$8$ people including $A,B,C$, and $D$ will be rearranged. In how many ways can they be rearranged such that...
up vote
2
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favorite
$8$ people including $A,B,C$, and $D$ will be rearranged. In how many ways can they be rearranged such that $B$ and $C$ will be between $A$ and $D$?
I tried to break this into cases. However, it didn't work.
Regards
combinatorics permutations
edited yesterday
Asaf Karagila♦
301k32422753
asked yesterday
Mr. Maxwell
325
add a comment |
up vote
2
down vote
favorite
$8$ people including $A,B,C$, and $D$ will be rearranged. In how many ways can they be rearranged such that $B$ and $C$ will be between $A$ and $D$?
I tried to break this into cases. However, it didn't work.
Regards
combinatorics permutations
edited yesterday
Asaf Karagila♦
301k32422753
asked yesterday
Mr. Maxwell
325
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$8$ people including $A,B,C$, and $D$ will be rearranged. In how many ways can they be rearranged such that $B$ and $C$ will be between $A$ and $D$?
I tried to break this into cases. However, it didn't work.
Regards
combinatorics permutations
edited yesterday
Asaf Karagila♦
301k32422753
asked yesterday
Mr. Maxwell
325
$8$ people including $A,B,C$, and $D$ will be rearranged. In how many ways can they be rearranged such that $B$ and $C$ will be between $A$ and $D$?
I tried to break this into cases. However, it didn't work.
Regards
combinatorics permutations
combinatorics permutations
edited yesterday
Asaf Karagila♦
301k32422753
asked yesterday
Mr. Maxwell
325
edited yesterday
Asaf Karagila♦
301k32422753
asked yesterday
Mr. Maxwell
325
edited yesterday
Asaf Karagila♦
301k32422753
edited yesterday
Asaf Karagila♦
301k32422753
edited yesterday
Asaf Karagila♦
301k32422753
301k32422753
asked yesterday
Mr. Maxwell
325
asked yesterday
Mr. Maxwell
325
asked yesterday
Mr. Maxwell
325
325
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
up vote
3
down vote
First choose 4 places, that you can do on ${8choose 4}$ ways, then on edge put $A$ and $D$, you can do that on 2 ways and in the midle $B$ and $C$, again on 2 ways. Then arrange all others on remaining places, that is on 4! ways. Now multiply all these. So the answer is $${8choose 4}cdot 2cdot 2cdot 4! = 8!/6$$
answered yesterday
greedoid
36.7k114593
That seems a bit unclear. Could you please make it more clear?
– Mr. Maxwell
yesterday
Which part of it?
– greedoid
yesterday
$${8choose 4}$$ And what's also the reason you put B and c in the middle?
– Mr. Maxwell
yesterday
What about it? You don't know what it means?
– greedoid
yesterday
1
@Mr.Maxwell You have to put $B,C$ between $A,D$. This doesn't mean $A,D$ have to sit on the edges. The arrangement can also look like this: _ _ A _ B _ C D
– Shubham Johri
yesterday
|
show 5 more comments
up vote
2
down vote
We first choose among the $8$ positions the $4$ ones for $A,B,C,D$. This can be done in $binom{8}{4}$ ways. These positions cab be filled in $4$ ways: $ABCD$, $ACBD$, $DBCA$, $DCBA$. The remanining $8-4$ positions can be filled in $4!$ ways. Hence the total number of arrangements is
$${8choose 4}cdot 4 cdot 4!$$
answered yesterday
Robert Z
92.5k1059130
Can you help me understand the first part. I'm a bit confused. When you do 8c4 doesn't that include instances such as: D_CAB_? If so, shouldn't you remove them because BC should be between AD. Thanks
– adhg
2 hours ago
add a comment |
up vote
0
down vote
This answer assumes that ABCD is unique from ACBD, though I can answer for the other case as well if that is the case.
N = number of spots (8 in your case)
2 * summation(n = 1, N - 3){n! * (N+1-n)!}
In your case, this simplifies to
Take the two boundary tokens and orient them on the edges. There are two "symmetric" orientations for each orientation of the boundary tokens (A******D is distinct from D******A, and there are only two orientations of two tokens with two spots.) From there, you have the inner spaces, designated as "Sep". For each value Sep, there are Sep! permutations inside the boundary tokens, and (N - (Sep + 2))! permutations outside the boundary, so for each value Sep, there are Sep! * (N - (Sep + 2))! * 2 * (number of "slide" positions). The "Slide" positions are the orientations you can bring A and D to through sliding; in simple terms, if the string was A****D**, Slide is 3, since you can have A****D**, *A****D*, **A****D. Factor the 2 out and do some refactoring, plug in N = 8, and you get 2976.
