Permutations with constraint












2















Q. Find the number of different ways that $6$ boys and $4$ girls can stand in a line if



(i) all $6$ boys stand next to each other,



(ii) no girl stands next to another girl.




A. (i) There are $5$ "patterns", because I can start with $0$ to $4$ Gs, then put $6$ Bs, and then the remaining $4$ to $0$ Gs. For each pattern, $6!$ orders of B and $4!$ of G, so the answer is $5.6!.4! = 86400$.



(ii) As before, the answer is $k.6!.4!$, but it is not so easy to find $k$, the number of patterns. I can enumerate them as xGBxGBxGBxGx where each x represents $0$ to $3$ Bs, and the total of the xs is $3$. With some work, I find $k=35$. But there should be an easier way, I think.






combinatorics permutations







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Q. Find the number of different ways that $6$ boys and $4$ girls can stand in a line if



(i) all $6$ boys stand next to each other,



(ii) no girl stands next to another girl.




A. (i) There are $5$ "patterns", because I can start with $0$ to $4$ Gs, then put $6$ Bs, and then the remaining $4$ to $0$ Gs. For each pattern, $6!$ orders of B and $4!$ of G, so the answer is $5.6!.4! = 86400$.



(ii) As before, the answer is $k.6!.4!$, but it is not so easy to find $k$, the number of patterns. I can enumerate them as xGBxGBxGBxGx where each x represents $0$ to $3$ Bs, and the total of the xs is $3$. With some work, I find $k=35$. But there should be an easier way, I think.






combinatorics permutations







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  • Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
    – N. F. Taussig
    Nov 18 at 16:41














2












2








2








Q. Find the number of different ways that $6$ boys and $4$ girls can stand in a line if



(i) all $6$ boys stand next to each other,



(ii) no girl stands next to another girl.




A. (i) There are $5$ "patterns", because I can start with $0$ to $4$ Gs, then put $6$ Bs, and then the remaining $4$ to $0$ Gs. For each pattern, $6!$ orders of B and $4!$ of G, so the answer is $5.6!.4! = 86400$.



(ii) As before, the answer is $k.6!.4!$, but it is not so easy to find $k$, the number of patterns. I can enumerate them as xGBxGBxGBxGx where each x represents $0$ to $3$ Bs, and the total of the xs is $3$. With some work, I find $k=35$. But there should be an easier way, I think.






combinatorics permutations







share|cite|improve this question
















Q. Find the number of different ways that $6$ boys and $4$ girls can stand in a line if



(i) all $6$ boys stand next to each other,



(ii) no girl stands next to another girl.




A. (i) There are $5$ "patterns", because I can start with $0$ to $4$ Gs, then put $6$ Bs, and then the remaining $4$ to $0$ Gs. For each pattern, $6!$ orders of B and $4!$ of G, so the answer is $5.6!.4! = 86400$.



(ii) As before, the answer is $k.6!.4!$, but it is not so easy to find $k$, the number of patterns. I can enumerate them as xGBxGBxGBxGx where each x represents $0$ to $3$ Bs, and the total of the xs is $3$. With some work, I find $k=35$. But there should be an easier way, I think.






combinatorics permutations




combinatorics permutations






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edited Nov 18 at 16:40









N. F. Taussig

43.5k93355




43.5k93355










asked Nov 18 at 16:24









Rob625

111




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  • Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
    – N. F. Taussig
    Nov 18 at 16:41


















  • Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
    – N. F. Taussig
    Nov 18 at 16:41
















Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Nov 18 at 16:41




Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Nov 18 at 16:41










2 Answers
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1














(ii)



You must find the number of sums $b_1+b_2+b_3+b_4+b_5=6$ where $b_1,b_5$ are nonnegative integers and $b_2,b_3,b_4$ are positive integers.



Here $b_1$ stands for the number of boys utmost left, $b_5$ for the number of boys utmost right and e.g. $b_2$ for the number of boys between the first and the second girl on the left.



Finding this number comes to the same as finding the number of sums $a_1+a_2+a_3+a_4+a_5=3$ where then $a_i$ are nonnegative numbers.



This arises if we substitute $b_1=a_1, b_5=a_5$ and $b_i=a_i+1$ for $i=2,3,4$.



Then with stars and bars we find $binom{3+4}4=35$ possibilities for that.



