No boy stands next to each others permutation problem
$begingroup$
The question is:
Find the number of ways that 5 boys and 6 girls can stand in a row if no boy stands next to another boy
In which the correct model would be:
$$square g square g square g square g square g square gsquare$$
Therefore:
$$6!cdot 7P5=1space 814space 400$$
However, I noticed that if you take one of the girls away (let's label her as $^a$), we have:
$$^asquare^ag^asquare^ag^asquare^ag^asquare^ag^asquare^ag^asquare^a$$
We see that no matter where girl$^a$ stands, the boys will still not stand next to each others:
$$5!cdot6P5cdot12=1space036space800$$
Update
Also, if we are to swap it around:
$$square b square b square b square b square b square$$
We now have:
$$5!cdot6!=86space 400 $$
What is wrong with the last two logic?
Thank you.
combinatorics permutations
asked 2 hours ago
Huy Tran VanHuy Tran Van
314
$endgroup$
add a comment |
$begingroup$
The question is:
Find the number of ways that 5 boys and 6 girls can stand in a row if no boy stands next to another boy
In which the correct model would be:
$$square g square g square g square g square g square gsquare$$
Therefore:
$$6!cdot 7P5=1space 814space 400$$
However, I noticed that if you take one of the girls away (let's label her as $^a$), we have:
$$^asquare^ag^asquare^ag^asquare^ag^asquare^ag^asquare^ag^asquare^a$$
We see that no matter where girl$^a$ stands, the boys will still not stand next to each others:
$$5!cdot6P5cdot12=1space036space800$$
Update
Also, if we are to swap it around:
$$square b square b square b square b square b square$$
We now have:
$$5!cdot6!=86space 400 $$
What is wrong with the last two logic?
Thank you.
combinatorics permutations
asked 2 hours ago
Huy Tran VanHuy Tran Van
314
$endgroup$
add a comment |
$begingroup$
The question is:
Find the number of ways that 5 boys and 6 girls can stand in a row if no boy stands next to another boy
In which the correct model would be:
$$square g square g square g square g square g square gsquare$$
Therefore:
$$6!cdot 7P5=1space 814space 400$$
However, I noticed that if you take one of the girls away (let's label her as $^a$), we have:
$$^asquare^ag^asquare^ag^asquare^ag^asquare^ag^asquare^ag^asquare^a$$
We see that no matter where girl$^a$ stands, the boys will still not stand next to each others:
$$5!cdot6P5cdot12=1space036space800$$
Update
Also, if we are to swap it around:
$$square b square b square b square b square b square$$
We now have:
$$5!cdot6!=86space 400 $$
What is wrong with the last two logic?
Thank you.
combinatorics permutations
asked 2 hours ago
Huy Tran VanHuy Tran Van
314
$endgroup$
The question is:
Find the number of ways that 5 boys and 6 girls can stand in a row if no boy stands next to another boy
In which the correct model would be:
$$square g square g square g square g square g square gsquare$$
Therefore:
$$6!cdot 7P5=1space 814space 400$$
However, I noticed that if you take one of the girls away (let's label her as $^a$), we have:
$$^asquare^ag^asquare^ag^asquare^ag^asquare^ag^asquare^ag^asquare^a$$
We see that no matter where girl$^a$ stands, the boys will still not stand next to each others:
$$5!cdot6P5cdot12=1space036space800$$
Update
Also, if we are to swap it around:
$$square b square b square b square b square b square$$
We now have:
$$5!cdot6!=86space 400 $$
What is wrong with the last two logic?
Thank you.
combinatorics permutations
combinatorics permutations
asked 2 hours ago
Huy Tran VanHuy Tran Van
314
asked 2 hours ago
Huy Tran VanHuy Tran Van
314
edited 32 mins ago
Huy Tran Van
asked 2 hours ago
Huy Tran VanHuy Tran Van
314
asked 2 hours ago
Huy Tran VanHuy Tran Van
314
asked 2 hours ago
Huy Tran VanHuy Tran Van
314
314
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Your first computation is correct.
The latter two computations undercount because they start off by taking one of the girls away. Let's say they always take girl #1 away.
Now if we number the girls, those solutions can contain as an example $1 2 b 3 b 4 b 5 b 6b$. But they can not contain $1b2b34b5b6b$. We've lost the ability to have two girls next to eachother without involving girl #1.
answered 1 hour ago
orlporlp
7,4211330
$endgroup$
$begingroup$
Now, I know the last two undercount somehow but I can't see why '$1b2b3b4b5b6b$' is not possible in the second case?
$endgroup$
– Huy Tran Van
1 hour ago
$begingroup$
@HuyTranVan You misread. I said $1b2b34b5b6b$. There is no $b$ between $3$ and $4$.
