Find the number of ways that 5 boys and 6 girls can stand in a row if no boy stands next to another boy
$begingroup$
We have:
$$square g square g square g square g square g square gsquare$$
Therefore:
$$6!cdot 7P5=1space 814space 400$$
However, I notice that if you take one of the girls away (let's label her as $^*$), we have:
$$^*square^*g^*square^*g^*square^*g^*square^*g^*square^*g^*square^*$$
We see that no matter where girl$^*$ stands, the boys will still not stand next to each others:
$$5!cdot6P5cdot12=1space036space800$$
Update
If we are to swap it around:
$$square b square b square b square b square b square$$
We now have:
$$5!cdot6!=86space 400 $$
So which is right? Is there anything wrong with each logic?
Thank you.
combinatorics permutations
$endgroup$
add a comment |
$begingroup$
We have:
$$square g square g square g square g square g square gsquare$$
Therefore:
$$6!cdot 7P5=1space 814space 400$$
However, I notice that if you take one of the girls away (let's label her as $^*$), we have:
$$^*square^*g^*square^*g^*square^*g^*square^*g^*square^*g^*square^*$$
We see that no matter where girl$^*$ stands, the boys will still not stand next to each others:
$$5!cdot6P5cdot12=1space036space800$$
Update
If we are to swap it around:
$$square b square b square b square b square b square$$
We now have:
$$5!cdot6!=86space 400 $$
So which is right? Is there anything wrong with each logic?
Thank you.
combinatorics permutations
$endgroup$
add a comment |
$begingroup$
We have:
$$square g square g square g square g square g square gsquare$$
Therefore:
$$6!cdot 7P5=1space 814space 400$$
However, I notice that if you take one of the girls away (let's label her as $^*$), we have:
$$^*square^*g^*square^*g^*square^*g^*square^*g^*square^*g^*square^*$$
We see that no matter where girl$^*$ stands, the boys will still not stand next to each others:
$$5!cdot6P5cdot12=1space036space800$$
Update
If we are to swap it around:
$$square b square b square b square b square b square$$
We now have:
$$5!cdot6!=86space 400 $$
So which is right? Is there anything wrong with each logic?
Thank you.
combinatorics permutations
$endgroup$
We have:
$$square g square g square g square g square g square gsquare$$
Therefore:
$$6!cdot 7P5=1space 814space 400$$
However, I notice that if you take one of the girls away (let's label her as $^*$), we have:
$$^*square^*g^*square^*g^*square^*g^*square^*g^*square^*g^*square^*$$
We see that no matter where girl$^*$ stands, the boys will still not stand next to each others:
$$5!cdot6P5cdot12=1space036space800$$
Update
If we are to swap it around:
$$square b square b square b square b square b square$$
We now have:
$$5!cdot6!=86space 400 $$
So which is right? Is there anything wrong with each logic?
Thank you.
combinatorics permutations
combinatorics permutations
edited 3 mins ago
Huy Tran Van
asked 1 hour 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.
$endgroup$
$begingroup$
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
43 mins ago
$begingroup$
@HuyTranVan You misread. I said $1b2b34b5b6b$. There is no $b$ between $3$ and $4$.
$endgroup$
– orlp
31 mins 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
25 mins 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.
$endgroup$
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
New contributor
$endgroup$
$begingroup$
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
56 mins ago
add a comment |
Your Answer
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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.
$endgroup$
$begingroup$
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
43 mins ago
$begingroup$
@HuyTranVan You misread. I said $1b2b34b5b6b$. There is no $b$ between $3$ and $4$.
$endgroup$
– orlp
31 mins 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
25 mins 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.
$endgroup$
$begingroup$
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
43 mins ago
$begingroup$
@HuyTranVan You misread. I said $1b2b34b5b6b$. There is no $b$ between $3$ and $4$.
$endgroup$
– orlp
31 mins 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
25 mins 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.
$endgroup$
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
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
43 mins ago
$begingroup$
@HuyTranVan You misread. I said $1b2b34b5b6b$. There is no $b$ between $3$ and $4$.
$endgroup$
– orlp
31 mins 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
25 mins 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
43 mins ago
$begingroup$
@HuyTranVan You misread. I said $1b2b34b5b6b$. There is no $b$ between $3$ and $4$.
$endgroup$
– orlp
31 mins 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
25 mins 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
43 mins 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
43 mins ago
$begingroup$
@HuyTranVan You misread. I said $1b2b34b5b6b$. There is no $b$ between $3$ and $4$.
$endgroup$
– orlp
31 mins ago
$begingroup$
@HuyTranVan You misread. I said $1b2b34b5b6b$. There is no $b$ between $3$ and $4$.
$endgroup$
– orlp
31 mins 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
25 mins 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
25 mins 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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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.
edited 1 hour ago
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
New contributor
$endgroup$
$begingroup$
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
56 mins 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
New contributor
$endgroup$
$begingroup$
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
56 mins 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
New contributor
$endgroup$
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
New contributor
New contributor
answered 58 mins ago
TucktuckgooseTucktuckgoose
11
11
New contributor
New contributor
$begingroup$
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
56 mins 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
56 mins 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
56 mins 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
56 mins ago
add a comment |
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