How many ways can 6 boys and 4 girls stand in a row if the girls must not be together?
How many ways can 6 boys and 4 girls stand in a row if the girls must not be together?
Arrange all 6 boys in a row . --> ways to do this is 6!
This leaves 7 gap in between the boys to place 4 girls in. -> $_7C_4$
The girls in between the boys can have 4! ways to arrange them.
therefore, answer is $6! cdot _7C_4 cdot 4! $
why am i wrong ?
the right answer is $6! cdot _7C_4 cdot 4! + 6! cdot _7C_2 cdot 4! + 6! cdot _7C_2 cdot 4!$
combinatorics permutations
edited Nov 25 at 11:31
N. F. Taussig
43.5k93355
asked Nov 24 at 18:28
Erikien
494
|
show 1 more comment
How many ways can 6 boys and 4 girls stand in a row if the girls must not be together?
Arrange all 6 boys in a row . --> ways to do this is 6!
This leaves 7 gap in between the boys to place 4 girls in. -> $_7C_4$
The girls in between the boys can have 4! ways to arrange them.
therefore, answer is $6! cdot _7C_4 cdot 4! $
why am i wrong ?
the right answer is $6! cdot _7C_4 cdot 4! + 6! cdot _7C_2 cdot 4! + 6! cdot _7C_2 cdot 4!$
combinatorics permutations
edited Nov 25 at 11:31
N. F. Taussig
43.5k93355
asked Nov 24 at 18:28
Erikien
494
It would help if you supplied the correct answer.
– herb steinberg
Nov 24 at 18:31
Maybe the boys (and girls) are not supposed to be distinguishable from eah other? What were you told the right answer is?
– Bram28
Nov 24 at 18:34
Your answer sounds correct to me, I do not know why you think it might be incorrect. Reiterating what herbsteinberg said before, please include why you think it is incorrect and what the "correct" answer is. It could simply be that the answers are the same but written in a different format.
– JMoravitz
Nov 24 at 18:35
It looks like "girls must not be together" is supposed to mean that you can't have all four girls together rather than the stronger restriction of no two girls together. Poorly worded question.
– Ned
Nov 24 at 20:17
@Ned so how does that explain the other 2 sets which are added to the working i have ?
– Erikien
Nov 25 at 12:56
|
show 1 more comment
How many ways can 6 boys and 4 girls stand in a row if the girls must not be together?
Arrange all 6 boys in a row . --> ways to do this is 6!
This leaves 7 gap in between the boys to place 4 girls in. -> $_7C_4$
The girls in between the boys can have 4! ways to arrange them.
therefore, answer is $6! cdot _7C_4 cdot 4! $
why am i wrong ?
the right answer is $6! cdot _7C_4 cdot 4! + 6! cdot _7C_2 cdot 4! + 6! cdot _7C_2 cdot 4!$
combinatorics permutations
edited Nov 25 at 11:31
N. F. Taussig
43.5k93355
asked Nov 24 at 18:28
Erikien
494
How many ways can 6 boys and 4 girls stand in a row if the girls must not be together?
Arrange all 6 boys in a row . --> ways to do this is 6!
This leaves 7 gap in between the boys to place 4 girls in. -> $_7C_4$
The girls in between the boys can have 4! ways to arrange them.
therefore, answer is $6! cdot _7C_4 cdot 4! $
why am i wrong ?
the right answer is $6! cdot _7C_4 cdot 4! + 6! cdot _7C_2 cdot 4! + 6! cdot _7C_2 cdot 4!$
combinatorics permutations
combinatorics permutations
edited Nov 25 at 11:31
N. F. Taussig
43.5k93355
asked Nov 24 at 18:28
Erikien
494
edited Nov 25 at 11:31
N. F. Taussig
43.5k93355
asked Nov 24 at 18:28
Erikien
494
edited Nov 25 at 11:31
N. F. Taussig
43.5k93355
edited Nov 25 at 11:31
N. F. Taussig
43.5k93355
edited Nov 25 at 11:31
N. F. Taussig
43.5k93355
43.5k93355
asked Nov 24 at 18:28
Erikien
494
asked Nov 24 at 18:28
Erikien
494
asked Nov 24 at 18:28
Erikien
494
494
It would help if you supplied the correct answer.
– herb steinberg
Nov 24 at 18:31
Maybe the boys (and girls) are not supposed to be distinguishable from eah other? What were you told the right answer is?
– Bram28
Nov 24 at 18:34
Your answer sounds correct to me, I do not know why you think it might be incorrect. Reiterating what herbsteinberg said before, please include why you think it is incorrect and what the "correct" answer is. It could simply be that the answers are the same but written in a different format.
– JMoravitz
Nov 24 at 18:35
It looks like "girls must not be together" is supposed to mean that you can't have all four girls together rather than the stronger restriction of no two girls together. Poorly worded question.
– Ned
Nov 24 at 20:17
@Ned so how does that explain the other 2 sets which are added to the working i have ?
– Erikien
Nov 25 at 12:56
|
show 1 more comment
It would help if you supplied the correct answer.
– herb steinberg
Nov 24 at 18:31
Maybe the boys (and girls) are not supposed to be distinguishable from eah other? What were you told the right answer is?
– Bram28
Nov 24 at 18:34
Your answer sounds correct to me, I do not know why you think it might be incorrect. Reiterating what herbsteinberg said before, please include why you think it is incorrect and what the "correct" answer is. It could simply be that the answers are the same but written in a different format.
