Mixing
Doubt regarding William Feller's combinatorics problem of indistinguishable objects
$begingroup$
In William Feller's Probability theory and applications he says:
...Suppose next that p among the flags are red and q are blue (where p + q = r) It is easily seen that every display of r numbered flags can be obtain by numbering the red flags from 1 to p and the blue flags p+1 to p+q. It follows that the number of different displays is now N/(p!q!).
It is not easily seen for me that every display of r numbered flags can be obtained from numbering the red flags in such a way. Can you please explain to me how is it that every possible arrangement of r flags can be obtained in such a way?
The way I would calculate the different ways I can arrange such objects would be
$binom{r}{p} * binom{r-p}{q}$.
Context: The original problem said that you are supposed to assumed that r flags can be displayed on n poles in N=n(n+1)...(n+r-1) ways
probability combinatorics
asked Jan 30 at 12:31
César D. VázquezCésar D. Vázquez
1768
$endgroup$
add a comment |
$begingroup$
In William Feller's Probability theory and applications he says:
...Suppose next that p among the flags are red and q are blue (where p + q = r) It is easily seen that every display of r numbered flags can be obtain by numbering the red flags from 1 to p and the blue flags p+1 to p+q. It follows that the number of different displays is now N/(p!q!).
It is not easily seen for me that every display of r numbered flags can be obtained from numbering the red flags in such a way. Can you please explain to me how is it that every possible arrangement of r flags can be obtained in such a way?
The way I would calculate the different ways I can arrange such objects would be
$binom{r}{p} * binom{r-p}{q}$.
Context: The original problem said that you are supposed to assumed that r flags can be displayed on n poles in N=n(n+1)...(n+r-1) ways
probability combinatorics
asked Jan 30 at 12:31
César D. VázquezCésar D. Vázquez
1768
$endgroup$
$begingroup$
You seem to have a typo in the last formula (a missing paren?) Also, for context, it might be good to explain what a "display" is, and what $N$ is defined to be.
$endgroup$
– John Hughes
Jan 30 at 12:41
$begingroup$
Done @JohnHughes. I'm sorry, I thought it was irrelevant for the problem. Thanks
$endgroup$
– César D. Vázquez
Feb 15 at 15:21
add a comment |
$begingroup$
In William Feller's Probability theory and applications he says:
...Suppose next that p among the flags are red and q are blue (where p + q = r) It is easily seen that every display of r numbered flags can be obtain by numbering the red flags from 1 to p and the blue flags p+1 to p+q. It follows that the number of different displays is now N/(p!q!).
It is not easily seen for me that every display of r numbered flags can be obtained from numbering the red flags in such a way. Can you please explain to me how is it that every possible arrangement of r flags can be obtained in such a way?
The way I would calculate the different ways I can arrange such objects would be
$binom{r}{p} * binom{r-p}{q}$.
Context: The original problem said that you are supposed to assumed that r flags can be displayed on n poles in N=n(n+1)...(n+r-1) ways
probability combinatorics
asked Jan 30 at 12:31
César D. VázquezCésar D. Vázquez
1768
$endgroup$
In William Feller's Probability theory and applications he says:
...Suppose next that p among the flags are red and q are blue (where p + q = r) It is easily seen that every display of r numbered flags can be obtain by numbering the red flags from 1 to p and the blue flags p+1 to p+q. It follows that the number of different displays is now N/(p!q!).
It is not easily seen for me that every display of r numbered flags can be obtained from numbering the red flags in such a way. Can you please explain to me how is it that every possible arrangement of r flags can be obtained in such a way?
The way I would calculate the different ways I can arrange such objects would be
$binom{r}{p} * binom{r-p}{q}$.
Context: The original problem said that you are supposed to assumed that r flags can be displayed on n poles in N=n(n+1)...(n+r-1) ways
probability combinatorics
probability combinatorics
asked Jan 30 at 12:31
César D. VázquezCésar D. Vázquez
1768
asked Jan 30 at 12:31
César D. VázquezCésar D. Vázquez
1768
edited Feb 15 at 15:20
César D. Vázquez
asked Jan 30 at 12:31
César D. VázquezCésar D. Vázquez
1768
asked Jan 30 at 12:31
César D. VázquezCésar D. Vázquez
1768
asked Jan 30 at 12:31
César D. VázquezCésar D. Vázquez
1768
1768
$begingroup$
You seem to have a typo in the last formula (a missing paren?) Also, for context, it might be good to explain what a "display" is, and what $N$ is defined to be.