$endgroup$
– orlp
1 hour ago
$begingroup$
I wanted to know why $12b3b4b5b6b$ is possible but then $1b2b34b5b6b$ (no b between 3 and 4) is not counted in the second case. According to you when one of the girls stands next to another, the question changed?
$endgroup$
– Huy Tran Van
1 hour ago
add a comment |
$begingroup$
In the last one you're obviously leaving some out, which could be gotten by swapping some of bg combinations in either sum.
In the second, also, you're not counting certain arrangements, which could be gotten by, again, swapping at least some bg combinations.
The right one is the first.
answered 1 hour ago
Chris CusterChris Custer
11.5k3824
$endgroup$
add a comment |
$begingroup$
You are on the right track with thinking that there are 7 possible slots with 5 boys to be arranged into those slots. This is a combination problem with can be treated as a 7 choose 5 or a $7 choose 5$.
A $7 choose 5$ is calculated as
$$
frac{7!}{5!*(7-5)!} = 21
$$
Therefore there are 21 combinations in which the boys could be arranged.
I have also used an excel spreadsheet to show the possible ways that they could be arranged which ends up equaling 21.
Picture of how the boys and girls could be arranged in 21 combinations
answered 1 hour ago
TucktuckgooseTucktuckgoose
11
New contributor
Tucktuckgoose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
My answer is predicated that this is a combination problem, not a permutation problem. As they are described as boys and girls, they are likely interchangeable and this is a combination. If the question was how many ways can Tom, Dick, Harry, Sam, and Bob be arranged, that would be a different question with a much larger answer.
$endgroup$
– Tucktuckgoose
1 hour ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your first computation is correct.
The latter two computations undercount because they start off by taking one of the girls away. Let's say they always take girl #1 away.
Now if we number the girls, those solutions can contain as an example $1 2 b 3 b 4 b 5 b 6b$. But they can not contain $1b2b34b5b6b$. We've lost the ability to have two girls next to eachother without involving girl #1.
answered 1 hour ago
orlporlp
7,4211330
$endgroup$
$begingroup$
Now, I know the last two undercount somehow but I can't see why '$1b2b3b4b5b6b$' is not possible in the second case?
$endgroup$
– Huy Tran Van
1 hour ago
$begingroup$
@HuyTranVan You misread. I said $1b2b34b5b6b$. There is no $b$ between $3$ and $4$.
$endgroup$
– orlp
1 hour ago
$begingroup$
I wanted to know why $12b3b4b5b6b$ is possible but then $1b2b34b5b6b$ (no b between 3 and 4) is not counted in the second case. According to you when one of the girls stands next to another, the question changed?
$endgroup$
– Huy Tran Van
1 hour ago
add a comment |
$begingroup$
Your first computation is correct.
The latter two computations undercount because they start off by taking one of the girls away. Let's say they always take girl #1 away.
Now if we number the girls, those solutions can contain as an example $1 2 b 3 b 4 b 5 b 6b$. But they can not contain $1b2b34b5b6b$. We've lost the ability to have two girls next to eachother without involving girl #1.
answered 1 hour ago
orlporlp
7,4211330
$endgroup$
$begingroup$
Now, I know the last two undercount somehow but I can't see why '$1b2b3b4b5b6b$' is not possible in the second case?
$endgroup$
– Huy Tran Van
1 hour ago
$begingroup$
@HuyTranVan You misread. I said $1b2b34b5b6b$. There is no $b$ between $3$ and $4$.
$endgroup$
– orlp
1 hour ago
$begingroup$
I wanted to know why $12b3b4b5b6b$ is possible but then $1b2b34b5b6b$ (no b between 3 and 4) is not counted in the second case. According to you when one of the girls stands next to another, the question changed?
$endgroup$
– Huy Tran Van
1 hour ago
add a comment |
$begingroup$
Your first computation is correct.
The latter two computations undercount because they start off by taking one of the girls away. Let's say they always take girl #1 away.
Now if we number the girls, those solutions can contain as an example $1 2 b 3 b 4 b 5 b 6b$. But they can not contain $1b2b34b5b6b$. We've lost the ability to have two girls next to eachother without involving girl #1.
answered 1 hour ago
orlporlp
7,4211330
$endgroup$
Your first computation is correct.
The latter two computations undercount because they start off by taking one of the girls away. Let's say they always take girl #1 away.
