Ordered group between boys and girls
up vote
1
down vote
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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.
combinatorics
asked Nov 17 at 21:42
Johnny
175
add a comment |
up vote
1
down vote
favorite
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
asked Nov 17 at 21:42
Johnny
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
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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
asked Nov 17 at 21:42
Johnny
175
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
combinatorics
asked Nov 17 at 21:42
Johnny
175
asked Nov 17 at 21:42
Johnny
175
asked Nov 17 at 21:42
Johnny
175
asked Nov 17 at 21:42
Johnny
175
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
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?
answered Nov 18 at 14:21
Na'omi
23310
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
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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?
answered Nov 18 at 14:21
Na'omi
23310
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
add a comment |
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?
answered Nov 18 at 14:21
Na'omi
23310
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
add a comment |
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?
answered Nov 18 at 14:21
Na'omi
23310
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?
answered Nov 18 at 14:21
Na'omi
23310
answered Nov 18 at 14:21
Na'omi
23310
answered Nov 18 at 14:21
Na'omi
23310
answered Nov 18 at 14:21
Na'omi
23310
23310
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
add a comment |
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
add a comment |
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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