How many different combinations

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Suppose there are 8 boys and 4 girls. Suppose they are lining up and entering the room in order e.g (b,g,b,b,g,b,g,b,b,g,b,b). The only rule is that the number of girls in the room cannot be greater than the number of boys. How many different combination of the entering order are there?



Editted (Attempts):
First i try to consider $(x,y)$ represent $x$ girls and $y$ boys. Then I convert the problem into how many ways to travel from $(0,0)$ to $(4,8)$ with moving up or right only and not going below the line $y=x$. then i am not sure what are the right way to calculate




combinatorics




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  • If there were equally many boys and girls, would you know the answer? (This is me trying to assess what you know, which is necessary since you haven't supplied any context on your own).
    – Arthur
    Aug 29 at 9:47











  • @Arthur Edited: I knew that, it has to be $16!over 9times8!8!$
    – user588102
    Aug 29 at 9:50











  • Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
    – José Carlos Santos
    Aug 29 at 9:51










  • Do you know the Catalan numbers?
    – ArsenBerk
    Aug 29 at 10:05















up vote
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Suppose there are 8 boys and 4 girls. Suppose they are lining up and entering the room in order e.g (b,g,b,b,g,b,g,b,b,g,b,b). The only rule is that the number of girls in the room cannot be greater than the number of boys. How many different combination of the entering order are there?



Editted (Attempts):
First i try to consider $(x,y)$ represent $x$ girls and $y$ boys. Then I convert the problem into how many ways to travel from $(0,0)$ to $(4,8)$ with moving up or right only and not going below the line $y=x$. then i am not sure what are the right way to calculate




combinatorics




share|cite|improve this question






















  • If there were equally many boys and girls, would you know the answer? (This is me trying to assess what you know, which is necessary since you haven't supplied any context on your own).
    – Arthur
    Aug 29 at 9:47











  • @Arthur Edited: I knew that, it has to be $16!over 9times8!8!$
    – user588102
    Aug 29 at 9:50











  • Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
    – José Carlos Santos
    Aug 29 at 9:51










  • Do you know the Catalan numbers?
    – ArsenBerk
    Aug 29 at 10:05













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Suppose there are 8 boys and 4 girls. Suppose they are lining up and entering the room in order e.g (b,g,b,b,g,b,g,b,b,g,b,b). The only rule is that the number of girls in the room cannot be greater than the number of boys. How many different combination of the entering order are there?



Editted (Attempts):
First i try to consider $(x,y)$ represent $x$ girls and $y$ boys. Then I convert the problem into how many ways to travel from $(0,0)$ to $(4,8)$ with moving up or right only and not going below the line $y=x$. then i am not sure what are the right way to calculate




combinatorics




share|cite|improve this question














Suppose there are 8 boys and 4 girls. Suppose they are lining up and entering the room in order e.g (b,g,b,b,g,b,g,b,b,g,b,b). The only rule is that the number of girls in the room cannot be greater than the number of boys. How many different combination of the entering order are there?



Editted (Attempts):
First i try to consider $(x,y)$ represent $x$ girls and $y$ boys. Then I convert the problem into how many ways to travel from $(0,0)$ to $(4,8)$ with moving up or right only and not going below the line $y=x$. then i am not sure what are the right way to calculate





combinatorics





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 29 at 9:54

























asked Aug 29 at 9:45









user588102

11




11











  • If there were equally many boys and girls, would you know the answer? (This is me trying to assess what you know, which is necessary since you haven't supplied any context on your own).
    – Arthur
    Aug 29 at 9:47











  • @Arthur Edited: I knew that, it has to be $16!over 9times8!8!$
    – user588102
    Aug 29 at 9:50











  • Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
    – José Carlos Santos
    Aug 29 at 9:51










  • Do you know the Catalan numbers?
    – ArsenBerk
    Aug 29 at 10:05

















  • If there were equally many boys and girls, would you know the answer? (This is me trying to assess what you know, which is necessary since you haven't supplied any context on your own).
    – Arthur
    Aug 29 at 9:47











  • @Arthur Edited: I knew that, it has to be $16!over 9times8!8!$
    – user588102
    Aug 29 at 9:50











  • Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
    – José Carlos Santos
    Aug 29 at 9:51










  • Do you know the Catalan numbers?
    – ArsenBerk
    Aug 29 at 10:05
















If there were equally many boys and girls, would you know the answer? (This is me trying to assess what you know, which is necessary since you haven't supplied any context on your own).
– Arthur
Aug 29 at 9:47





If there were equally many boys and girls, would you know the answer? (This is me trying to assess what you know, which is necessary since you haven't supplied any context on your own).
– Arthur
Aug 29 at 9:47













@Arthur Edited: I knew that, it has to be $16!over 9times8!8!$
– user588102
Aug 29 at 9:50





@Arthur Edited: I knew that, it has to be $16!over 9times8!8!$
– user588102
Aug 29 at 9:50













Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 29 at 9:51




Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 29 at 9:51












Do you know the Catalan numbers?
– ArsenBerk
Aug 29 at 10:05





Do you know the Catalan numbers?
– ArsenBerk
Aug 29 at 10:05











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Let us not work with $8$ boys but with $9$ boys and let us demand that at any stage there must be more boys than girls in the room.



That gives evidently the same number of possibilities.



There are $binom134$ possibilities of lining up if the demand is neglected.



According to Bertrand's ballot theorem the number of possibilities in which at any stage there are more boys than girls in the room is: $$frac9-49+4binom134=275$$






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    up vote
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    Let us not work with $8$ boys but with $9$ boys and let us demand that at any stage there must be more boys than girls in the room.



    That gives evidently the same number of possibilities.



    There are $binom134$ possibilities of lining up if the demand is neglected.



    According to Bertrand's ballot theorem the number of possibilities in which at any stage there are more boys than girls in the room is: $$frac9-49+4binom134=275$$






    share|cite|improve this answer
























      up vote
      0
      down vote













      Let us not work with $8$ boys but with $9$ boys and let us demand that at any stage there must be more boys than girls in the room.



      That gives evidently the same number of possibilities.



      There are $binom134$ possibilities of lining up if the demand is neglected.



      According to Bertrand's ballot theorem the number of possibilities in which at any stage there are more boys than girls in the room is: $$frac9-49+4binom134=275$$






      share|cite|improve this answer






















        up vote
        0
        down vote










        up vote
        0
        down vote









        Let us not work with $8$ boys but with $9$ boys and let us demand that at any stage there must be more boys than girls in the room.



        That gives evidently the same number of possibilities.



        There are $binom134$ possibilities of lining up if the demand is neglected.



        According to Bertrand's ballot theorem the number of possibilities in which at any stage there are more boys than girls in the room is: $$frac9-49+4binom134=275$$






        share|cite|improve this answer












        Let us not work with $8$ boys but with $9$ boys and let us demand that at any stage there must be more boys than girls in the room.



        That gives evidently the same number of possibilities.



        There are $binom134$ possibilities of lining up if the demand is neglected.



        According to Bertrand's ballot theorem the number of possibilities in which at any stage there are more boys than girls in the room is: $$frac9-49+4binom134=275$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 29 at 11:33









        drhab

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