Now if we number the girls, those solutions can contain as an example $1 2 b 3 b 4 b 5 b 6b$. But they can not contain $1b2b34b5b6b$. We've lost the ability to have two girls next to eachother without involving girl #1.
answered 1 hour ago
orlporlp
7,4211330
answered 1 hour ago
orlporlp
7,4211330
answered 1 hour ago
orlporlp
7,4211330
answered 1 hour ago
orlporlp
7,4211330
7,4211330
$begingroup$
Now, I know the last two undercount somehow but I can't see why '$1b2b3b4b5b6b$' is not possible in the second case?
$endgroup$
– Huy Tran Van
1 hour ago
$begingroup$
@HuyTranVan You misread. I said $1b2b34b5b6b$. There is no $b$ between $3$ and $4$.
$endgroup$
– orlp
1 hour ago
$begingroup$
I wanted to know why $12b3b4b5b6b$ is possible but then $1b2b34b5b6b$ (no b between 3 and 4) is not counted in the second case. According to you when one of the girls stands next to another, the question changed?
$endgroup$
– Huy Tran Van
1 hour ago
add a comment |
$begingroup$
Now, I know the last two undercount somehow but I can't see why '$1b2b3b4b5b6b$' is not possible in the second case?
$endgroup$
– Huy Tran Van
1 hour ago
$begingroup$
@HuyTranVan You misread. I said $1b2b34b5b6b$. There is no $b$ between $3$ and $4$.
$endgroup$
– orlp
1 hour ago
$begingroup$
I wanted to know why $12b3b4b5b6b$ is possible but then $1b2b34b5b6b$ (no b between 3 and 4) is not counted in the second case. According to you when one of the girls stands next to another, the question changed?
$endgroup$
– Huy Tran Van
1 hour ago
$begingroup$
Now, I know the last two undercount somehow but I can't see why '$1b2b3b4b5b6b$' is not possible in the second case?
$endgroup$
– Huy Tran Van
1 hour ago
$begingroup$
Now, I know the last two undercount somehow but I can't see why '$1b2b3b4b5b6b$' is not possible in the second case?
$endgroup$
– Huy Tran Van
1 hour ago
$begingroup$
@HuyTranVan You misread. I said $1b2b34b5b6b$. There is no $b$ between $3$ and $4$.
$endgroup$
– orlp
1 hour ago
$begingroup$
@HuyTranVan You misread. I said $1b2b34b5b6b$. There is no $b$ between $3$ and $4$.
$endgroup$
– orlp
1 hour ago
$begingroup$
I wanted to know why $12b3b4b5b6b$ is possible but then $1b2b34b5b6b$ (no b between 3 and 4) is not counted in the second case. According to you when one of the girls stands next to another, the question changed?
$endgroup$
– Huy Tran Van
1 hour ago
$begingroup$
I wanted to know why $12b3b4b5b6b$ is possible but then $1b2b34b5b6b$ (no b between 3 and 4) is not counted in the second case. According to you when one of the girls stands next to another, the question changed?
$endgroup$
– Huy Tran Van
1 hour ago
add a comment |
$begingroup$
In the last one you're obviously leaving some out, which could be gotten by swapping some of bg combinations in either sum.
In the second, also, you're not counting certain arrangements, which could be gotten by, again, swapping at least some bg combinations.
The right one is the first.
answered 1 hour ago
Chris CusterChris Custer
11.5k3824
$endgroup$
add a comment |
$begingroup$
In the last one you're obviously leaving some out, which could be gotten by swapping some of bg combinations in either sum.
In the second, also, you're not counting certain arrangements, which could be gotten by, again, swapping at least some bg combinations.
The right one is the first.
answered 1 hour ago
Chris CusterChris Custer
11.5k3824
$endgroup$
add a comment |
$begingroup$
In the last one you're obviously leaving some out, which could be gotten by swapping some of bg combinations in either sum.
In the second, also, you're not counting certain arrangements, which could be gotten by, again, swapping at least some bg combinations.
The right one is the first.
answered 1 hour ago
Chris CusterChris Custer
11.5k3824
$endgroup$
In the last one you're obviously leaving some out, which could be gotten by swapping some of bg combinations in either sum.
In the second, also, you're not counting certain arrangements, which could be gotten by, again, swapping at least some bg combinations.
The right one is the first.
answered 1 hour ago
Chris CusterChris Custer
11.5k3824
edited 1 hour ago
answered 1 hour ago
Chris CusterChris Custer
11.5k3824
answered 1 hour ago
Chris CusterChris Custer
11.5k3824
answered 1 hour ago
Chris CusterChris Custer
11.5k3824
11.5k3824
add a comment |
add a comment |
$begingroup$
You are on the right track with thinking that there are 7 possible slots with 5 boys to be arranged into those slots. This is a combination problem with can be treated as a 7 choose 5 or a $7 choose 5$.
