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Consider two scenarios:

  1. 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.

  2. 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 :)

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  • $\begingroup$ Those answers are both correct. They are the same. $\endgroup$ Sep 16 at 2:23
  • $\begingroup$ 90C4 * 10 / 100C5 is equal to 90P4 * 10 / 100P5. Either way of counting gives the same answer. $\endgroup$ Sep 16 at 2:25
  • $\begingroup$ yes I do realize that both answers in the end come out to be correct but that's only because the denominators cancel out. But it's technically incorrect to use combination instead of permutation and I am trying to understand why. And I still don't know what the solution for the bulbs question should be. To me it just seems like 9C5 but if order is important in Beatles songs question, it should be important here as well in which case it'd be 9P5. $\endgroup$ Sep 16 at 2:27
  • $\begingroup$ "but its technically incorrect to use combination instead of permutation" Says who? For probability scenarios we get to choose what sample space we use to describe the scenario. It is a choice. So long as we chose appropriately, there can be multiple choices. $\endgroup$
    – JMoravitz
    Sep 16 at 2:40
  • $\begingroup$ Umm okay so, in words, probability for scenario 1, is total ways in which the first Beatles song is the 5th song / total ways of choosing 5 from 100 What I just said is permutation or combination? Why? That's what I am trying to understand as in, even though they both yield the same answer, what's the difference conceptually? Can anyone put the two solutions in actual words highlighting the conceptual difference please? I am sorry I am not able to exactly construct the question I have $\endgroup$ Sep 16 at 2:47
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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.

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  • $\begingroup$ This makes much more sense to me. I am just not following why is it 4! and 5! in T and not just 4 and 5. Could you walk me through that part please? $\endgroup$ Sep 16 at 4:32
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    $\begingroup$ @ShivamUpadhyay The Permutation is the Combination times the ways to arrange the selection (which is the factorial). ${^nP_r} = {^nC_r}\cdot r!$ $\endgroup$ Sep 16 at 4:34
  • $\begingroup$ @ShivamUpadhyay In general, for $A, B \in \Bbb{Z^+} ~: ~A > B$, you have that $${}^AP_B = A \times (A-1) \times \cdots \times (A + 1 - B).$$ This equals $$\frac{A!}{(A - B)!} = \binom{A}{B} \times B!.$$ $\endgroup$ Sep 16 at 4:37
  • $\begingroup$ I understood the formula I think. But can anyone actually put it in words what the factorial part represents? 90C4 is the ways to choose 4 out of 90. What is the factorial component interpreted as? @GrahamKemp I did not full follow what you meant sorry $\endgroup$ Sep 16 at 4:40
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    $\begingroup$ @ShivamUpadhyay: The intution behind that is you choose any $r$ objects out of the $n$ total, this is $\binom nr$. Now, you have $r$ "slots", you can put the first object in any.of the $r$ slots, the second in the remaining $r-1$ and so on. Thus, for permuting $r$ objects from the $n$, you get $$\binom nr\times\Big(r(r-1)(r-2)\cdots3\cdot2\cdot1\Big)=\binom nr \cdot r!$$ $\endgroup$ Sep 16 at 5:20
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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.

$$\begin{align}\dfrac{{^{90}P_4}~{^{10}P_1}}{^{100}P_5} &= \dfrac{{^{90}C_4~4!}~{^{10}C_1~1!}}{^{100}C_5~5!}\\[1ex]&=\dfrac{{^{90}C_4}~{^{10}C_1 }}{^{100}C_5 ~{^5C_1}}\\[1ex]&=\dfrac{{^{90}C_4}~{^{10}C_1 }}{^{100}C_4 ~{^{96}C_1}}\end{align}$$

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