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












2












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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






combinatorics permutations







share|cite|improve this question











$endgroup$

















    2












    $begingroup$



    $8$ 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







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$



      $8$ 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







      share|cite|improve this question











      $endgroup$





      $8$ 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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 16 '18 at 14:03









      Asaf Karagila♦

      302k32429760




      302k32429760










      asked Dec 16 '18 at 10:33









      Mr. MaxwellMr. Maxwell

      325




      325






















          3 Answers
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          active

          oldest

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          3












          $begingroup$

          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









          $endgroup$













          • $begingroup$
            That seems a bit unclear. Could you please make it more clear?
            $endgroup$
            – Mr. Maxwell
            Dec 16 '18 at 10:43










          • $begingroup$
            Which part of it?
            $endgroup$
            – greedoid
            Dec 16 '18 at 10:43










          • $begingroup$
            $${8choose 4}$$ And what's also the reason you put B and c in the middle?
            $endgroup$
            – Mr. Maxwell
            Dec 16 '18 at 10:44












          • $begingroup$
            What about it? You don't know what it means?
            $endgroup$
            – greedoid
            Dec 16 '18 at 10:45






          • 1




            $begingroup$
            @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
            $endgroup$
            – Shubham Johri
            Dec 16 '18 at 11:12



















          3












          $begingroup$

          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









          $endgroup$













          • $begingroup$
            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
            $endgroup$
            – adhg
            Dec 18 '18 at 1:39






          • 1




            $begingroup$
            No. With $binom{8}{4}$ we choose just the positions where we are going to put the letters $A$, $B$, $C$, $D$ (no order is specified). In the complement we will put the other letters. Then we place in chosen positions the letters $A$, $B$, $C$, $D$ in one of the $4$ orders and the remaining letters in the remaining positions in one of the $4!$ orders.
            $endgroup$
            – Robert Z
            Dec 18 '18 at 5:21










          • $begingroup$
            ohh. Right! Got it
            $endgroup$
            – adhg
            Dec 18 '18 at 13:11



















          0












          $begingroup$

          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











          $endgroup$













            Your Answer





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            3 Answers
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            3 Answers
            3






            active

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            active

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            active

            oldest

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            3












            $begingroup$

            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









            $endgroup$













            • $begingroup$
              That seems a bit unclear. Could you please make it more clear?
              $endgroup$
              – Mr. Maxwell
              Dec 16 '18 at 10:43










            • $begingroup$
              Which part of it?
              $endgroup$
              – greedoid
              Dec 16 '18 at 10:43










            • $begingroup$
              $${8choose 4}$$ And what's also the reason you put B and c in the middle?
              $endgroup$
              – Mr. Maxwell
              Dec 16 '18 at 10:44












            • $begingroup$
              What about it? You don't know what it means?
              $endgroup$
              – greedoid
              Dec 16 '18 at 10:45






            • 1




              $begingroup$
              @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
              $endgroup$
              – Shubham Johri
              Dec 16 '18 at 11:12
















            3












            $begingroup$

            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









            $endgroup$













            • $begingroup$
              That seems a bit unclear. Could you please make it more clear?
              $endgroup$
              – Mr. Maxwell
              Dec 16 '18 at 10:43










            • $begingroup$
              Which part of it?
              $endgroup$
              – greedoid
              Dec 16 '18 at 10:43










            • $begingroup$
              $${8choose 4}$$ And what's also the reason you put B and c in the middle?
              $endgroup$
              – Mr. Maxwell
              Dec 16 '18 at 10:44












            • $begingroup$
              What about it? You don't know what it means?
              $endgroup$
              – greedoid
              Dec 16 '18 at 10:45






            • 1




              $begingroup$
              @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
              $endgroup$
              – Shubham Johri
              Dec 16 '18 at 11:12














            3












            3








            3





            $begingroup$

            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









            $endgroup$



            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 Dec 16 '18 at 10:37









            greedoidgreedoid

            38.7k114797




            38.7k114797












            • $begingroup$
              That seems a bit unclear. Could you please make it more clear?
              $endgroup$
              – Mr. Maxwell
              Dec 16 '18 at 10:43










            • $begingroup$
              Which part of it?
              $endgroup$
              – greedoid
              Dec 16 '18 at 10:43










            • $begingroup$
              $${8choose 4}$$ And what's also the reason you put B and c in the middle?
              $endgroup$
              – Mr. Maxwell
              Dec 16 '18 at 10:44












            • $begingroup$
              What about it? You don't know what it means?
              $endgroup$
              – greedoid
              Dec 16 '18 at 10:45






            • 1




              $begingroup$
              @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
              $endgroup$
              – Shubham Johri
              Dec 16 '18 at 11:12


















            • $begingroup$
              That seems a bit unclear. Could you please make it more clear?
              $endgroup$
              – Mr. Maxwell
              Dec 16 '18 at 10:43










            • $begingroup$
              Which part of it?
              $endgroup$
              – greedoid
              Dec 16 '18 at 10:43










            • $begingroup$
              $${8choose 4}$$ And what's also the reason you put B and c in the middle?
              $endgroup$
              – Mr. Maxwell
              Dec 16 '18 at 10:44












            • $begingroup$
              What about it? You don't know what it means?
              $endgroup$
              – greedoid
              Dec 16 '18 at 10:45






            • 1




              $begingroup$
              @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
              $endgroup$
              – Shubham Johri
              Dec 16 '18 at 11:12
















            $begingroup$
            That seems a bit unclear. Could you please make it more clear?
            $endgroup$
            – Mr. Maxwell
            Dec 16 '18 at 10:43




            $begingroup$
            That seems a bit unclear. Could you please make it more clear?
            $endgroup$
            – Mr. Maxwell
            Dec 16 '18 at 10:43