A $7 choose 5$ is calculated as
$$
frac{7!}{5!*(7-5)!} = 21
$$
Therefore there are 21 combinations in which the boys could be arranged.
I have also used an excel spreadsheet to show the possible ways that they could be arranged which ends up equaling 21.
Picture of how the boys and girls could be arranged in 21 combinations
answered 1 hour ago
TucktuckgooseTucktuckgoose
11
New contributor
Tucktuckgoose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
My answer is predicated that this is a combination problem, not a permutation problem. As they are described as boys and girls, they are likely interchangeable and this is a combination. If the question was how many ways can Tom, Dick, Harry, Sam, and Bob be arranged, that would be a different question with a much larger answer.
$endgroup$
– Tucktuckgoose
1 hour ago
add a comment |
$begingroup$
You are on the right track with thinking that there are 7 possible slots with 5 boys to be arranged into those slots. This is a combination problem with can be treated as a 7 choose 5 or a $7 choose 5$.
A $7 choose 5$ is calculated as
$$
frac{7!}{5!*(7-5)!} = 21
$$
Therefore there are 21 combinations in which the boys could be arranged.
I have also used an excel spreadsheet to show the possible ways that they could be arranged which ends up equaling 21.
Picture of how the boys and girls could be arranged in 21 combinations
answered 1 hour ago
TucktuckgooseTucktuckgoose
11
New contributor
Tucktuckgoose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
My answer is predicated that this is a combination problem, not a permutation problem. As they are described as boys and girls, they are likely interchangeable and this is a combination. If the question was how many ways can Tom, Dick, Harry, Sam, and Bob be arranged, that would be a different question with a much larger answer.
$endgroup$
– Tucktuckgoose
1 hour ago
add a comment |
$begingroup$
You are on the right track with thinking that there are 7 possible slots with 5 boys to be arranged into those slots. This is a combination problem with can be treated as a 7 choose 5 or a $7 choose 5$.
A $7 choose 5$ is calculated as
$$
frac{7!}{5!*(7-5)!} = 21
$$
Therefore there are 21 combinations in which the boys could be arranged.
I have also used an excel spreadsheet to show the possible ways that they could be arranged which ends up equaling 21.
Picture of how the boys and girls could be arranged in 21 combinations
answered 1 hour ago
TucktuckgooseTucktuckgoose
11
New contributor
Tucktuckgoose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
You are on the right track with thinking that there are 7 possible slots with 5 boys to be arranged into those slots. This is a combination problem with can be treated as a 7 choose 5 or a $7 choose 5$.
A $7 choose 5$ is calculated as
$$
frac{7!}{5!*(7-5)!} = 21
$$
Therefore there are 21 combinations in which the boys could be arranged.
I have also used an excel spreadsheet to show the possible ways that they could be arranged which ends up equaling 21.
Picture of how the boys and girls could be arranged in 21 combinations
answered 1 hour ago
TucktuckgooseTucktuckgoose
11
New contributor
Tucktuckgoose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
TucktuckgooseTucktuckgoose
11
New contributor
Tucktuckgoose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
TucktuckgooseTucktuckgoose
11
answered 1 hour ago
TucktuckgooseTucktuckgoose
11
11
New contributor
Tucktuckgoose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Tucktuckgoose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Tucktuckgoose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
My answer is predicated that this is a combination problem, not a permutation problem. As they are described as boys and girls, they are likely interchangeable and this is a combination. If the question was how many ways can Tom, Dick, Harry, Sam, and Bob be arranged, that would be a different question with a much larger answer.
$endgroup$
– Tucktuckgoose
1 hour ago
add a comment |
$begingroup$
My answer is predicated that this is a combination problem, not a permutation problem. As they are described as boys and girls, they are likely interchangeable and this is a combination. If the question was how many ways can Tom, Dick, Harry, Sam, and Bob be arranged, that would be a different question with a much larger answer.
$endgroup$
– Tucktuckgoose
1 hour ago
$begingroup$
My answer is predicated that this is a combination problem, not a permutation problem. As they are described as boys and girls, they are likely interchangeable and this is a combination. If the question was how many ways can Tom, Dick, Harry, Sam, and Bob be arranged, that would be a different question with a much larger answer.
$endgroup$
– Tucktuckgoose
1 hour ago
$begingroup$
My answer is predicated that this is a combination problem, not a permutation problem. As they are described as boys and girls, they are likely interchangeable and this is a combination. If the question was how many ways can Tom, Dick, Harry, Sam, and Bob be arranged, that would be a different question with a much larger answer.
$endgroup$
– Tucktuckgoose
1 hour ago
add a comment |
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