8 people including A,B,C, and D will be rearranged. In how many ways can they be rearranged such that B and C...
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 2 hours ago
greedoid
36.7k114592
asked 2 hours ago
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 2 hours ago
greedoid
36.7k114592
asked 2 hours ago
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 2 hours ago
greedoid
36.7k114592
asked 2 hours ago
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 2 hours ago
greedoid
36.7k114592
asked 2 hours ago
Mr. Maxwell
325
edited 2 hours ago
greedoid
36.7k114592
asked 2 hours ago
Mr. Maxwell
325
edited 2 hours ago
greedoid
36.7k114592
edited 2 hours ago
greedoid
36.7k114592
edited 2 hours ago
greedoid
36.7k114592
36.7k114592
asked 2 hours ago
Mr. Maxwell
325
asked 2 hours ago
Mr. Maxwell
325
asked 2 hours ago
Mr. Maxwell
325
325
add a comment |
add a comment |
2 Answers
2
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 2 hours ago
greedoid
36.7k114592
That seems a bit unclear. Could you please make it more clear?
– Mr. Maxwell
2 hours ago
Which part of it?
– greedoid
2 hours ago
$${8choose 4}$$ And what's also the reason you put B and c in the middle?
– Mr. Maxwell
2 hours ago
What about it? You don't know what it means?
– greedoid
2 hours ago
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
2 hours ago
|
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 2 hours ago
Robert Z
92.3k1058129
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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 2 hours ago
greedoid
36.7k114592
That seems a bit unclear. Could you please make it more clear?
– Mr. Maxwell
2 hours ago
Which part of it?
– greedoid
2 hours ago
$${8choose 4}$$ And what's also the reason you put B and c in the middle?
– Mr. Maxwell
2 hours ago
What about it? You don't know what it means?
– greedoid
2 hours ago
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
2 hours ago
|
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 2 hours ago
greedoid
36.7k114592
That seems a bit unclear. Could you please make it more clear?
– Mr. Maxwell
2 hours ago
Which part of it?
– greedoid
2 hours ago
$${8choose 4}$$ And what's also the reason you put B and c in the middle?
– Mr. Maxwell
2 hours ago
What about it? You don't know what it means?
– greedoid
2 hours ago
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
2 hours ago
|
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 2 hours ago
greedoid
36.7k114592
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 2 hours ago
greedoid
36.7k114592
answered 2 hours ago
greedoid
36.7k114592
answered 2 hours ago
greedoid
36.7k114592
answered 2 hours ago
greedoid
36.7k114592
36.7k114592
That seems a bit unclear. Could you please make it more clear?
– Mr. Maxwell
2 hours ago
Which part of it?
– greedoid
2 hours ago
$${8choose 4}$$ And what's also the reason you put B and c in the middle?
– Mr. Maxwell
2 hours ago
What about it? You don't know what it means?
– greedoid
2 hours ago
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
2 hours ago
|
show 5 more comments
That seems a bit unclear. Could you please make it more clear?
– Mr. Maxwell
2 hours ago
Which part of it?
– greedoid
2 hours ago
$${8choose 4}$$ And what's also the reason you put B and c in the middle?
– Mr. Maxwell
2 hours ago
What about it? You don't know what it means?
– greedoid
2 hours ago
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
2 hours ago
That seems a bit unclear. Could you please make it more clear?
– Mr. Maxwell
2 hours ago
That seems a bit unclear. Could you please make it more clear?
– Mr. Maxwell
2 hours ago
Which part of it?
– greedoid
2 hours ago
Which part of it?
– greedoid
2 hours ago
$${8choose 4}$$ And what's also the reason you put B and c in the middle?
– Mr. Maxwell
2 hours ago
$${8choose 4}$$ And what's also the reason you put B and c in the middle?
– Mr. Maxwell
2 hours ago
What about it? You don't know what it means?
– greedoid
2 hours ago
What about it? You don't know what it means?
– greedoid
2 hours ago
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
2 hours ago
@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
2 hours ago
|
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 2 hours ago
Robert Z
92.3k1058129
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 2 hours ago
Robert Z
92.3k1058129
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 2 hours ago
Robert Z
92.3k1058129
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 2 hours ago
Robert Z
92.3k1058129
answered 2 hours ago
Robert Z
92.3k1058129
answered 2 hours ago
Robert Z
92.3k1058129
answered 2 hours ago
Robert Z
92.3k1058129
92.3k1058129
add a comment |
add a comment |
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