$8$ people including $A,B,C$, and $D$ will be rearranged. In how many ways can they be rearranged such that...











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$8$ 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










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    up vote
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    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










    share|cite|improve this question


























      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










      share|cite|improve this question
















      $8$ 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






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      edited yesterday









      Asaf Karagila

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      asked yesterday









      Mr. Maxwell

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          3 Answers
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          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 $${8choose 4}cdot 2cdot 2cdot 4! = 8!/6$$






          share|cite|improve this answer





















          • 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


















          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
          $${8choose 4}cdot 4 cdot 4!$$






          share|cite|improve this answer





















          • 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


















          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.






          share|cite|improve this answer










          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.


















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            3 Answers
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            up vote
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            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 $${8choose 4}cdot 2cdot 2cdot 4! = 8!/6$$






            share|cite|improve this answer





















            • 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















            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 $${8choose 4}cdot 2cdot 2cdot 4! = 8!/6$$






            share|cite|improve this answer





















            • 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













            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 $${8choose 4}cdot 2cdot 2cdot 4! = 8!/6$$






            share|cite|improve this answer












            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 $${8choose 4}cdot 2cdot 2cdot 4! = 8!/6$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            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


















            • 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










            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
            $${8choose 4}cdot 4 cdot 4!$$






            share|cite|improve this answer





















            • 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















            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
            $${8choose 4}cdot 4 cdot 4!$$






            share|cite|improve this answer





















            • 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













            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
            $${8choose 4}cdot 4 cdot 4!$$






            share|cite|improve this answer












            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
            $${8choose 4}cdot 4 cdot 4!$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            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


















            • 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










            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.






            share|cite|improve this answer










            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.






















              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.






              share|cite|improve this answer










              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.




















                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.






                share|cite|improve this answer










                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.







                share|cite|improve this answer










                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.









                share|cite|improve this answer



                share|cite|improve this answer








                edited yesterday





















                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




                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.






























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