Question about permutation?

Question

Can someone help me with this permutation exercises...

1) In a NY parade, there are 8 floats and 3 orchestras. How many possible orders are there?

Ok. Tell me if I'm right. There is no repetition in here. So if I'm going to line them up this will be a $P(11,11)=\frac{11!}{(11-11)!}=\frac{11!}{0!}=\frac{11!}{1}=39,916,800$. But I don't know this is a huge number. So I don't know if make sense at all..

2) A Labrador Retriever, a Siberian Husky, and a German Sheperd are raffled at a pet store. If there are 10 participants, how many ways can the prizes be allocated?

This one is also without replacement, so it would be

$P(10,3)=\frac{10!}{(10-3)!}=\frac{10!}{7!}=10(9)(8)=720$.

Your help will be appreciated.

score 2 · Answer 1 · answered Dec 14 '18 at 16:51

Your first is correct if you regard the floats and orchestras as distinct. I would read the problem to consider all the floats as interchangeable and the orchestras likewise. Now you just need to choose three places in line to put orchestras.

In the second the dogs are clearly distinguishable and we assume the people are, too. That makes your answer correct.

score 1 · Answer 2 · answered Dec 14 '18 at 16:52

Hints

For (1) you are assuming the floats and the orchestras are all different. If that were the case then your answer of $11!$ is right. But the question suggests that the orchestras are interchangeable, as are the floats. So find the number of positions for the orchestras.

(In the future, one question per post please.)

score 1 · Answer 3 · answered Dec 14 '18 at 16:52

1) In a NY parade, there are 8 floats and 3 orchestras. How many possible orders are there?

Ok. Tell me if I'm right. There is no repetition in here. So if I'm going to line them up this will be a $P(11,11)=\frac{11!}{(11-11)!}=\frac{11!}{0!}=\frac{11!}{1}=39,916,800$. But I don't know this is a huge number. So I don't know if make sense at all..

This is alright if all the floats and orchestras are distinguishable. If all 11 'objects' are, then you can permute them in $11!$ ways. I assume they mean that all the floats aren't distinguishable, the same for the orchestras. In that case you have to compensate for the fact that permuting the floats ($8!$ ways) and the orchestras ($3!$ ways) don't change the line-up: $$\frac{11!}{8!\;3!} = \ldots$$

2) A Labrador Retriever, a Siberian Husky, and a German Sheperd are raffled at a pet store. If there are 10 participants, how many ways can the prizes be allocated?

This one is also without replacement, so it would be

$P(10,3)=\frac{10!}{(10-3)!}=\frac{10!}{7!}=10(9)(8)=720$.

Looks good, unless a person can win more than one prize (dog)...?

Q2: Then you're right, but note that it makes a difference and that's it not that obvious that this is the correct interpretation: one person could have three winning raffle tickets, for example! You can't really tell without knowing how the raffle is organised. — StackTD, Dec 14 '18 at 17:35
Q1: that you can tell the floats (or the orchestras) apart; or to put it differently: would swapping two floats (or two orchestras) count as a different line-up, or not? If so, your work is fine. If not - which is more likely in this context - see my answer (and the other ones!). — StackTD, Dec 14 '18 at 17:36

score 0 · Answer 4 · answered Dec 14 '18 at 16:57

1) is correct ... assuming one distinguishes between the floats (and same for the orchestras) .... for example, I can see how someone would be merely interested in the distribution of the orchestras among the floats (e.g. they want to make sure there are no two orchestras next to each other ... personally, I couldn't take two orchestras in a row ...), and from that perspective they might have no interested in distinguishing between the different orchestras, or between the different floats.

For 2) .. couldn't someone win two dogs? Or all? How is this raffle organized?

Question about permutation?

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