Why can use a combinations method? A particular iPod playlist contains $100$ songs, $10$ of which are by the Beatles. Suppose the shuffle feature is used to play the songs in random order. What is the probability that the first Beatles song heard is the fifth song played?
The answer is $\frac{{95 \choose 9}}{{100 \choose 10}}$. I don't understand why we can use combinations when the ordering matters (I could choose the first $10$ songs to be Beatles songs but if I reorder those $10$ its a different outcome).
 A: All that matters for this problem is, out of the 100 songs, which of them are Beatle's song.  While we could think about having 100 unique songs, 10 of which are Beatle's songs and 90 of which are not, if we say which of the 100 songs are Beatle's songs, then there are exactly 10!90! ways to assign the songs consistent with that observation.  Because this factor does not depend on which slots are Beatle's songs, it will cancel out when we calculate the overall probability.
There are $\binom{100}{10}$ ways to pick which of the $100$ songs in the shuffled list are Beatle's songs.  How many ways are there to have the first Beatle's song be the fifth song played?  We know that the first 4 songs are not Beatle's songs, but the fifth one is, but we have 95 slots left, 9 on which will be Beatle's songs. There are therefore $\binom{95}{9}$ choices satisfying this condition.
Finally, taking the quotient of the number of choices we want over the number of choices there are yields
$$\frac{\binom{95}{9}}{\binom{100}{10}}$$

Alternatively, one can instead try to count the total number of sequences satisfying the requirement and divide by $100!$, the total number of sequences of songs.
We can generate the sequences according to the following process.  First, we pick which $10$ places in our list have Beatle's songs.  Then we pick an ordering for the Beatle's songs.  Then we pick an ordering for the other $90$ songs.  If we have no conditions for which songs are which, there are $\binom{100}{10}10!90!=100!$ ways to do this, which agrees with just taking all 100 songs and shuffling them.  However, if we insist that the first 4 songs are non-Beatles and the 5th song is Beatles, then the same argument as before shows there are $\binom{95}{9}$ ways to pick where the Beatles songs go, yielding $\binom{95}{9}10!90!$ ways to order the songs.  Calculating the quotient to find the probability, we get
$$\frac{\binom{95}{9}10!90!}{100!}$$
which, after dividing the numerator and denominator both by $10!90!$, yields the same answer as before.

One can also analyze the problem sequentially instead.  The probability that the first song isn't a Beatle's song is 90/100.  For the second song, there are now 10 Beatles songs and 89 non-Beatles songs to choose from, so the probability the second song isn't a Beatle's song is 89/99.  Now there are only 98 songs left, 88 of which are aren't Beatle's songs.  So the probability the third song isn't a Beatle's song is 88/98.  Now there are 97 songs left, 87 of which aren't Beatle's songs.  So the probability that the 4th song isn't a Beatle's song is 87/97.  Now there are 96 songs left, 10 of which are Beatle's songs.  So the probability that this IS a Beatle's son is 10/96.  This gives an overall probability of $$\frac{90}{100}\frac{89}{99}\frac{88}{98}\frac{87}{97}\frac{10}{96}.$$
One can verify that this is equal to the other answer above.  However, there are plenty of probability problems where a sequential process can quickly become too complicated to analyze sequentially, and having a birds eye view or viewing it as equivalent to a different process (e.g., first pick which songs will be Beatle's songs, then put the 10 songs in some order, then put the other 90 songs in some order) yields a much cleaner analysis.
A: This is not actually randomly picking a song from a list each time, instead it only plays each song once in a random arrangement.
So there are $100\choose10$ ways to arrange the ten Beatles songs within this "random" order, which includes playing a Beatles first, or all ten right at the end. There is only one way of playing four non-Beatles followed by one Beatles song. Then there are $95\choose9$ ways of ordering the remaining nine Beatles songs among the remaining 95 songs after the first four non-Beatles and then the one Beatles song have played.
The answer of $${2581\over38024}\approx0.068$$ is about what you would expect from a genuinely random selection where the probability of four non-Beatles followed by one Beatles song is $0.9^4\times0.1=0.06561$.