So the final number of possibilities equals $35times4!times6!$.






share|cite|improve this answer





























    1














    The answers you have found are correct but there is simpler way to approach this.



    (i) Consider the 6 boys to be a single person. Now the number of permutations will be $5!$.
    However the boys themselves can be arranged in $6!$ ways. So, by principle of multiplication, the number of permutations if all the 6 boys stand together is $5!.6!=86400$



    (ii) You need to visualize this in a different way to make things simpler.



    Consider 6 boys standing with spaces between them i.e.



    $ text{ X B X B X B X B X B X B X}$



    where the $text{X}$s denote the empty spaces.



    We need to place the girls in place of $text{X}$ which is the same as "no girl stands next to another girl".



    This can be done in $P(7,4)$ or $binom{7}{4}.4!$ ways as there are 7 empty spaces or $text{X}$s and 4 girls.



    Also, the boys themselves can be arranged in $6!$ ways.



    Therefore, by principle of multiplication, the number of permutations that no girl stands next to another girl is $6!.binom{7}{4}.4!=604800$






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

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      active

      oldest

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      1














      (ii)



      You must find the number of sums $b_1+b_2+b_3+b_4+b_5=6$ where $b_1,b_5$ are nonnegative integers and $b_2,b_3,b_4$ are positive integers.



      Here $b_1$ stands for the number of boys utmost left, $b_5$ for the number of boys utmost right and e.g. $b_2$ for the number of boys between the first and the second girl on the left.



      Finding this number comes to the same as finding the number of sums $a_1+a_2+a_3+a_4+a_5=3$ where then $a_i$ are nonnegative numbers.



      This arises if we substitute $b_1=a_1, b_5=a_5$ and $b_i=a_i+1$ for $i=2,3,4$.



      Then with stars and bars we find $binom{3+4}4=35$ possibilities for that.



      So the final number of possibilities equals $35times4!times6!$.






      share|cite|improve this answer


























        1














        (ii)



        You must find the number of sums $b_1+b_2+b_3+b_4+b_5=6$ where $b_1,b_5$ are nonnegative integers and $b_2,b_3,b_4$ are positive integers.



        Here $b_1$ stands for the number of boys utmost left, $b_5$ for the number of boys utmost right and e.g. $b_2$ for the number of boys between the first and the second girl on the left.



        Finding this number comes to the same as finding the number of sums $a_1+a_2+a_3+a_4+a_5=3$ where then $a_i$ are nonnegative numbers.



        This arises if we substitute $b_1=a_1, b_5=a_5$ and $b_i=a_i+1$ for $i=2,3,4$.



        Then with stars and bars we find $binom{3+4}4=35$ possibilities for that.



        So the final number of possibilities equals $35times4!times6!$.






        share|cite|improve this answer
























          1












          1








          1






          (ii)



          You must find the number of sums $b_1+b_2+b_3+b_4+b_5=6$ where $b_1,b_5$ are nonnegative integers and $b_2,b_3,b_4$ are positive integers.



          Here $b_1$ stands for the number of boys utmost left, $b_5$ for the number of boys utmost right and e.g. $b_2$ for the number of boys between the first and the second girl on the left.



          Finding this number comes to the same as finding the number of sums $a_1+a_2+a_3+a_4+a_5=3$ where then $a_i$ are nonnegative numbers.



          This arises if we substitute $b_1=a_1, b_5=a_5$ and $b_i=a_i+1$ for $i=2,3,4$.



          Then with stars and bars we find $binom{3+4}4=35$ possibilities for that.



          So the final number of possibilities equals $35times4!times6!$.






          share|cite|improve this answer












          (ii)



          You must find the number of sums $b_1+b_2+b_3+b_4+b_5=6$ where $b_1,b_5$ are nonnegative integers and $b_2,b_3,b_4$ are positive integers.



          Here $b_1$ stands for the number of boys utmost left, $b_5$ for the number of boys utmost right and e.g. $b_2$ for the number of boys between the first and the second girl on the left.



          Finding this number comes to the same as finding the number of sums $a_1+a_2+a_3+a_4+a_5=3$ where then $a_i$ are nonnegative numbers.



          This arises if we substitute $b_1=a_1, b_5=a_5$ and $b_i=a_i+1$ for $i=2,3,4$.