            $begingroup$
            Which part of it?
            $endgroup$
            – greedoid
            Dec 16 '18 at 10:43




            $begingroup$
            Which part of it?
            $endgroup$
            – greedoid
            Dec 16 '18 at 10:43












            $begingroup$
            $${8choose 4}$$ And what's also the reason you put B and c in the middle?
            $endgroup$
            – Mr. Maxwell
            Dec 16 '18 at 10:44






            $begingroup$
            $${8choose 4}$$ And what's also the reason you put B and c in the middle?
            $endgroup$
            – Mr. Maxwell
            Dec 16 '18 at 10:44














            $begingroup$
            What about it? You don't know what it means?
            $endgroup$
            – greedoid
            Dec 16 '18 at 10:45




            $begingroup$
            What about it? You don't know what it means?
            $endgroup$
            – greedoid
            Dec 16 '18 at 10:45




            1




            1




            $begingroup$
            @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
            $endgroup$
            – Shubham Johri
            Dec 16 '18 at 11:12




            $begingroup$
            @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
            $endgroup$
            – Shubham Johri
            Dec 16 '18 at 11:12











            3












            $begingroup$

            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









            $endgroup$













            • $begingroup$
              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
              $endgroup$
              – adhg
              Dec 18 '18 at 1:39






            • 1




              $begingroup$
              No. With $binom{8}{4}$ we choose just the positions where we are going to put the letters $A$, $B$, $C$, $D$ (no order is specified). In the complement we will put the other letters. Then we place in chosen positions the letters $A$, $B$, $C$, $D$ in one of the $4$ orders and the remaining letters in the remaining positions in one of the $4!$ orders.
              $endgroup$
              – Robert Z
              Dec 18 '18 at 5:21










            • $begingroup$
              ohh. Right! Got it
              $endgroup$
              – adhg
              Dec 18 '18 at 13:11
















            3












            $begingroup$

            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









            $endgroup$













            • $begingroup$
              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
              $endgroup$
              – adhg
              Dec 18 '18 at 1:39






            • 1




              $begingroup$
              No. With $binom{8}{4}$ we choose just the positions where we are going to put the letters $A$, $B$, $C$, $D$ (no order is specified). In the complement we will put the other letters. Then we place in chosen positions the letters $A$, $B$, $C$, $D$ in one of the $4$ orders and the remaining letters in the remaining positions in one of the $4!$ orders.
              $endgroup$
              – Robert Z
              Dec 18 '18 at 5:21










            • $begingroup$
              ohh. Right! Got it
              $endgroup$
              – adhg
              Dec 18 '18 at 13:11














            3












            3








            3





            $begingroup$

            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









            $endgroup$



            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 Dec 16 '18 at 10:50









            Robert ZRobert Z

            94.4k1063134




            94.4k1063134












            • $begingroup$
              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
              $endgroup$
              – adhg
              Dec 18 '18 at 1:39






            • 1




              $begingroup$
              No. With $binom{8}{4}$ we choose just the positions where we are going to put the letters $A$, $B$, $C$, $D$ (no order is specified). In the complement we will put the other letters. Then we place in chosen positions the letters $A$, $B$, $C$, $D$ in one of the $4$ orders and the remaining letters in the remaining positions in one of the $4!$ orders.
              $endgroup$
              – Robert Z
              Dec 18 '18 at 5:21










            • $begingroup$
              ohh. Right! Got it
              $endgroup$
              – adhg
              Dec 18 '18 at 13:11


















            • $begingroup$
              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
              $endgroup$
              – adhg
              Dec 18 '18 at 1:39






            • 1




              $begingroup$
              No. With $binom{8}{4}$ we choose just the positions where we are going to put the letters $A$, $B$, $C$, $D$ (no order is specified). In the complement we will put the other letters. Then we place in chosen positions the letters $A$, $B$, $C$, $D$ in one of the $4$ orders and the remaining letters in the remaining positions in one of the $4!$ orders.
              $endgroup$
              – Robert Z
              Dec 18 '18 at 5:21










            • $begingroup$
              ohh. Right! Got it
              $endgroup$
              – adhg
              Dec 18 '18 at 13:11
















            $begingroup$
            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
            $endgroup$
            – adhg
            Dec 18 '18 at 1:39




            $begingroup$
            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
            $endgroup$
            – adhg
            Dec 18 '18 at 1:39




            1




            1




            $begingroup$
            No. With $binom{8}{4}$ we choose just the positions where we are going to put the letters $A$, $B$, $C$, $D$ (no order is specified). In the complement we will put the other letters. Then we place in chosen positions the letters $A$, $B$, $C$, $D$ in one of the $4$ orders and the remaining letters in the remaining positions in one of the $4!$ orders.
            $endgroup$
            – Robert Z
            Dec 18 '18 at 5:21




            $begingroup$
            No. With $binom{8}{4}$ we choose just the positions where we are going to put the letters $A$, $B$, $C$, $D$ (no order is specified). In the complement we will put the other letters. Then we place in chosen positions the letters $A$, $B$, $C$, $D$ in one of the $4$ orders and the remaining letters in the remaining positions in one of the $4!$ orders.
            $endgroup$
            – Robert Z
            Dec 18 '18 at 5:21












            $begingroup$
            ohh. Right! Got it
            $endgroup$
            – adhg
            Dec 18 '18 at 13:11




            $begingroup$
            ohh. Right! Got it
            $endgroup$
            – adhg
            Dec 18 '18 at 13:11











            0












            $begingroup$

            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











            $endgroup$


















              0












              $begingroup$

              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











              $endgroup$
















                0












                0








                0





                $begingroup$

                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











                $endgroup$



                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














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 16 '18 at 14:55

























                answered Dec 16 '18 at 14:50









                tom riddletom riddle

                11




                11






























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