Ordered group between boys and girls











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Lets say I have 20k people.(k>0)
Lets say 15k boys 5k girls.
If I want to make k groups of 15 boys and 5 girls, where inner ordering
of the group does matter(it can be girl boy boy ... or whatever) - point is
the boys and girls are mixed within a group. How do I do it?



From my understanding 15 boys,then 5 girls in the following way
15k,15k-1,...15k-14,5k,5k-1,...5k-4,...
and this way we arrange a row of 15 boys followed by 5 girls.
In total its (15k)!(5k)!.
Now we can divide by 15! and 5! to get rid of the inner ordering
between boys and girls respectively, and then divide by k! to get
rid of between the groups.
But how do I go from here - or in a completely different way - and
make order in each group between the 15 boys and girls.










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  • 1




    For the first group: $C^{15}_{15k}times C^5_{5k}$ (where $C$ indicates combination) For the second: $C^{15}_{15(k-1)}times C^5_{5(k-1)}$ What do you think?
    – Na'omi
    Nov 17 at 21:48

















up vote
1
down vote

favorite
1












Lets say I have 20k people.(k>0)
Lets say 15k boys 5k girls.
If I want to make k groups of 15 boys and 5 girls, where inner ordering
of the group does matter(it can be girl boy boy ... or whatever) - point is
the boys and girls are mixed within a group. How do I do it?



From my understanding 15 boys,then 5 girls in the following way
15k,15k-1,...15k-14,5k,5k-1,...5k-4,...
and this way we arrange a row of 15 boys followed by 5 girls.
In total its (15k)!(5k)!.
Now we can divide by 15! and 5! to get rid of the inner ordering
between boys and girls respectively, and then divide by k! to get
rid of between the groups.
But how do I go from here - or in a completely different way - and
make order in each group between the 15 boys and girls.










share|cite|improve this question


















  • 1




    For the first group: $C^{15}_{15k}times C^5_{5k}$ (where $C$ indicates combination) For the second: $C^{15}_{15(k-1)}times C^5_{5(k-1)}$ What do you think?
    – Na'omi
    Nov 17 at 21:48















up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





Lets say I have 20k people.(k>0)
Lets say 15k boys 5k girls.
If I want to make k groups of 15 boys and 5 girls, where inner ordering
of the group does matter(it can be girl boy boy ... or whatever) - point is
the boys and girls are mixed within a group. How do I do it?



From my understanding 15 boys,then 5 girls in the following way
15k,15k-1,...15k-14,5k,5k-1,...5k-4,...
and this way we arrange a row of 15 boys followed by 5 girls.
In total its (15k)!(5k)!.
Now we can divide by 15! and 5! to get rid of the inner ordering
between boys and girls respectively, and then divide by k! to get
rid of between the groups.
But how do I go from here - or in a completely different way - and
make order in each group between the 15 boys and girls.










share|cite|improve this question













Lets say I have 20k people.(k>0)
Lets say 15k boys 5k girls.
If I want to make k groups of 15 boys and 5 girls, where inner ordering
of the group does matter(it can be girl boy boy ... or whatever) - point is
the boys and girls are mixed within a group. How do I do it?



From my understanding 15 boys,then 5 girls in the following way
15k,15k-1,...15k-14,5k,5k-1,...5k-4,...
and this way we arrange a row of 15 boys followed by 5 girls.
In total its (15k)!(5k)!.
Now we can divide by 15! and 5! to get rid of the inner ordering
between boys and girls respectively, and then divide by k! to get
rid of between the groups.
But how do I go from here - or in a completely different way - and
make order in each group between the 15 boys and girls.







combinatorics






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asked Nov 17 at 21:42









Johnny

175




175








  • 1




    For the first group: $C^{15}_{15k}times C^5_{5k}$ (where $C$ indicates combination) For the second: $C^{15}_{15(k-1)}times C^5_{5(k-1)}$ What do you think?
    – Na'omi
    Nov 17 at 21:48
















  • 1




    For the first group: $C^{15}_{15k}times C^5_{5k}$ (where $C$ indicates combination) For the second: $C^{15}_{15(k-1)}times C^5_{5(k-1)}$ What do you think?
    – Na'omi
    Nov 17 at 21:48










1




1




For the first group: $C^{15}_{15k}times C^5_{5k}$ (where $C$ indicates combination) For the second: $C^{15}_{15(k-1)}times C^5_{5(k-1)}$ What do you think?
– Na'omi
Nov 17 at 21:48






For the first group: $C^{15}_{15k}times C^5_{5k}$ (where $C$ indicates combination) For the second: $C^{15}_{15(k-1)}times C^5_{5(k-1)}$ What do you think?
– Na'omi
Nov 17 at 21:48












1 Answer
1






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oldest

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up vote
0
down vote













For the first group:



$C^{15}_{15k}times C^{5}_{5k}$ (where $C$ indicates combination)



I think this is as you did.