$endgroup$
– John Hughes
Jan 30 at 12:41
$begingroup$
Done @JohnHughes. I'm sorry, I thought it was irrelevant for the problem. Thanks
$endgroup$
– César D. Vázquez
Feb 15 at 15:21
add a comment |
$begingroup$
You seem to have a typo in the last formula (a missing paren?) Also, for context, it might be good to explain what a "display" is, and what $N$ is defined to be.
$endgroup$
– John Hughes
Jan 30 at 12:41
$begingroup$
Done @JohnHughes. I'm sorry, I thought it was irrelevant for the problem. Thanks
$endgroup$
– César D. Vázquez
Feb 15 at 15:21
$begingroup$
You seem to have a typo in the last formula (a missing paren?) Also, for context, it might be good to explain what a "display" is, and what $N$ is defined to be.
$endgroup$
– John Hughes
Jan 30 at 12:41
$begingroup$
You seem to have a typo in the last formula (a missing paren?) Also, for context, it might be good to explain what a "display" is, and what $N$ is defined to be.
$endgroup$
– John Hughes
Jan 30 at 12:41
$begingroup$
Done @JohnHughes. I'm sorry, I thought it was irrelevant for the problem. Thanks
$endgroup$
– César D. Vázquez
Feb 15 at 15:21
$begingroup$
Done @JohnHughes. I'm sorry, I thought it was irrelevant for the problem. Thanks
$endgroup$
– César D. Vázquez
Feb 15 at 15:21
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let's use numbers for the red flags, and negative numbers for the blue. And let's look at 4 flags, and start with $p = 3, q = 1$.
A "display" of the flags is, I'm guessing, a pattern of red and blue, so we can draw the possibilities as
+++-
++-+
+-++
-+++
in this case: the blue flag can go in only one of four places. If we actually number the flags, the first display shows up with many possible numbering schemes:
1 2 3 -1
1 3 2 -1
2 1 3 -1
2 3 1 -1
3 1 2 -1
3 2 1 -1
There are $3!$ arrangements of the red flags, and $1! = 1$ arrangements of the blue, all corresponding to one "display". If we wrote out all four displays, we'd have a total of $24$ arrangements. I'm assuming that whatever Feller is calling $N$ amounts to $24$ in this situation, and that by dividing $N$ by $3! 1$, we get the four "displays."
answered Jan 30 at 12:40
John HughesJohn Hughes
65.2k24293
$endgroup$
add a comment |
$begingroup$
What is meant here is not that you obtain a different pattern of red vs blue, with each of these numberings, certainly this is not the case.
What is meant is that:
- each of the arrangements of the red flags within the list of red flags gives the same pattern.
- as does the each arrangement of the blue ones.
The number of these arrangements is indeed the number of ways of ordering numbers that correspond to them. I think this is unhelpfully worded as it implies something clever is going on. All it means is the ways of arranging a given number of objects is the same no matter what you call them.
Perhaps a clearer way to put the whole thing would be:
To count the ways of arranging $p$ read flags and $q$ blue flags is the same as the number of ways of arranging $p+q$ flags if they where all unique, divided by the number of orderings of the unique flags that give the same pattern.
The clever bit is to notice that this ordering of the red and blue flags are independent so it suffices to know how many ordering of each independently and multiply them together.
So the formula becomes:
The number of ways of arranging $p + q$ objects, divided by the number of ways of arranging $p$ objects times the number of ways of arranging $q$ objects.
answered Jan 30 at 16:18
drjpizzledrjpizzle
1712
$endgroup$
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%2f3093464%2fdoubt-regarding-william-fellers-combinatorics-problem-of-indistinguishable-obje%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's use numbers for the red flags, and negative numbers for the blue. And let's look at 4 flags, and start with $p = 3, q = 1$.
