and expresses the amount that you receive upon winning, not including the return of your original bet.
Casinos like to seduce the betting public, who are generally non-Mathematicians. $6$ for $1$ sounds more attractive than its mathematically equivalent expression of $5$ to $1$.
When reading a book on odds, you have to be careful to decipher the author's intent. Authors (unfortunately) may be careless in not explicitly specifying whether the quoted odds of $n::1$ represent $n$ to $1$ or $n$ for $1$.
Addendum
You are assuming that what you read accurately expressed the intent. It is my contention that the text that you quoted intends a Mathematically valid exchange but is very poorly written.
It is my contention that the most likely intent is that you are the bookmaker. The bettor automatically gives you either $bS$ or $(1-b)S$, before the outcome of the event is determined.
Then, if the bettor wins, you pay the bettor $S$.
This means that if the bettor wins, he will receive $S$ back total after first giving you either $bS$ or $(1-b)S$. This is in fact standard practice with sportsbooks in Casinos. That is, the Sportsbooks never trust the bettor, but instead make the bettor put up the money that he is risking, in advance.
As an example, suppose that a sportbook requires you to lay $11$ to $10$ on a pick-em bet. This means that regardless of which side that you bet, you have to put up $11\$$ to win $10\$$.
Then, if you win, you receive $21\$$, which would represent a net profit to you of $10\$$, if you win. Horse racing is the exact same. You have to purchase the ticket in advance. If the horse that you bet on wins, then you can cash your ticket.
So, my answer reflected my surmise that the author that you quoted is not Mathematically inept, but (more likely) simply writes very poorly (or carelessly).
That is, I am surmising that the intent was that the bettor gives you either $bS$ or $(1-b)S$, in advance. Then, if he wins, he gets back $S$. Otherwise, he gets back nothing.
Assuming that my surmise is correct, the effect is the same as the bettor giving you the $bS$ or $(1-b)S$ only if he loses, and collecting from you $(1-b)S$ or $bS$ if he wins.
This is because (for example) the bettor paying $bS$ and then getting back $S$ is equivalent to the bettor winning $S - bS = (1-b)S.$
So, the theory indicated in my original answer is tenable and likely. That is, it is more likely that the author writes poorly, than that the author does not understand arithmetic.
Best Answer
For $3$ particular songs to appear in the first $6$ songs in a shuffle, all we need is to choose rest of the $3$ songs from the remaining $152$ songs.
If you knew before the first shuffle that you were looking for $3$ particular songs - $A, B, C$, then the probability that you get these $3$ songs in both the shuffles is,
$\left[\displaystyle {152 \choose 3}/{155 \choose 6}\right]^2$
But if you marked $3$ particular songs out of $6$ as $A, B, C$ after the first shuffle (which is what the question seems to suggest), then it is just,
$\displaystyle {152 \choose 3}/{155 \choose 6}$