For the second:



$C^{15}_{15(k-1)}times C^{5}_{5(k-1)}$



...



For the last:



$C^{15}_{15(k-(k-1))}times C^{5}_{5(k-(k-1))}=C^{15}_{15}times C^{5}_{5}=1$



Then, as you said, divide by $k!$ because the ordering of the groups don't matter.



So, you have



$dfrac{1}{k!}prod_{i=0}^{k-1}C^{15}_{15(k-i)}times C^{5}_{5(k-i)}$ manners to get this groups...



What do you think?






share|cite|improve this answer





















  • I am not entirely sure. If I understand correctly, you are making k groups with boys and girls, but you dont make them ordered within the group. I think if we multiply your answer by (20!)^k to make them ordered it would be correct. Not sure.
    – Johnny
    Nov 18 at 16:53










  • Sorry, I think I didn't understand your question right. Does the order of people matter? Or the order of the groups? In fact I've did that any ordering doesn't matter.
    – Na'omi
    Nov 19 at 13:53








  • 1




    The order within the group matters - the order of people, but the order of the groups doesnt.
    – Johnny
    Nov 19 at 15:11










  • Oh, now I understand... I think you're right. Taking the groups as above, the order in a group can be take in $20!$ different manners. So, multiply by $(20!)^k$. I think is it... Thank you.
    – Na'omi
    Nov 19 at 15:36











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up vote
0
down vote













For the first group:



$C^{15}_{15k}times C^{5}_{5k}$ (where $C$ indicates combination)



I think this is as you did.



For the second:



$C^{15}_{15(k-1)}times C^{5}_{5(k-1)}$



...



For the last:



$C^{15}_{15(k-(k-1))}times C^{5}_{5(k-(k-1))}=C^{15}_{15}times C^{5}_{5}=1$



Then, as you said, divide by $k!$ because the ordering of the groups don't matter.



So, you have



$dfrac{1}{k!}prod_{i=0}^{k-1}C^{15}_{15(k-i)}times C^{5}_{5(k-i)}$ manners to get this groups...



What do you think?






share|cite|improve this answer





















  • I am not entirely sure. If I understand correctly, you are making k groups with boys and girls, but you dont make them ordered within the group. I think if we multiply your answer by (20!)^k to make them ordered it would be correct. Not sure.
    – Johnny
    Nov 18 at 16:53










  • Sorry, I think I didn't understand your question right. Does the order of people matter? Or the order of the groups? In fact I've did that any ordering doesn't matter.
    – Na'omi
    Nov 19 at 13:53








  • 1




    The order within the group matters - the order of people, but the order of the groups doesnt.
    – Johnny
    Nov 19 at 15:11










  • Oh, now I understand... I think you're right. Taking the groups as above, the order in a group can be take in $20!$ different manners. So, multiply by $(20!)^k$. I think is it... Thank you.
    – Na'omi
    Nov 19 at 15:36















up vote
0
down vote













For the first group:



$C^{15}_{15k}times C^{5}_{5k}$ (where $C$ indicates combination)



I think this is as you did.



For the second:



$C^{15}_{15(k-1)}times C^{5}_{5(k-1)}$



...



For the last:



$C^{15}_{15(k-(k-1))}times C^{5}_{5(k-(k-1))}=C^{15}_{15}times C^{5}_{5}=1$



Then, as you said, divide by $k!$ because the ordering of the groups don't matter.



So, you have



$dfrac{1}{k!}prod_{i=0}^{k-1}C^{15}_{15(k-i)}times C^{5}_{5(k-i)}$ manners to get this groups...



What do you think?






share|cite|improve this answer





















  • I am not entirely sure. If I understand correctly, you are making k groups with boys and girls, but you dont make them ordered within the group. I think if we multiply your answer by (20!)^k to make them ordered it would be correct. Not sure.
    – Johnny
    Nov 18 at 16:53










  • Sorry, I think I didn't understand your question right. Does the order of people matter? Or the order of the groups? In fact I've did that any ordering doesn't matter.
    – Na'omi
    Nov 19 at 13:53








  • 1




    The order within the group matters - the order of people, but the order of the groups doesnt.
    – Johnny
    Nov 19 at 15:11










  • Oh, now I understand... I think you're right. Taking the groups as above, the order in a group can be take in $20!$ different manners. So, multiply by $(20!)^k$. I think is it... Thank you.
    – Na'omi
    Nov 19 at 15:36













up vote
0
down vote










up vote
0
down vote









For the first group:



$C^{15}_{15k}times C^{5}_{5k}$ (where $C$ indicates combination)



I think this is as you did.