A "display" of the flags is, I'm guessing, a pattern of red and blue, so we can draw the possibilities as
+++-
++-+
+-++
-+++
in this case: the blue flag can go in only one of four places. If we actually number the flags, the first display shows up with many possible numbering schemes:
1 2 3 -1
1 3 2 -1
2 1 3 -1
2 3 1 -1
3 1 2 -1
3 2 1 -1
There are $3!$ arrangements of the red flags, and $1! = 1$ arrangements of the blue, all corresponding to one "display". If we wrote out all four displays, we'd have a total of $24$ arrangements. I'm assuming that whatever Feller is calling $N$ amounts to $24$ in this situation, and that by dividing $N$ by $3! 1$, we get the four "displays."
answered Jan 30 at 12:40
John HughesJohn Hughes
65.2k24293
$endgroup$
add a comment |
$begingroup$
Let's use numbers for the red flags, and negative numbers for the blue. And let's look at 4 flags, and start with $p = 3, q = 1$.
A "display" of the flags is, I'm guessing, a pattern of red and blue, so we can draw the possibilities as
+++-
++-+
+-++
-+++
in this case: the blue flag can go in only one of four places. If we actually number the flags, the first display shows up with many possible numbering schemes:
1 2 3 -1
1 3 2 -1
2 1 3 -1
2 3 1 -1
3 1 2 -1
3 2 1 -1
There are $3!$ arrangements of the red flags, and $1! = 1$ arrangements of the blue, all corresponding to one "display". If we wrote out all four displays, we'd have a total of $24$ arrangements. I'm assuming that whatever Feller is calling $N$ amounts to $24$ in this situation, and that by dividing $N$ by $3! 1$, we get the four "displays."
answered Jan 30 at 12:40
John HughesJohn Hughes
65.2k24293
$endgroup$
add a comment |
$begingroup$
Let's use numbers for the red flags, and negative numbers for the blue. And let's look at 4 flags, and start with $p = 3, q = 1$.
A "display" of the flags is, I'm guessing, a pattern of red and blue, so we can draw the possibilities as
+++-
++-+
+-++
-+++
in this case: the blue flag can go in only one of four places. If we actually number the flags, the first display shows up with many possible numbering schemes:
1 2 3 -1
1 3 2 -1
2 1 3 -1
2 3 1 -1
3 1 2 -1
3 2 1 -1
There are $3!$ arrangements of the red flags, and $1! = 1$ arrangements of the blue, all corresponding to one "display". If we wrote out all four displays, we'd have a total of $24$ arrangements. I'm assuming that whatever Feller is calling $N$ amounts to $24$ in this situation, and that by dividing $N$ by $3! 1$, we get the four "displays."
answered Jan 30 at 12:40
John HughesJohn Hughes
65.2k24293
$endgroup$
Let's use numbers for the red flags, and negative numbers for the blue. And let's look at 4 flags, and start with $p = 3, q = 1$.
A "display" of the flags is, I'm guessing, a pattern of red and blue, so we can draw the possibilities as
+++-
++-+
+-++
-+++
in this case: the blue flag can go in only one of four places. If we actually number the flags, the first display shows up with many possible numbering schemes:
1 2 3 -1
1 3 2 -1
2 1 3 -1
2 3 1 -1
3 1 2 -1
3 2 1 -1
There are $3!$ arrangements of the red flags, and $1! = 1$ arrangements of the blue, all corresponding to one "display". If we wrote out all four displays, we'd have a total of $24$ arrangements. I'm assuming that whatever Feller is calling $N$ amounts to $24$ in this situation, and that by dividing $N$ by $3! 1$, we get the four "displays."
answered Jan 30 at 12:40
John HughesJohn Hughes
65.2k24293
answered Jan 30 at 12:40
John HughesJohn Hughes
65.2k24293
answered Jan 30 at 12:40
John HughesJohn Hughes
65.2k24293
answered Jan 30 at 12:40
John HughesJohn Hughes
65.2k24293
65.2k24293
add a comment |
add a comment |
$begingroup$
What is meant here is not that you obtain a different pattern of red vs blue, with each of these numberings, certainly this is not the case.
What is meant is that:
- each of the arrangements of the red flags within the list of red flags gives the same pattern.
- as does the each arrangement of the blue ones.
The number of these arrangements is indeed the number of ways of ordering numbers that correspond to them. I think this is unhelpfully worded as it implies something clever is going on. All it means is the ways of arranging a given number of objects is the same no matter what you call them.
Perhaps a clearer way to put the whole thing would be:
To count the ways of arranging $p$ read flags and $q$ blue flags is the same as the number of ways of arranging $p+q$ flags if they where all unique, divided by the number of orderings of the unique flags that give the same pattern.
