Question about permutation?
$begingroup$
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.
permutations
asked Dec 14 '18 at 16:46
gi2302gi2302
103
$endgroup$
add a comment |
$begingroup$
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.
permutations
asked Dec 14 '18 at 16:46
gi2302gi2302
103
$endgroup$
$begingroup$
1) is correct, assumingone 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), 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?
$endgroup$
– Bram28
Dec 14 '18 at 16:52
add a comment |
$begingroup$
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.
permutations
asked Dec 14 '18 at 16:46
gi2302gi2302
103
$endgroup$
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.
permutations
permutations
asked Dec 14 '18 at 16:46
gi2302gi2302
103
asked Dec 14 '18 at 16:46
gi2302gi2302
103
asked Dec 14 '18 at 16:46
gi2302gi2302
103
asked Dec 14 '18 at 16:46
gi2302gi2302
103
asked Dec 14 '18 at 16:46
gi2302gi2302
103
103
$begingroup$
1) is correct, assumingone 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), 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?
$endgroup$
– Bram28
Dec 14 '18 at 16:52
add a comment |
$begingroup$
1) is correct, assumingone 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), 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?
$endgroup$
– Bram28
Dec 14 '18 at 16:52
$begingroup$
1) is correct, assumingone 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), 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?
$endgroup$
– Bram28
Dec 14 '18 at 16:52
$begingroup$
1) is correct, assumingone 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), 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?
$endgroup$
– Bram28
Dec 14 '18 at 16:52
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
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.
answered Dec 14 '18 at 16:51
Ross MillikanRoss Millikan
295k23198371
$endgroup$
add a comment |
$begingroup$
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.)
answered Dec 14 '18 at 16:52
Ethan BolkerEthan Bolker
42.7k549113
$endgroup$
add a comment |
$begingroup$
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)...?
answered Dec 14 '18 at 16:52
StackTDStackTD
22.7k2050
$endgroup$
$begingroup$
I'm assuming, a person can't win more than one dog.
$endgroup$
– gi2302
Dec 14 '18 at 17:33
$begingroup$
What u mean with distinguishable...?
$endgroup$
– gi2302
Dec 14 '18 at 17:35
$begingroup$
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.
$endgroup$
– StackTD
Dec 14 '18 at 17:35
$begingroup$
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!).
$endgroup$
– StackTD
Dec 14 '18 at 17:36
add a comment |
$begingroup$
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?
answered Dec 14 '18 at 16:57
Bram28Bram28
62.1k44793
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3039605%2fquestion-about-permutation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Dec 14 '18 at 16:51
Ross MillikanRoss Millikan
295k23198371
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 14 '18 at 16:51
Ross MillikanRoss Millikan
295k23198371
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 14 '18 at 16:51
Ross MillikanRoss Millikan
295k23198371
$endgroup$
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.
answered Dec 14 '18 at 16:51
Ross MillikanRoss Millikan
295k23198371
answered Dec 14 '18 at 16:51
Ross MillikanRoss Millikan
295k23198371
answered Dec 14 '18 at 16:51
Ross MillikanRoss Millikan
295k23198371
answered Dec 14 '18 at 16:51
Ross MillikanRoss Millikan
295k23198371
295k23198371
add a comment |
add a comment |
$begingroup$
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.)
answered Dec 14 '18 at 16:52
Ethan BolkerEthan Bolker
42.7k549113
$endgroup$
add a comment |
$begingroup$
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.)
answered Dec 14 '18 at 16:52
Ethan BolkerEthan Bolker
42.7k549113
$endgroup$
add a comment |
$begingroup$
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.)
answered Dec 14 '18 at 16:52
Ethan BolkerEthan Bolker
42.7k549113
$endgroup$
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.)
answered Dec 14 '18 at 16:52
Ethan BolkerEthan Bolker
42.7k549113
answered Dec 14 '18 at 16:52
Ethan BolkerEthan Bolker
42.7k549113
answered Dec 14 '18 at 16:52
Ethan BolkerEthan Bolker
42.7k549113
answered Dec 14 '18 at 16:52
Ethan BolkerEthan Bolker
42.7k549113
42.7k549113
add a comment |
add a comment |
$begingroup$
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)...?
answered Dec 14 '18 at 16:52
StackTDStackTD
22.7k2050
$endgroup$
$begingroup$
I'm assuming, a person can't win more than one dog.
