How many ways are there to arrange $5$ red, $5$ blue, and $5$ green balls in a row so that no two blue balls lie next to each other?
Clash Royale CLAN TAG#URR8PPP
$begingroup$
Um I know that there are $largefrac15!5!5!5!$ combinations but I'm kinda stumped after that.
I tried doing the space thing and I got $11 choose 5^2$ after my answer.
I don't really know what to do.
combinatorics combinations
edited Feb 25 at 2:30
stressed out
6,5831939
asked Feb 24 at 23:43
Wesley WangWesley Wang
91
$endgroup$
add a comment |
$begingroup$
Um I know that there are $largefrac15!5!5!5!$ combinations but I'm kinda stumped after that.
I tried doing the space thing and I got $11 choose 5^2$ after my answer.
I don't really know what to do.
combinatorics combinations
edited Feb 25 at 2:30
stressed out
6,5831939
asked Feb 24 at 23:43
Wesley WangWesley Wang
91
$endgroup$
$begingroup$
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Feb 25 at 1:10
$begingroup$
Hint: for the purpose of placing the blue balls only, red and green are equivalent. So when it comes to placing blue balls, you have 5 blue and 10 non-blue balls. Easier now? (Red and green are still distinct for the purposes of placing them)
$endgroup$
– smci
Feb 25 at 3:00
add a comment |
$begingroup$
Um I know that there are $largefrac15!5!5!5!$ combinations but I'm kinda stumped after that.
I tried doing the space thing and I got $11 choose 5^2$ after my answer.
I don't really know what to do.
combinatorics combinations
edited Feb 25 at 2:30
stressed out
6,5831939
asked Feb 24 at 23:43
Wesley WangWesley Wang
91
$endgroup$
Um I know that there are $largefrac15!5!5!5!$ combinations but I'm kinda stumped after that.
I tried doing the space thing and I got $11 choose 5^2$ after my answer.
I don't really know what to do.
combinatorics combinations
combinatorics combinations
edited Feb 25 at 2:30
stressed out
6,5831939
asked Feb 24 at 23:43
Wesley WangWesley Wang
91
edited Feb 25 at 2:30
stressed out
6,5831939
asked Feb 24 at 23:43
Wesley WangWesley Wang
91
edited Feb 25 at 2:30
stressed out
6,5831939
edited Feb 25 at 2:30
stressed out
6,5831939
edited Feb 25 at 2:30
stressed out
6,5831939
6,5831939
asked Feb 24 at 23:43
Wesley WangWesley Wang
91
asked Feb 24 at 23:43
Wesley WangWesley Wang
91
asked Feb 24 at 23:43
Wesley WangWesley Wang
91
91
$begingroup$
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Feb 25 at 1:10
$begingroup$
Hint: for the purpose of placing the blue balls only, red and green are equivalent. So when it comes to placing blue balls, you have 5 blue and 10 non-blue balls. Easier now? (Red and green are still distinct for the purposes of placing them)
$endgroup$
– smci
Feb 25 at 3:00
add a comment |
$begingroup$
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Feb 25 at 1:10
$begingroup$
Hint: for the purpose of placing the blue balls only, red and green are equivalent. So when it comes to placing blue balls, you have 5 blue and 10 non-blue balls. Easier now? (Red and green are still distinct for the purposes of placing them)
$endgroup$
– smci
Feb 25 at 3:00
$begingroup$
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Feb 25 at 1:10
$begingroup$
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Feb 25 at 1:10
$begingroup$
Hint: for the purpose of placing the blue balls only, red and green are equivalent. So when it comes to placing blue balls, you have 5 blue and 10 non-blue balls. Easier now? (Red and green are still distinct for the purposes of placing them)
$endgroup$
– smci
Feb 25 at 3:00
$begingroup$
Hint: for the purpose of placing the blue balls only, red and green are equivalent. So when it comes to placing blue balls, you have 5 blue and 10 non-blue balls. Easier now? (Red and green are still distinct for the purposes of placing them)
$endgroup$
– smci
Feb 25 at 3:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Arrange the red and green balls first, which can be done in $ 10choose 5 $ ways. The blue balls can then only be placed one at either end of the row, or in a space between two of the red and green balls. There are thus exactly 11 places where they can be put, and this can be done in $ 11choose 5 $ ways. Therefore, there are $ 10choose 511choose 5 $ ways of arranging the balls so that no two blue ones lie next to each other.
answered Feb 25 at 0:10
lonza leggieralonza leggiera
1,14328
$endgroup$
$begingroup$
Thanks what i did was 11 choose 5 squared not 10 choose 5 * 11 choose 5. Thank You!
