How many ways are there to arrange $5$ red, $5$ blue, and $5$ green balls in a row so that no two blue balls...
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Um I know that there are $largefrac{15!}{5!5!5!}$ 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 40 secs ago
stressed out
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asked 2 hours ago
Wesley WangWesley Wang
91
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add a comment |
$begingroup$
Um I know that there are $largefrac{15!}{5!5!5!}$ 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 40 secs ago
stressed out
5,7421738
asked 2 hours ago
Wesley WangWesley Wang
91
New contributor
Wesley Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
1 hour ago
add a comment |
$begingroup$
Um I know that there are $largefrac{15!}{5!5!5!}$ 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 40 secs ago
stressed out
5,7421738
asked 2 hours ago
Wesley WangWesley Wang
91
New contributor
Wesley Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Um I know that there are $largefrac{15!}{5!5!5!}$ 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 40 secs ago
stressed out
5,7421738
asked 2 hours ago
Wesley WangWesley Wang
91
New contributor
Wesley Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 40 secs ago
stressed out
5,7421738
asked 2 hours ago
Wesley WangWesley Wang
91
New contributor
Wesley Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 40 secs ago
stressed out
5,7421738
edited 40 secs ago
stressed out
5,7421738
edited 40 secs ago
stressed out
5,7421738
5,7421738
asked 2 hours ago
Wesley WangWesley Wang
91
New contributor
Wesley Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Wesley WangWesley Wang
91
asked 2 hours ago
Wesley WangWesley Wang
91
91
New contributor
Wesley Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Wesley Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Wesley Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
1 hour ago
add a comment |
$begingroup$
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
1 hour ago
$begingroup$
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
1 hour ago
$begingroup$
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
1 hour ago
add a comment |
1 Answer
1
active
oldest
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$begingroup$
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 5}{11choose 5} $ ways of arranging the balls so that no two blue ones lie next to each other.
answered 2 hours ago
lonza leggieralonza leggiera
87117
$endgroup$
$begingroup$
Thanks what i did was 11 choose 5 squared not 10 choose 5 * 11 choose 5. Thank You!
$endgroup$
– Wesley Wang
2 hours ago
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
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 5}{11choose 5} $ ways of arranging the balls so that no two blue ones lie next to each other.
answered 2 hours ago
lonza leggieralonza leggiera
87117
$endgroup$
$begingroup$
Thanks what i did was 11 choose 5 squared not 10 choose 5 * 11 choose 5. Thank You!
$endgroup$
– Wesley Wang
2 hours ago
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 5}{11choose 5} $ ways of arranging the balls so that no two blue ones lie next to each other.
answered 2 hours ago
lonza leggieralonza leggiera
87117
$endgroup$
$begingroup$
Thanks what i did was 11 choose 5 squared not 10 choose 5 * 11 choose 5. Thank You!
$endgroup$
– Wesley Wang
2 hours ago
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 5}{11choose 5} $ ways of arranging the balls so that no two blue ones lie next to each other.
answered 2 hours ago
lonza leggieralonza leggiera
87117
$endgroup$
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 5}{11choose 5} $ ways of arranging the balls so that no two blue ones lie next to each other.
answered 2 hours ago
lonza leggieralonza leggiera
87117
answered 2 hours ago
lonza leggieralonza leggiera
87117
answered 2 hours ago
lonza leggieralonza leggiera
87117
answered 2 hours ago
lonza leggieralonza leggiera
87117
87117
$begingroup$
Thanks what i did was 11 choose 5 squared not 10 choose 5 * 11 choose 5. Thank You!
$endgroup$
– Wesley Wang
2 hours ago
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
2 hours ago
$begingroup$
Thanks what i did was 11 choose 5 squared not 10 choose 5 * 11 choose 5. Thank You!
$endgroup$
– Wesley Wang
2 hours ago
$begingroup$
Thanks what i did was 11 choose 5 squared not 10 choose 5 * 11 choose 5. Thank You!
$endgroup$
– Wesley Wang
2 hours ago
add a comment |
Wesley Wang is a new contributor. Be nice, and check out our Code of Conduct.
Wesley Wang is a new contributor. Be nice, and check out our Code of Conduct.
Wesley Wang is a new contributor. Be nice, and check out our Code of Conduct.
Wesley Wang is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
1 hour ago