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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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$begingroup$


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.










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  • $begingroup$
    Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
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1












$begingroup$


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.










share|cite|improve this question









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














1












1








1





$begingroup$


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.










share|cite|improve this question









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 $11 choose 5$ squared after my answer.



I don't really know what to do.







combinatorics combinations






share|cite|improve this question









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.











share|cite|improve this question









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.









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edited 33 secs ago









stressed out

5,7421738




5,7421738






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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.
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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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















5












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks what i did was 11 choose 5 squared not 10 choose 5 * 11 choose 5. Thank You!
    $endgroup$
    – Wesley Wang
    2 hours ago











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks what i did was 11 choose 5 squared not 10 choose 5 * 11 choose 5. Thank You!
    $endgroup$
    – Wesley Wang
    2 hours ago
















5












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks what i did was 11 choose 5 squared not 10 choose 5 * 11 choose 5. Thank You!
    $endgroup$
    – Wesley Wang
    2 hours ago














5












5








5





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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










Wesley Wang is a new contributor. Be nice, and check out our Code of Conduct.










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