Mixing
Number of ways of putting $n$ balls into $k$ boxes, with a twist…
$begingroup$
Usually, when counting the number of ways in which $n$ balls can be put into $k$ boxes, two cases are considered:
- Boxes cannot be empty
- Boxes can be empty
Respectively, each of the listed cases has a formula:
$$C_1={n-1choose k-1}$$
$$C_2={n+k-1choose n}$$
This however doesn't suffice my needs. I want the boxes to be all not empty, and be countable, just like in the first case... But I want the boxes to be indistinguishable.
That is, as long as you have 4 balls in box 1, 2 balls in box 2, and 1 ball in box 3, it doesn't matter how the boxes themselves are configured or labeled.
For example, if there are 7 balls and 3 boxes, there would be four such ways to arrange them:
- O|O|OOOOO
- O|OO|OOOO
- O|OOO|OOO
- OO|OO|OOO
I have no ideas how to construct a formula for this
probability combinatorics
asked Jan 26 at 4:10
KKZiomekKKZiomek
2,2541641
$endgroup$
add a comment |
$begingroup$
Usually, when counting the number of ways in which $n$ balls can be put into $k$ boxes, two cases are considered:
- Boxes cannot be empty
- Boxes can be empty
Respectively, each of the listed cases has a formula:
$$C_1={n-1choose k-1}$$
$$C_2={n+k-1choose n}$$
This however doesn't suffice my needs. I want the boxes to be all not empty, and be countable, just like in the first case... But I want the boxes to be indistinguishable.
That is, as long as you have 4 balls in box 1, 2 balls in box 2, and 1 ball in box 3, it doesn't matter how the boxes themselves are configured or labeled.
For example, if there are 7 balls and 3 boxes, there would be four such ways to arrange them:
- O|O|OOOOO
- O|OO|OOOO
- O|OOO|OOO
- OO|OO|OOO
I have no ideas how to construct a formula for this
probability combinatorics
asked Jan 26 at 4:10
KKZiomekKKZiomek
2,2541641
$endgroup$
3
$begingroup$
What you are looking for is the number of partitions of $n$ into $k$ parts: whitman.edu/mathematics/cgt_online/book/section03.03.html. This is complicated to solve. I don't believe there is a nice formula for it, but there is a recurrence relation $p_k(n)=p_k(n-k)+p_{k-1}(n-1)$.
$endgroup$
– kccu
Jan 26 at 4:22
$begingroup$
@kccu I didn't recognize this as partitions, wow. But, didn't Ramanujan find an infinite series that converges to the solution of specific partitions? As far as I can remember, that series was very crazy and messy
$endgroup$
– KKZiomek
Jan 26 at 4:28
$begingroup$
I'm not familiar with that, but I wouldn't be surprised if Ramanujan had some sort of result about partitions.
$endgroup$
– kccu
Jan 26 at 4:29
$begingroup$
The generating function for partitions into $k$ parts goes back to Euler. Ramanujan did discover a considerable number of theorems about partitions and partition generating functions.
$endgroup$
– Peter Taylor
Jan 28 at 10:58
add a comment |
$begingroup$
Usually, when counting the number of ways in which $n$ balls can be put into $k$ boxes, two cases are considered:
- Boxes cannot be empty
- Boxes can be empty
Respectively, each of the listed cases has a formula:
$$C_1={n-1choose k-1}$$
$$C_2={n+k-1choose n}$$
This however doesn't suffice my needs. I want the boxes to be all not empty, and be countable, just like in the first case... But I want the boxes to be indistinguishable.
That is, as long as you have 4 balls in box 1, 2 balls in box 2, and 1 ball in box 3, it doesn't matter how the boxes themselves are configured or labeled.
For example, if there are 7 balls and 3 boxes, there would be four such ways to arrange them:
- O|O|OOOOO
- O|OO|OOOO
- O|OOO|OOO
- OO|OO|OOO
I have no ideas how to construct a formula for this
probability combinatorics
asked Jan 26 at 4:10
KKZiomekKKZiomek
2,2541641
$endgroup$
Usually, when counting the number of ways in which $n$ balls can be put into $k$ boxes, two cases are considered:
- Boxes cannot be empty
- Boxes can be empty
Respectively, each of the listed cases has a formula:
$$C_1={n-1choose k-1}$$
$$C_2={n+k-1choose n}$$
This however doesn't suffice my needs. I want the boxes to be all not empty, and be countable, just like in the first case... But I want the boxes to be indistinguishable.
That is, as long as you have 4 balls in box 1, 2 balls in box 2, and 1 ball in box 3, it doesn't matter how the boxes themselves are configured or labeled.
For example, if there are 7 balls and 3 boxes, there would be four such ways to arrange them:
- O|O|OOOOO
- O|OO|OOOO
- O|OOO|OOO
- OO|OO|OOO
I have no ideas how to construct a formula for this
probability combinatorics
probability combinatorics
asked Jan 26 at 4:10
KKZiomekKKZiomek
2,2541641
asked Jan 26 at 4:10
KKZiomekKKZiomek
2,2541641
asked Jan 26 at 4:10
KKZiomekKKZiomek
2,2541641
asked Jan 26 at 4:10
KKZiomekKKZiomek
2,2541641
asked Jan 26 at 4:10
KKZiomekKKZiomek
2,2541641
2,2541641
3
$begingroup$
What you are looking for is the number of partitions of $n$ into $k$ parts: whitman.edu/mathematics/cgt_online/book/section03.03.html. This is complicated to solve. I don't believe there is a nice formula for it, but there is a recurrence relation $p_k(n)=p_k(n-k)+p_{k-1}(n-1)$.