          Then with stars and bars we find $binom{3+4}4=35$ possibilities for that.



          So the final number of possibilities equals $35times4!times6!$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 18 at 16:39









          drhab

          97.8k544129




          97.8k544129























              1














              The answers you have found are correct but there is simpler way to approach this.



              (i) Consider the 6 boys to be a single person. Now the number of permutations will be $5!$.
              However the boys themselves can be arranged in $6!$ ways. So, by principle of multiplication, the number of permutations if all the 6 boys stand together is $5!.6!=86400$



              (ii) You need to visualize this in a different way to make things simpler.



              Consider 6 boys standing with spaces between them i.e.



              $ text{ X B X B X B X B X B X B X}$



              where the $text{X}$s denote the empty spaces.



              We need to place the girls in place of $text{X}$ which is the same as "no girl stands next to another girl".



              This can be done in $P(7,4)$ or $binom{7}{4}.4!$ ways as there are 7 empty spaces or $text{X}$s and 4 girls.



              Also, the boys themselves can be arranged in $6!$ ways.



              Therefore, by principle of multiplication, the number of permutations that no girl stands next to another girl is $6!.binom{7}{4}.4!=604800$






              share|cite|improve this answer


























                1














                The answers you have found are correct but there is simpler way to approach this.



                (i) Consider the 6 boys to be a single person. Now the number of permutations will be $5!$.
                However the boys themselves can be arranged in $6!$ ways. So, by principle of multiplication, the number of permutations if all the 6 boys stand together is $5!.6!=86400$



                (ii) You need to visualize this in a different way to make things simpler.



                Consider 6 boys standing with spaces between them i.e.



                $ text{ X B X B X B X B X B X B X}$



                where the $text{X}$s denote the empty spaces.



                We need to place the girls in place of $text{X}$ which is the same as "no girl stands next to another girl".



                This can be done in $P(7,4)$ or $binom{7}{4}.4!$ ways as there are 7 empty spaces or $text{X}$s and 4 girls.



                Also, the boys themselves can be arranged in $6!$ ways.



                Therefore, by principle of multiplication, the number of permutations that no girl stands next to another girl is $6!.binom{7}{4}.4!=604800$






                share|cite|improve this answer
























                  1












                  1








                  1






                  The answers you have found are correct but there is simpler way to approach this.



                  (i) Consider the 6 boys to be a single person. Now the number of permutations will be $5!$.
                  However the boys themselves can be arranged in $6!$ ways. So, by principle of multiplication, the number of permutations if all the 6 boys stand together is $5!.6!=86400$



                  (ii) You need to visualize this in a different way to make things simpler.



                  Consider 6 boys standing with spaces between them i.e.



                  $ text{ X B X B X B X B X B X B X}$



                  where the $text{X}$s denote the empty spaces.



                  We need to place the girls in place of $text{X}$ which is the same as "no girl stands next to another girl".



                  This can be done in $P(7,4)$ or $binom{7}{4}.4!$ ways as there are 7 empty spaces or $text{X}$s and 4 girls.



                  Also, the boys themselves can be arranged in $6!$ ways.



                  Therefore, by principle of multiplication, the number of permutations that no girl stands next to another girl is $6!.binom{7}{4}.4!=604800$






                  share|cite|improve this answer












                  The answers you have found are correct but there is simpler way to approach this.



                  (i) Consider the 6 boys to be a single person. Now the number of permutations will be $5!$.
                  However the boys themselves can be arranged in $6!$ ways. So, by principle of multiplication, the number of permutations if all the 6 boys stand together is $5!.6!=86400$



                  (ii) You need to visualize this in a different way to make things simpler.



                  Consider 6 boys standing with spaces between them i.e.



                  $ text{ X B X B X B X B X B X B X}$



                  where the $text{X}$s denote the empty spaces.



                  We need to place the girls in place of $text{X}$ which is the same as "no girl stands next to another girl".



                  This can be done in $P(7,4)$ or $binom{7}{4}.4!$ ways as there are 7 empty spaces or $text{X}$s and 4 girls.



                  Also, the boys themselves can be arranged in $6!$ ways.



                  Therefore, by principle of multiplication, the number of permutations that no girl stands next to another girl is $6!.binom{7}{4}.4!=604800$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 18 at 16:55









                  Vaibhav

                  588




                  588






























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