How many ways are there to arrange $5$ red, $5$ blue, and $5$ green balls in a row so that no two blue balls...
$begingroup$
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 4 hours ago
stressed out
5,7521738
asked 6 hours ago
Wesley WangWesley Wang
91
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$endgroup$
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 4 hours ago
stressed out
5,7521738
asked 6 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
5 hours ago
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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
3 hours 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 4 hours ago
stressed out
5,7521738
asked 6 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 4 hours ago
stressed out
5,7521738
asked 6 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 4 hours ago
stressed out
5,7521738
asked 6 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 4 hours ago
stressed out
5,7521738
edited 4 hours ago
stressed out
5,7521738
edited 4 hours ago
stressed out
5,7521738
5,7521738
asked 6 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 6 hours ago
Wesley WangWesley Wang
91
asked 6 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
5 hours ago
$begingroup$
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
3 hours ago
add a comment |
$begingroup$
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
5 hours ago
$begingroup$
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
3 hours ago
$begingroup$
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
5 hours ago
$begingroup$
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
5 hours ago
$begingroup$
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
3 hours ago
$begingroup$
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
3 hours ago
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 5}{11choose 5} $ ways of arranging the balls so that no two blue ones lie next to each other.
answered 6 hours ago
lonza leggieralonza leggiera
88117
$endgroup$
$begingroup$
Thanks what i did was 11 choose 5 squared not 10 choose 5 * 11 choose 5. Thank You!
$endgroup$
– Wesley Wang
6 hours ago
add a comment |
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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 6 hours ago
lonza leggieralonza leggiera
88117
$endgroup$
$begingroup$
Thanks what i did was 11 choose 5 squared not 10 choose 5 * 11 choose 5. Thank You!
$endgroup$
– Wesley Wang
6 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 6 hours ago
lonza leggieralonza leggiera
88117
$endgroup$
$begingroup$
Thanks what i did was 11 choose 5 squared not 10 choose 5 * 11 choose 5. Thank You!
$endgroup$
– Wesley Wang
6 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 6 hours ago
lonza leggieralonza leggiera
88117
$endgroup$
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 6 hours ago
lonza leggieralonza leggiera
88117
answered 6 hours ago
lonza leggieralonza leggiera
88117
answered 6 hours ago
lonza leggieralonza leggiera
88117
answered 6 hours ago
lonza leggieralonza leggiera
88117
88117
$begingroup$
Thanks what i did was 11 choose 5 squared not 10 choose 5 * 11 choose 5. Thank You!
$endgroup$
– Wesley Wang
6 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
6 hours ago
$begingroup$
Thanks what i did was 11 choose 5 squared not 10 choose 5 * 11 choose 5. Thank You!
$endgroup$
– Wesley Wang
6 hours ago
$begingroup$
Thanks what i did was 11 choose 5 squared not 10 choose 5 * 11 choose 5. Thank You!
$endgroup$
– Wesley Wang
6 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
5 hours ago
$begingroup$
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
3 hours ago