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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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?
How many ways are there for 8 men and 5 women to stand in a line so that no two women stand next to each other?put 4 red, blue and green triangles in a rowIn how many ways can the alphabet be ordered such that no four or five vowels are next to each other?In how many ways can 4 couples sit in a row if no 2 women sit next to each other?Ways to place 3 red, 4 blue and 5 green wagons such that no 2 blue wagons were standing next to each other$3$ red balls and $7$ blue balls are randomly placed in a row. What is the probability that no two red balls are adjacent?There are $10$ marbles, $4$ blue and $6$ redHow many ways are there to choose $4$ balls from $3$ black, $2$ green, and $1$ red ball?Number of arrangements of four blue, three green, and two red balls in which no two blue balls are adjacentA bag contains 3 blue, 3 green and 3 red balls. You pick two balls at random
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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 $11 choose 5$ squared after my answer.
I don't really know what to do.
combinatorics combinations
New contributor
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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 $11 choose 5$ squared after my answer.
I don't really know what to do.
combinatorics combinations
New contributor
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$begingroup$
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
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– 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 $11 choose 5$ squared after my answer.
I don't really know what to do.
combinatorics combinations
New contributor
$endgroup$
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 $11 choose 5$ squared after my answer.
I don't really know what to do.
combinatorics combinations
combinatorics combinations
New contributor
New contributor
edited 33 secs ago
stressed out
5,7421738
5,7421738
New contributor
asked 2 hours ago
Wesley WangWesley Wang
91
91
New contributor
New contributor
$begingroup$
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
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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.
$endgroup$
$begingroup$
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 |
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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.
$endgroup$
$begingroup$
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.
$endgroup$
$begingroup$
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.
$endgroup$
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
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