EDIT: My final sum of 2976 was a quick desmos plugin, so if anyone would like to plug in the equation I provided above to verify, you can go ahead and do that.
answered yesterday
tom riddle
11
New contributor
tom riddle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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3 Answers
3
active
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3 Answers
3
active
oldest
votes
active
oldest
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active
oldest
votes
up vote
3
down vote
First choose 4 places, that you can do on ${8choose 4}$ ways, then on edge put $A$ and $D$, you can do that on 2 ways and in the midle $B$ and $C$, again on 2 ways. Then arrange all others on remaining places, that is on 4! ways. Now multiply all these. So the answer is $${8choose 4}cdot 2cdot 2cdot 4! = 8!/6$$
answered yesterday
greedoid
36.7k114593
That seems a bit unclear. Could you please make it more clear?
– Mr. Maxwell
yesterday
Which part of it?
– greedoid
yesterday
$${8choose 4}$$ And what's also the reason you put B and c in the middle?
– Mr. Maxwell
yesterday
What about it? You don't know what it means?
– greedoid
yesterday
1
@Mr.Maxwell You have to put $B,C$ between $A,D$. This doesn't mean $A,D$ have to sit on the edges. The arrangement can also look like this: _ _ A _ B _ C D
– Shubham Johri
yesterday
|
show 5 more comments
up vote
3
down vote
First choose 4 places, that you can do on ${8choose 4}$ ways, then on edge put $A$ and $D$, you can do that on 2 ways and in the midle $B$ and $C$, again on 2 ways. Then arrange all others on remaining places, that is on 4! ways. Now multiply all these. So the answer is $${8choose 4}cdot 2cdot 2cdot 4! = 8!/6$$
answered yesterday
greedoid
36.7k114593
That seems a bit unclear. Could you please make it more clear?
– Mr. Maxwell
yesterday
Which part of it?
– greedoid
yesterday
$${8choose 4}$$ And what's also the reason you put B and c in the middle?
– Mr. Maxwell
yesterday
What about it? You don't know what it means?
– greedoid
yesterday
1
@Mr.Maxwell You have to put $B,C$ between $A,D$. This doesn't mean $A,D$ have to sit on the edges. The arrangement can also look like this: _ _ A _ B _ C D
– Shubham Johri
yesterday
|
show 5 more comments
up vote
3
down vote
up vote
3
down vote
First choose 4 places, that you can do on ${8choose 4}$ ways, then on edge put $A$ and $D$, you can do that on 2 ways and in the midle $B$ and $C$, again on 2 ways. Then arrange all others on remaining places, that is on 4! ways. Now multiply all these. So the answer is $${8choose 4}cdot 2cdot 2cdot 4! = 8!/6$$
answered yesterday
greedoid
36.7k114593
First choose 4 places, that you can do on ${8choose 4}$ ways, then on edge put $A$ and $D$, you can do that on 2 ways and in the midle $B$ and $C$, again on 2 ways. Then arrange all others on remaining places, that is on 4! ways. Now multiply all these. So the answer is $${8choose 4}cdot 2cdot 2cdot 4! = 8!/6$$
answered yesterday
greedoid
36.7k114593
answered yesterday
greedoid
36.7k114593
answered yesterday
greedoid
36.7k114593
answered yesterday
greedoid
36.7k114593
36.7k114593
That seems a bit unclear. Could you please make it more clear?
– Mr. Maxwell
yesterday
Which part of it?
– greedoid
yesterday
$${8choose 4}$$ And what's also the reason you put B and c in the middle?
– Mr. Maxwell
yesterday
What about it? You don't know what it means?
– greedoid
yesterday
1
@Mr.Maxwell You have to put $B,C$ between $A,D$. This doesn't mean $A,D$ have to sit on the edges. The arrangement can also look like this: _ _ A _ B _ C D
– Shubham Johri
yesterday
|
show 5 more comments
That seems a bit unclear. Could you please make it more clear?
– Mr. Maxwell
yesterday
Which part of it?
– greedoid
yesterday
$${8choose 4}$$ And what's also the reason you put B and c in the middle?
– Mr. Maxwell
yesterday
What about it? You don't know what it means?
– greedoid
yesterday
1
@Mr.Maxwell You have to put $B,C$ between $A,D$. This doesn't mean $A,D$ have to sit on the edges. The arrangement can also look like this: _ _ A _ B _ C D
– Shubham Johri
yesterday
That seems a bit unclear. Could you please make it more clear?
– Mr. Maxwell
yesterday
That seems a bit unclear. Could you please make it more clear?