$endgroup$
– Wesley Wang
Feb 25 at 0:13
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%2f3125416%2fhow-many-ways-are-there-to-arrange-5-red-5-blue-and-5-green-balls-in-a-r%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Arrange the red and green balls first, which can be done in $ 10choose 5 $ ways. The blue balls can then only be placed one at either end of the row, or in a space between two of the red and green balls. There are thus exactly 11 places where they can be put, and this can be done in $ 11choose 5 $ ways. Therefore, there are $ 10choose 511choose 5 $ ways of arranging the balls so that no two blue ones lie next to each other.
answered Feb 25 at 0:10
lonza leggieralonza leggiera
1,14328
$endgroup$
$begingroup$
Thanks what i did was 11 choose 5 squared not 10 choose 5 * 11 choose 5. Thank You!
$endgroup$
– Wesley Wang
Feb 25 at 0:13
add a comment |
$begingroup$
Arrange the red and green balls first, which can be done in $ 10choose 5 $ ways. The blue balls can then only be placed one at either end of the row, or in a space between two of the red and green balls. There are thus exactly 11 places where they can be put, and this can be done in $ 11choose 5 $ ways. Therefore, there are $ 10choose 511choose 5 $ ways of arranging the balls so that no two blue ones lie next to each other.
answered Feb 25 at 0:10
lonza leggieralonza leggiera
1,14328
$endgroup$
$begingroup$
Thanks what i did was 11 choose 5 squared not 10 choose 5 * 11 choose 5. Thank You!
$endgroup$
– Wesley Wang
Feb 25 at 0:13
add a comment |
$begingroup$
Arrange the red and green balls first, which can be done in $ 10choose 5 $ ways. The blue balls can then only be placed one at either end of the row, or in a space between two of the red and green balls. There are thus exactly 11 places where they can be put, and this can be done in $ 11choose 5 $ ways. Therefore, there are $ 10choose 511choose 5 $ ways of arranging the balls so that no two blue ones lie next to each other.
answered Feb 25 at 0:10
lonza leggieralonza leggiera
1,14328
$endgroup$
Arrange the red and green balls first, which can be done in $ 10choose 5 $ ways. The blue balls can then only be placed one at either end of the row, or in a space between two of the red and green balls. There are thus exactly 11 places where they can be put, and this can be done in $ 11choose 5 $ ways. Therefore, there are $ 10choose 511choose 5 $ ways of arranging the balls so that no two blue ones lie next to each other.
answered Feb 25 at 0:10
lonza leggieralonza leggiera
1,14328
answered Feb 25 at 0:10
lonza leggieralonza leggiera
1,14328
answered Feb 25 at 0:10
lonza leggieralonza leggiera
1,14328
answered Feb 25 at 0:10
lonza leggieralonza leggiera
1,14328
1,14328
$begingroup$
Thanks what i did was 11 choose 5 squared not 10 choose 5 * 11 choose 5. Thank You!
$endgroup$
– Wesley Wang
Feb 25 at 0:13
add a comment |
$begingroup$
Thanks what i did was 11 choose 5 squared not 10 choose 5 * 11 choose 5. Thank You!
$endgroup$
– Wesley Wang
Feb 25 at 0:13
$begingroup$
Thanks what i did was 11 choose 5 squared not 10 choose 5 * 11 choose 5. Thank You!
$endgroup$
– Wesley Wang
Feb 25 at 0:13
$begingroup$
Thanks what i did was 11 choose 5 squared not 10 choose 5 * 11 choose 5. Thank You!
$endgroup$
– Wesley Wang
Feb 25 at 0:13
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%2f3125416%2fhow-many-ways-are-there-to-arrange-5-red-5-blue-and-5-green-balls-in-a-r%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
$begingroup$
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Feb 25 at 1:10
$begingroup$
Hint: for the purpose of placing the blue balls only, red and green are equivalent. So when it comes to placing blue balls, you have 5 blue and 10 non-blue balls. Easier now? (Red and green are still distinct for the purposes of placing them)
$endgroup$
– smci
Feb 25 at 3:00