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











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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.



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combinatorics permutations







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    up vote
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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


























      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







      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




      combinatorics permutations






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      share|cite|improve this question













      share|cite|improve this question




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      edited 2 hours ago









      greedoid

      36.7k114592




      36.7k114592










      asked 2 hours ago









      Mr. Maxwell

      325




      325






















          2 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
            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


















          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





















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            2 Answers
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            2 Answers
            2






            active

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            active

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            active

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            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
              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















            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
              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













            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 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


















            • 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










            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

























              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























                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 2 hours ago









                Robert Z

                92.3k1058129




                92.3k1058129






























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