– Mr. Maxwell
yesterday
Which part of it?
– greedoid
yesterday
Which part of it?
– greedoid
yesterday
$${8choose 4}$$ And what's also the reason you put B and c in the middle?
– Mr. Maxwell
yesterday
$${8choose 4}$$ And what's also the reason you put B and c in the middle?
– Mr. Maxwell
yesterday
What about it? You don't know what it means?
– greedoid
yesterday
What about it? You don't know what it means?
– greedoid
yesterday
1
1
@Mr.Maxwell You have to put $B,C$ between $A,D$. This doesn't mean $A,D$ have to sit on the edges. The arrangement can also look like this: _ _ A _ B _ C D
– Shubham Johri
yesterday
@Mr.Maxwell You have to put $B,C$ between $A,D$. This doesn't mean $A,D$ have to sit on the edges. The arrangement can also look like this: _ _ A _ B _ C D
– Shubham Johri
yesterday
|
show 5 more comments
up vote
2
down vote
We first choose among the $8$ positions the $4$ ones for $A,B,C,D$. This can be done in $binom{8}{4}$ ways. These positions cab be filled in $4$ ways: $ABCD$, $ACBD$, $DBCA$, $DCBA$. The remanining $8-4$ positions can be filled in $4!$ ways. Hence the total number of arrangements is
$${8choose 4}cdot 4 cdot 4!$$
answered yesterday
Robert Z
92.5k1059130
Can you help me understand the first part. I'm a bit confused. When you do 8c4 doesn't that include instances such as: D_CAB_? If so, shouldn't you remove them because BC should be between AD. Thanks
– adhg
2 hours ago
add a comment |
up vote
2
down vote
We first choose among the $8$ positions the $4$ ones for $A,B,C,D$. This can be done in $binom{8}{4}$ ways. These positions cab be filled in $4$ ways: $ABCD$, $ACBD$, $DBCA$, $DCBA$. The remanining $8-4$ positions can be filled in $4!$ ways. Hence the total number of arrangements is
$${8choose 4}cdot 4 cdot 4!$$
answered yesterday
Robert Z
92.5k1059130
Can you help me understand the first part. I'm a bit confused. When you do 8c4 doesn't that include instances such as: D_CAB_? If so, shouldn't you remove them because BC should be between AD. Thanks
– adhg
2 hours ago
add a comment |
up vote
2
down vote
up vote
2
down vote
We first choose among the $8$ positions the $4$ ones for $A,B,C,D$. This can be done in $binom{8}{4}$ ways. These positions cab be filled in $4$ ways: $ABCD$, $ACBD$, $DBCA$, $DCBA$. The remanining $8-4$ positions can be filled in $4!$ ways. Hence the total number of arrangements is
$${8choose 4}cdot 4 cdot 4!$$
answered yesterday
Robert Z
92.5k1059130
We first choose among the $8$ positions the $4$ ones for $A,B,C,D$. This can be done in $binom{8}{4}$ ways. These positions cab be filled in $4$ ways: $ABCD$, $ACBD$, $DBCA$, $DCBA$. The remanining $8-4$ positions can be filled in $4!$ ways. Hence the total number of arrangements is
$${8choose 4}cdot 4 cdot 4!$$
answered yesterday
Robert Z
92.5k1059130
answered yesterday
Robert Z
92.5k1059130
answered yesterday
Robert Z
92.5k1059130
answered yesterday
Robert Z
92.5k1059130
92.5k1059130
Can you help me understand the first part. I'm a bit confused. When you do 8c4 doesn't that include instances such as: D_CAB_? If so, shouldn't you remove them because BC should be between AD. Thanks
– adhg
2 hours ago
add a comment |
Can you help me understand the first part. I'm a bit confused. When you do 8c4 doesn't that include instances such as: D_CAB_? If so, shouldn't you remove them because BC should be between AD. Thanks
– adhg
2 hours ago
Can you help me understand the first part. I'm a bit confused. When you do 8c4 doesn't that include instances such as: D_CAB_? If so, shouldn't you remove them because BC should be between AD. Thanks
– adhg
2 hours ago
Can you help me understand the first part. I'm a bit confused. When you do 8c4 doesn't that include instances such as: D_CAB_? If so, shouldn't you remove them because BC should be between AD. Thanks
– adhg
2 hours ago
add a comment |
up vote
0
down vote
This answer assumes that ABCD is unique from ACBD, though I can answer for the other case as well if that is the case.