– JMoravitz
Nov 24 at 18:35
It looks like "girls must not be together" is supposed to mean that you can't have all four girls together rather than the stronger restriction of no two girls together. Poorly worded question.
– Ned
Nov 24 at 20:17
@Ned so how does that explain the other 2 sets which are added to the working i have ?
– Erikien
Nov 25 at 12:56
It would help if you supplied the correct answer.
– herb steinberg
Nov 24 at 18:31
It would help if you supplied the correct answer.
– herb steinberg
Nov 24 at 18:31
Maybe the boys (and girls) are not supposed to be distinguishable from eah other? What were you told the right answer is?
– Bram28
Nov 24 at 18:34
Maybe the boys (and girls) are not supposed to be distinguishable from eah other? What were you told the right answer is?
– Bram28
Nov 24 at 18:34
Your answer sounds correct to me, I do not know why you think it might be incorrect. Reiterating what herbsteinberg said before, please include why you think it is incorrect and what the "correct" answer is. It could simply be that the answers are the same but written in a different format.
– JMoravitz
Nov 24 at 18:35
Your answer sounds correct to me, I do not know why you think it might be incorrect. Reiterating what herbsteinberg said before, please include why you think it is incorrect and what the "correct" answer is. It could simply be that the answers are the same but written in a different format.
– JMoravitz
Nov 24 at 18:35
It looks like "girls must not be together" is supposed to mean that you can't have all four girls together rather than the stronger restriction of no two girls together. Poorly worded question.
– Ned
Nov 24 at 20:17
It looks like "girls must not be together" is supposed to mean that you can't have all four girls together rather than the stronger restriction of no two girls together. Poorly worded question.
– Ned
Nov 24 at 20:17
@Ned so how does that explain the other 2 sets which are added to the working i have ?
– Erikien
Nov 25 at 12:56
@Ned so how does that explain the other 2 sets which are added to the working i have ?
– Erikien
Nov 25 at 12:56
|
show 1 more comment
1 Answer
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Say we have girls $A,B,C,D$. The numbers that $X,Y$ are together is $2cdot 9!$, then $X,Y$ and $Y,Z$ are together is $2cdot 8!$. Now we use PIE.
Then at least two are together is $${4choose 2}cdot 2cdot 9!-24cdot 8! = 84cdot 8!$$
So the answer is $$10!-84cdot 8! = 8! (90-84) = 6cdot 8!$$
answered Nov 24 at 18:38
greedoid
37.7k114794
add a comment |
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1 Answer
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Say we have girls $A,B,C,D$. The numbers that $X,Y$ are together is $2cdot 9!$, then $X,Y$ and $Y,Z$ are together is $2cdot 8!$. Now we use PIE.
Then at least two are together is $${4choose 2}cdot 2cdot 9!-24cdot 8! = 84cdot 8!$$
So the answer is $$10!-84cdot 8! = 8! (90-84) = 6cdot 8!$$
answered Nov 24 at 18:38
greedoid
37.7k114794
add a comment |
Say we have girls $A,B,C,D$. The numbers that $X,Y$ are together is $2cdot 9!$, then $X,Y$ and $Y,Z$ are together is $2cdot 8!$. Now we use PIE.
Then at least two are together is $${4choose 2}cdot 2cdot 9!-24cdot 8! = 84cdot 8!$$
So the answer is $$10!-84cdot 8! = 8! (90-84) = 6cdot 8!$$
answered Nov 24 at 18:38
greedoid
37.7k114794
add a comment |
Say we have girls $A,B,C,D$. The numbers that $X,Y$ are together is $2cdot 9!$, then $X,Y$ and $Y,Z$ are together is $2cdot 8!$. Now we use PIE.
Then at least two are together is $${4choose 2}cdot 2cdot 9!-24cdot 8! = 84cdot 8!$$
So the answer is $$10!-84cdot 8! = 8! (90-84) = 6cdot 8!$$
answered Nov 24 at 18:38
greedoid
37.7k114794
Say we have girls $A,B,C,D$. The numbers that $X,Y$ are together is $2cdot 9!$, then $X,Y$ and $Y,Z$ are together is $2cdot 8!$. Now we use PIE.
Then at least two are together is $${4choose 2}cdot 2cdot 9!-24cdot 8! = 84cdot 8!$$
So the answer is $$10!-84cdot 8! = 8! (90-84) = 6cdot 8!$$
answered Nov 24 at 18:38
greedoid
37.7k114794
answered Nov 24 at 18:38
greedoid
37.7k114794
answered Nov 24 at 18:38
greedoid
37.7k114794
answered Nov 24 at 18:38
greedoid
37.7k114794
37.7k114794
add a comment |
add a comment |
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It would help if you supplied the correct answer.
– herb steinberg
Nov 24 at 18:31
Maybe the boys (and girls) are not supposed to be distinguishable from eah other? What were you told the right answer is?
– Bram28
Nov 24 at 18:34
Your answer sounds correct to me, I do not know why you think it might be incorrect. Reiterating what herbsteinberg said before, please include why you think it is incorrect and what the "correct" answer is. It could simply be that the answers are the same but written in a different format.
– JMoravitz
Nov 24 at 18:35
It looks like "girls must not be together" is supposed to mean that you can't have all four girls together rather than the stronger restriction of no two girls together. Poorly worded question.
– Ned
Nov 24 at 20:17
@Ned so how does that explain the other 2 sets which are added to the working i have ?
– Erikien
Nov 25 at 12:56