For the second:



$C^{15}_{15(k-1)}times C^{5}_{5(k-1)}$



...



For the last:



$C^{15}_{15(k-(k-1))}times C^{5}_{5(k-(k-1))}=C^{15}_{15}times C^{5}_{5}=1$



Then, as you said, divide by $k!$ because the ordering of the groups don't matter.



So, you have



$dfrac{1}{k!}prod_{i=0}^{k-1}C^{15}_{15(k-i)}times C^{5}_{5(k-i)}$ manners to get this groups...



What do you think?






share|cite|improve this answer












For the first group:



$C^{15}_{15k}times C^{5}_{5k}$ (where $C$ indicates combination)



I think this is as you did.



For the second:



$C^{15}_{15(k-1)}times C^{5}_{5(k-1)}$



...



For the last:



$C^{15}_{15(k-(k-1))}times C^{5}_{5(k-(k-1))}=C^{15}_{15}times C^{5}_{5}=1$



Then, as you said, divide by $k!$ because the ordering of the groups don't matter.



So, you have



$dfrac{1}{k!}prod_{i=0}^{k-1}C^{15}_{15(k-i)}times C^{5}_{5(k-i)}$ manners to get this groups...



What do you think?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 18 at 14:21









Na'omi

24011




24011












  • I am not entirely sure. If I understand correctly, you are making k groups with boys and girls, but you dont make them ordered within the group. I think if we multiply your answer by (20!)^k to make them ordered it would be correct. Not sure.
    – Johnny
    Nov 18 at 16:53










  • Sorry, I think I didn't understand your question right. Does the order of people matter? Or the order of the groups? In fact I've did that any ordering doesn't matter.
    – Na'omi
    Nov 19 at 13:53








  • 1




    The order within the group matters - the order of people, but the order of the groups doesnt.
    – Johnny
    Nov 19 at 15:11










  • Oh, now I understand... I think you're right. Taking the groups as above, the order in a group can be take in $20!$ different manners. So, multiply by $(20!)^k$. I think is it... Thank you.
    – Na'omi
    Nov 19 at 15:36


















  • I am not entirely sure. If I understand correctly, you are making k groups with boys and girls, but you dont make them ordered within the group. I think if we multiply your answer by (20!)^k to make them ordered it would be correct. Not sure.
    – Johnny
    Nov 18 at 16:53










  • Sorry, I think I didn't understand your question right. Does the order of people matter? Or the order of the groups? In fact I've did that any ordering doesn't matter.
    – Na'omi
    Nov 19 at 13:53








  • 1




    The order within the group matters - the order of people, but the order of the groups doesnt.
    – Johnny
    Nov 19 at 15:11










  • Oh, now I understand... I think you're right. Taking the groups as above, the order in a group can be take in $20!$ different manners. So, multiply by $(20!)^k$. I think is it... Thank you.
    – Na'omi
    Nov 19 at 15:36
















I am not entirely sure. If I understand correctly, you are making k groups with boys and girls, but you dont make them ordered within the group. I think if we multiply your answer by (20!)^k to make them ordered it would be correct. Not sure.
– Johnny
Nov 18 at 16:53




I am not entirely sure. If I understand correctly, you are making k groups with boys and girls, but you dont make them ordered within the group. I think if we multiply your answer by (20!)^k to make them ordered it would be correct. Not sure.
– Johnny
Nov 18 at 16:53












Sorry, I think I didn't understand your question right. Does the order of people matter? Or the order of the groups? In fact I've did that any ordering doesn't matter.
– Na'omi
Nov 19 at 13:53






Sorry, I think I didn't understand your question right. Does the order of people matter? Or the order of the groups? In fact I've did that any ordering doesn't matter.
– Na'omi
Nov 19 at 13:53






1




1




The order within the group matters - the order of people, but the order of the groups doesnt.
– Johnny
Nov 19 at 15:11




The order within the group matters - the order of people, but the order of the groups doesnt.
– Johnny
Nov 19 at 15:11












Oh, now I understand... I think you're right. Taking the groups as above, the order in a group can be take in $20!$ different manners. So, multiply by $(20!)^k$. I think is it... Thank you.
– Na'omi
Nov 19 at 15:36




Oh, now I understand... I think you're right. Taking the groups as above, the order in a group can be take in $20!$ different manners. So, multiply by $(20!)^k$. I think is it... Thank you.
– Na'omi
Nov 19 at 15:36


















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