The clever bit is to notice that this ordering of the red and blue flags are independent so it suffices to know how many ordering of each independently and multiply them together.
So the formula becomes:
The number of ways of arranging $p + q$ objects, divided by the number of ways of arranging $p$ objects times the number of ways of arranging $q$ objects.
answered Jan 30 at 16:18
drjpizzledrjpizzle
1712
$endgroup$
add a comment |
$begingroup$
What is meant here is not that you obtain a different pattern of red vs blue, with each of these numberings, certainly this is not the case.
What is meant is that:
- each of the arrangements of the red flags within the list of red flags gives the same pattern.
- as does the each arrangement of the blue ones.
The number of these arrangements is indeed the number of ways of ordering numbers that correspond to them. I think this is unhelpfully worded as it implies something clever is going on. All it means is the ways of arranging a given number of objects is the same no matter what you call them.
Perhaps a clearer way to put the whole thing would be:
To count the ways of arranging $p$ read flags and $q$ blue flags is the same as the number of ways of arranging $p+q$ flags if they where all unique, divided by the number of orderings of the unique flags that give the same pattern.
The clever bit is to notice that this ordering of the red and blue flags are independent so it suffices to know how many ordering of each independently and multiply them together.
So the formula becomes:
The number of ways of arranging $p + q$ objects, divided by the number of ways of arranging $p$ objects times the number of ways of arranging $q$ objects.
answered Jan 30 at 16:18
drjpizzledrjpizzle
1712
$endgroup$
add a comment |
$begingroup$
What is meant here is not that you obtain a different pattern of red vs blue, with each of these numberings, certainly this is not the case.
What is meant is that:
- each of the arrangements of the red flags within the list of red flags gives the same pattern.
- as does the each arrangement of the blue ones.
The number of these arrangements is indeed the number of ways of ordering numbers that correspond to them. I think this is unhelpfully worded as it implies something clever is going on. All it means is the ways of arranging a given number of objects is the same no matter what you call them.
Perhaps a clearer way to put the whole thing would be:
To count the ways of arranging $p$ read flags and $q$ blue flags is the same as the number of ways of arranging $p+q$ flags if they where all unique, divided by the number of orderings of the unique flags that give the same pattern.
The clever bit is to notice that this ordering of the red and blue flags are independent so it suffices to know how many ordering of each independently and multiply them together.
So the formula becomes:
The number of ways of arranging $p + q$ objects, divided by the number of ways of arranging $p$ objects times the number of ways of arranging $q$ objects.
answered Jan 30 at 16:18
drjpizzledrjpizzle
1712
$endgroup$
What is meant here is not that you obtain a different pattern of red vs blue, with each of these numberings, certainly this is not the case.
What is meant is that:
- each of the arrangements of the red flags within the list of red flags gives the same pattern.
- as does the each arrangement of the blue ones.
The number of these arrangements is indeed the number of ways of ordering numbers that correspond to them. I think this is unhelpfully worded as it implies something clever is going on. All it means is the ways of arranging a given number of objects is the same no matter what you call them.
Perhaps a clearer way to put the whole thing would be:
To count the ways of arranging $p$ read flags and $q$ blue flags is the same as the number of ways of arranging $p+q$ flags if they where all unique, divided by the number of orderings of the unique flags that give the same pattern.
The clever bit is to notice that this ordering of the red and blue flags are independent so it suffices to know how many ordering of each independently and multiply them together.
So the formula becomes:
The number of ways of arranging $p + q$ objects, divided by the number of ways of arranging $p$ objects times the number of ways of arranging $q$ objects.
answered Jan 30 at 16:18
drjpizzledrjpizzle
1712
answered Jan 30 at 16:18
drjpizzledrjpizzle
1712
answered Jan 30 at 16:18
drjpizzledrjpizzle
1712
answered Jan 30 at 16:18
drjpizzledrjpizzle
1712
1712
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%2f3093464%2fdoubt-regarding-william-fellers-combinatorics-problem-of-indistinguishable-obje%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
$begingroup$
You seem to have a typo in the last formula (a missing paren?) Also, for context, it might be good to explain what a "display" is, and what $N$ is defined to be.
$endgroup$
– John Hughes
Jan 30 at 12:41
$begingroup$
Done @JohnHughes. I'm sorry, I thought it was irrelevant for the problem. Thanks
$endgroup$
– César D. Vázquez
Feb 15 at 15:21