$endgroup$
– kccu
Jan 26 at 4:22
$begingroup$
@kccu I didn't recognize this as partitions, wow. But, didn't Ramanujan find an infinite series that converges to the solution of specific partitions? As far as I can remember, that series was very crazy and messy
$endgroup$
– KKZiomek
Jan 26 at 4:28
$begingroup$
I'm not familiar with that, but I wouldn't be surprised if Ramanujan had some sort of result about partitions.
$endgroup$
– kccu
Jan 26 at 4:29
$begingroup$
The generating function for partitions into $k$ parts goes back to Euler. Ramanujan did discover a considerable number of theorems about partitions and partition generating functions.
$endgroup$
– Peter Taylor
Jan 28 at 10:58
add a comment |
3
$begingroup$
What you are looking for is the number of partitions of $n$ into $k$ parts: whitman.edu/mathematics/cgt_online/book/section03.03.html. This is complicated to solve. I don't believe there is a nice formula for it, but there is a recurrence relation $p_k(n)=p_k(n-k)+p_{k-1}(n-1)$.
$endgroup$
– kccu
Jan 26 at 4:22
$begingroup$
@kccu I didn't recognize this as partitions, wow. But, didn't Ramanujan find an infinite series that converges to the solution of specific partitions? As far as I can remember, that series was very crazy and messy
$endgroup$
– KKZiomek
Jan 26 at 4:28
$begingroup$
I'm not familiar with that, but I wouldn't be surprised if Ramanujan had some sort of result about partitions.
$endgroup$
– kccu
Jan 26 at 4:29
$begingroup$
The generating function for partitions into $k$ parts goes back to Euler. Ramanujan did discover a considerable number of theorems about partitions and partition generating functions.
$endgroup$
– Peter Taylor
Jan 28 at 10:58
3
3
$begingroup$
What you are looking for is the number of partitions of $n$ into $k$ parts: whitman.edu/mathematics/cgt_online/book/section03.03.html. This is complicated to solve. I don't believe there is a nice formula for it, but there is a recurrence relation $p_k(n)=p_k(n-k)+p_{k-1}(n-1)$.
$endgroup$
– kccu
Jan 26 at 4:22
$begingroup$
What you are looking for is the number of partitions of $n$ into $k$ parts: whitman.edu/mathematics/cgt_online/book/section03.03.html. This is complicated to solve. I don't believe there is a nice formula for it, but there is a recurrence relation $p_k(n)=p_k(n-k)+p_{k-1}(n-1)$.
$endgroup$
– kccu
Jan 26 at 4:22
$begingroup$
@kccu I didn't recognize this as partitions, wow. But, didn't Ramanujan find an infinite series that converges to the solution of specific partitions? As far as I can remember, that series was very crazy and messy
$endgroup$
– KKZiomek
Jan 26 at 4:28
$begingroup$
@kccu I didn't recognize this as partitions, wow. But, didn't Ramanujan find an infinite series that converges to the solution of specific partitions? As far as I can remember, that series was very crazy and messy
$endgroup$
– KKZiomek
Jan 26 at 4:28
$begingroup$
I'm not familiar with that, but I wouldn't be surprised if Ramanujan had some sort of result about partitions.
$endgroup$
– kccu
Jan 26 at 4:29
$begingroup$
I'm not familiar with that, but I wouldn't be surprised if Ramanujan had some sort of result about partitions.
$endgroup$
– kccu
Jan 26 at 4:29
$begingroup$
The generating function for partitions into $k$ parts goes back to Euler. Ramanujan did discover a considerable number of theorems about partitions and partition generating functions.
$endgroup$
– Peter Taylor
Jan 28 at 10:58
$begingroup$
The generating function for partitions into $k$ parts goes back to Euler. Ramanujan did discover a considerable number of theorems about partitions and partition generating functions.
$endgroup$
– Peter Taylor
Jan 28 at 10:58
add a comment |
0
active
oldest
votes
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%2f3087889%2fnumber-of-ways-of-putting-n-balls-into-k-boxes-with-a-twist%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3087889%2fnumber-of-ways-of-putting-n-balls-into-k-boxes-with-a-twist%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
3
$begingroup$
What you are looking for is the number of partitions of $n$ into $k$ parts: whitman.edu/mathematics/cgt_online/book/section03.03.html. This is complicated to solve. I don't believe there is a nice formula for it, but there is a recurrence relation $p_k(n)=p_k(n-k)+p_{k-1}(n-1)$.
$endgroup$
– kccu
Jan 26 at 4:22
$begingroup$
@kccu I didn't recognize this as partitions, wow. But, didn't Ramanujan find an infinite series that converges to the solution of specific partitions? As far as I can remember, that series was very crazy and messy
$endgroup$
– KKZiomek
Jan 26 at 4:28
$begingroup$
I'm not familiar with that, but I wouldn't be surprised if Ramanujan had some sort of result about partitions.
$endgroup$
– kccu
Jan 26 at 4:29
$begingroup$
The generating function for partitions into $k$ parts goes back to Euler. Ramanujan did discover a considerable number of theorems about partitions and partition generating functions.
$endgroup$
– Peter Taylor
Jan 28 at 10:58