N = number of spots (8 in your case)
2 * summation(n = 1, N - 3){n! * (N+1-n)!}
In your case, this simplifies to
Take the two boundary tokens and orient them on the edges. There are two "symmetric" orientations for each orientation of the boundary tokens (A******D is distinct from D******A, and there are only two orientations of two tokens with two spots.) From there, you have the inner spaces, designated as "Sep". For each value Sep, there are Sep! permutations inside the boundary tokens, and (N - (Sep + 2))! permutations outside the boundary, so for each value Sep, there are Sep! * (N - (Sep + 2))! * 2 * (number of "slide" positions). The "Slide" positions are the orientations you can bring A and D to through sliding; in simple terms, if the string was A****D**, Slide is 3, since you can have A****D**, *A****D*, **A****D. Factor the 2 out and do some refactoring, plug in N = 8, and you get 2976.
EDIT: My final sum of 2976 was a quick desmos plugin, so if anyone would like to plug in the equation I provided above to verify, you can go ahead and do that.
answered yesterday
tom riddle
11
New contributor
tom riddle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
This answer assumes that ABCD is unique from ACBD, though I can answer for the other case as well if that is the case.
N = number of spots (8 in your case)
2 * summation(n = 1, N - 3){n! * (N+1-n)!}
In your case, this simplifies to
Take the two boundary tokens and orient them on the edges. There are two "symmetric" orientations for each orientation of the boundary tokens (A******D is distinct from D******A, and there are only two orientations of two tokens with two spots.) From there, you have the inner spaces, designated as "Sep". For each value Sep, there are Sep! permutations inside the boundary tokens, and (N - (Sep + 2))! permutations outside the boundary, so for each value Sep, there are Sep! * (N - (Sep + 2))! * 2 * (number of "slide" positions). The "Slide" positions are the orientations you can bring A and D to through sliding; in simple terms, if the string was A****D**, Slide is 3, since you can have A****D**, *A****D*, **A****D. Factor the 2 out and do some refactoring, plug in N = 8, and you get 2976.
EDIT: My final sum of 2976 was a quick desmos plugin, so if anyone would like to plug in the equation I provided above to verify, you can go ahead and do that.
answered yesterday
tom riddle
11
New contributor
tom riddle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
up vote
0
down vote
This answer assumes that ABCD is unique from ACBD, though I can answer for the other case as well if that is the case.
N = number of spots (8 in your case)
2 * summation(n = 1, N - 3){n! * (N+1-n)!}
In your case, this simplifies to
Take the two boundary tokens and orient them on the edges. There are two "symmetric" orientations for each orientation of the boundary tokens (A******D is distinct from D******A, and there are only two orientations of two tokens with two spots.) From there, you have the inner spaces, designated as "Sep". For each value Sep, there are Sep! permutations inside the boundary tokens, and (N - (Sep + 2))! permutations outside the boundary, so for each value Sep, there are Sep! * (N - (Sep + 2))! * 2 * (number of "slide" positions). The "Slide" positions are the orientations you can bring A and D to through sliding; in simple terms, if the string was A****D**, Slide is 3, since you can have A****D**, *A****D*, **A****D. Factor the 2 out and do some refactoring, plug in N = 8, and you get 2976.
EDIT: My final sum of 2976 was a quick desmos plugin, so if anyone would like to plug in the equation I provided above to verify, you can go ahead and do that.
answered yesterday
tom riddle
11
New contributor
tom riddle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
This answer assumes that ABCD is unique from ACBD, though I can answer for the other case as well if that is the case.
N = number of spots (8 in your case)
2 * summation(n = 1, N - 3){n! * (N+1-n)!}
In your case, this simplifies to
Take the two boundary tokens and orient them on the edges. There are two "symmetric" orientations for each orientation of the boundary tokens (A******D is distinct from D******A, and there are only two orientations of two tokens with two spots.) From there, you have the inner spaces, designated as "Sep". For each value Sep, there are Sep! permutations inside the boundary tokens, and (N - (Sep + 2))! permutations outside the boundary, so for each value Sep, there are Sep! * (N - (Sep + 2))! * 2 * (number of "slide" positions). The "Slide" positions are the orientations you can bring A and D to through sliding; in simple terms, if the string was A****D**, Slide is 3, since you can have A****D**, *A****D*, **A****D. Factor the 2 out and do some refactoring, plug in N = 8, and you get 2976.
EDIT: My final sum of 2976 was a quick desmos plugin, so if anyone would like to plug in the equation I provided above to verify, you can go ahead and do that.
answered yesterday
tom riddle
11
New contributor
tom riddle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
answered yesterday
tom riddle
11
New contributor
tom riddle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
tom riddle
11
answered yesterday
tom riddle
11
11
New contributor
tom riddle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
tom riddle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
tom riddle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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