$endgroup$
– gi2302
Dec 14 '18 at 17:33
$begingroup$
What u mean with distinguishable...?
$endgroup$
– gi2302
Dec 14 '18 at 17:35
$begingroup$
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.
$endgroup$
– StackTD
Dec 14 '18 at 17:35
$begingroup$
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!).
$endgroup$
– StackTD
Dec 14 '18 at 17:36
add a comment |
$begingroup$
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)...?
answered Dec 14 '18 at 16:52
StackTDStackTD
22.7k2050
$endgroup$
$begingroup$
I'm assuming, a person can't win more than one dog.
$endgroup$
– gi2302
Dec 14 '18 at 17:33
$begingroup$
What u mean with distinguishable...?
$endgroup$
– gi2302
Dec 14 '18 at 17:35
$begingroup$
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.
$endgroup$
– StackTD
Dec 14 '18 at 17:35
$begingroup$
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!).
$endgroup$
– StackTD
Dec 14 '18 at 17:36
add a comment |
$begingroup$
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)...?
answered Dec 14 '18 at 16:52
StackTDStackTD
22.7k2050
$endgroup$
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)...?
answered Dec 14 '18 at 16:52
StackTDStackTD
22.7k2050
answered Dec 14 '18 at 16:52
StackTDStackTD
22.7k2050
answered Dec 14 '18 at 16:52
StackTDStackTD
22.7k2050
answered Dec 14 '18 at 16:52
StackTDStackTD
22.7k2050
22.7k2050
$begingroup$
I'm assuming, a person can't win more than one dog.
$endgroup$
– gi2302
Dec 14 '18 at 17:33
$begingroup$
What u mean with distinguishable...?
$endgroup$
– gi2302
Dec 14 '18 at 17:35
$begingroup$
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.
$endgroup$
– StackTD
Dec 14 '18 at 17:35
$begingroup$
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!).
$endgroup$
– StackTD
Dec 14 '18 at 17:36
add a comment |
$begingroup$
I'm assuming, a person can't win more than one dog.
$endgroup$
– gi2302
Dec 14 '18 at 17:33
$begingroup$
What u mean with distinguishable...?
$endgroup$
– gi2302
Dec 14 '18 at 17:35
$begingroup$
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.
$endgroup$
– StackTD
Dec 14 '18 at 17:35
$begingroup$
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!).
$endgroup$
– StackTD
Dec 14 '18 at 17:36
$begingroup$
I'm assuming, a person can't win more than one dog.
$endgroup$
– gi2302
Dec 14 '18 at 17:33
$begingroup$
I'm assuming, a person can't win more than one dog.
$endgroup$
– gi2302
Dec 14 '18 at 17:33
$begingroup$
What u mean with distinguishable...?
$endgroup$
– gi2302
Dec 14 '18 at 17:35
$begingroup$
What u mean with distinguishable...?
$endgroup$
– gi2302
Dec 14 '18 at 17:35
$begingroup$
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.
$endgroup$
– StackTD
Dec 14 '18 at 17:35
$begingroup$
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.
$endgroup$
– StackTD
Dec 14 '18 at 17:35
$begingroup$
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!).
$endgroup$
– StackTD
Dec 14 '18 at 17:36
$begingroup$
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!).
$endgroup$
– StackTD
Dec 14 '18 at 17:36
add a comment |
$begingroup$
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?
answered Dec 14 '18 at 16:57
Bram28Bram28
62.1k44793
$endgroup$
add a comment |
$begingroup$
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?
answered Dec 14 '18 at 16:57
Bram28Bram28
62.1k44793
$endgroup$
add a comment |
$begingroup$
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?
answered Dec 14 '18 at 16:57
Bram28Bram28
62.1k44793
$endgroup$
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?
answered Dec 14 '18 at 16:57
Bram28Bram28
62.1k44793
answered Dec 14 '18 at 16:57
Bram28Bram28
62.1k44793
answered Dec 14 '18 at 16:57
Bram28Bram28
62.1k44793
answered Dec 14 '18 at 16:57
Bram28Bram28
62.1k44793
62.1k44793
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3039605%2fquestion-about-permutation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
1) is correct, assumingone 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), 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?
$endgroup$
– Bram28
Dec 14 '18 at 16:52