Why does order matter in first scenario but not in second Consider two scenarios:

*

*A playlist is created by a random selection of 100 songs that have 10 Beatles songs. What is the probability that the first beatle song played is the 5th song.


*From a total of 15 bulbs, in which 6 are 75 W bulb, if 6 bulbs are drawn what is the probability that the first 5 bulbs would not be 75W bulb while the 6th bulb could be or could not be 75W bulb
To me, both the scenarios seem extremely similar. In Beatles problem, you have to fill 4 slots with random songs but not Beatles while in the bulbs, you have to fill 5 slots with random bulbs but not 75W
It doesn't matter what the 4 non-beatle songs are or what the 5 bulbs are.
Why is order important here? I have looked at the solution and yes Beatles problem is indeed a permutation problem (although I am not sure whether the bulbs one is permutation or combination) but I can't wrap my head around why is it permutation.
My answer would have been 90C4 * 10 / 100C5 but it's actually 90P4 * 10 / 100P5
Can someone please help?
P.S. I am very new to this site and I know similar questions have been asked and it'd have been better to comment over there but it's not letting me comment due to low reputation. Sorry :)
 A: I disagree with the comments.
Let $S = \displaystyle \frac{\binom{90}{4} \times \binom{10}{1}}{\binom{100}{5}}.$
Let $T = \displaystyle \frac{\left[\binom{90}{4} \times 4!\right] \times \left[\binom{10}{1} \times 1!\right]}{\binom{100}{5} \times 5!}.$
Then, $S$ is not equal to $T$.
Order is important, so $T$ is correct.  $T$ computes the probability that the 1st $4$ specifically are not Beatles songs, while the $5$-th song specifically is a Beatles song.
$S$, which is incorrect, computes the probability that when $5$ songs are selected at random, exactly one of the $5$ will be a Beatles song.  This is not correct, because it allows the Beatles song to occur in any of the positions $1$ through $5$.  The problem specifically requires that the Beatles song occur in position $5$ exactly.
A: 
A playlist is created by a random selection of 100 songs that have 10 Beatles songs. What is the probability that the first beatle song played is the 5th song.

When counting permutations we seek the probability for obtaining an arrangement of 4 from 90 other songs then 1 from 10 beetle song (in fifth position), when selecting an arrangement of any 5 from 10 songs."$$\dfrac{{^{90}P_4}~{^{10}P_1}}{^{100}P_5}$$
When counting combinations, we need to treat the fifth place as special: So we seek the probability for obtaining 4 from 90 other songs and 1 from 10 beetle song and placing that in the fifth position, when selecting 4 from 90 songs and 1 from the remaining 96 to place in fifth position.
$$\dfrac{{^{90}C_4}~{^{10}C_1}}{{^{100}C_4}~{^{96}C_1}}$$
We could also seek the probability for obtaining 4 from 90 other songs and 1 from 10 beetle song to place in the fifth position, when selecting 5 from 90 songs choosing 1 among those to place in fifth position.
$$\dfrac{{^{90}C_4}~{^{10}C_1}}{{^{100}C_5}~{^{5}C_1}}$$

Of course, all of these probabilities are equal.
$$\begin{align}\dfrac{{^{90}P_4}~{^{10}P_1}}{^{100}P_5} &= \dfrac{{^{90}C_4~4!}~{^{10}C_1~1!}}{^{100}C_5~5!}\\[1ex]&=\dfrac{{^{90}C_4}~{^{10}C_1 }}{^{100}C_5 ~{^5C_1}}\\[1ex]&=\dfrac{{^{90}C_4}~{^{10}C_1 }}{^{100}C_4 ~{^{96}C_1}}\end{align}$$
