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There are 12 girls and 15 boys at a scool ball. In how many

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kedar1972
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There are 12 girls and 15 boys at a scool ball. In how many [#permalink] New post   09 Sep 2004, 02:38
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There are 12 girls and 15 boys at a scool ball. In how many ways can we form 4 paris in a dance?

Please state your explanation.



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Re: There are 12 girls and 15 boys at a scool ball. In how many [#permalink] New post   09 Sep 2004, 03:12
This is a tough one for me, but I will give it a try.

12C4 = 495 combinations to choose 4 girls from 12 girls in total.
15C4 = 1365 combinations to choose 4 boys from 15 boys in total.

Total ways to form 4 pairs from both combinations is:
495 * 1365 = 675675

I am not sure if this answer is correct because the total ways is huge.
Besides I assumed that an pair consists of only 1 boy and 1 girl.

Please correct me if I am wrong.

Regards,

Alex
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stuti
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Re: There are 12 girls and 15 boys at a scool ball. In how many [#permalink] New post   09 Sep 2004, 05:29
kedar1972 wrote:
There are 12 girls and 15 boys at a scool ball. In how many ways can we form 4 paris in a dance?

Please state your explanation.


is "how many ways can we form 4 pairs" same as "number of pairs of 4 that can be formed?"
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Re: There are 12 girls and 15 boys at a scool ball. In how many [#permalink] New post   09 Sep 2004, 06:03
In my approach, I assumed:

(x1,y1) (x2,y2) (x3,y3) (x4,y4)

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Alex
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Re: There are 12 girls and 15 boys at a scool ball. In how many [#permalink] New post   09 Sep 2004, 06:22
It should be 12C4*15C4*4!. I cant calculate the net number. But this is the logic..

4 girls can be selected in 12C4 ways.
4 boys can be selected in 15C4 ways.
Having selected 4 boys and 4 girls, they can be paired in 4! ways.

Hence the total no. of pairs possible = 12C4*15C4*4!.
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Re: There are 12 girls and 15 boys at a scool ball. In how many [#permalink] New post   09 Sep 2004, 06:40
venksune wrote:
It should be 12C4*15C4*4!. I cant calculate the net number. But this is the logic..

4 girls can be selected in 12C4 ways.
4 boys can be selected in 15C4 ways.
Having selected 4 boys and 4 girls, they can be paired in 4! ways.

Hence the total no. of pairs possible = 12C4*15C4*4!.


shouldnt it be 12C4*15C4*4!*4!

since there are 4 boys and 4 girls
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Re: There are 12 girls and 15 boys at a scool ball. In how many [#permalink] New post   09 Sep 2004, 07:23
Smadalika,
Its the '4 pairs' that can be arranged in 4! ways.
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Re: There are 12 girls and 15 boys at a scool ball. In how many [#permalink] New post   09 Sep 2004, 08:58
venksune wrote:
Smadalika,
Its the '4 pairs' that can be arranged in 4! ways.


but in that case you are considering only one pairing
what I mean by that is that

if you have G1,G2,G3,G4
and B1,B2,B3,B4

you are considering only one combination B1G1, B2G2, B3,G3 , B4,G4

and then you arrange them in 4! ways but what about other arrangements like

B1G2, B3G1 etc....
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Re: There are 12 girls and 15 boys at a scool ball. In how many [#permalink] New post   09 Sep 2004, 09:57
I think smandalinka is right!

I understood 15C4 * 12C4 but got stuck with the 4! * 4! part.

Both the girls and the boys are interchangable.

Regards,

Alex
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Re: There are 12 girls and 15 boys at a scool ball. In how many [#permalink] New post   09 Sep 2004, 10:11
OOOOOps I have the feeling my point of view is too complex

Selection of 8 apropriate dancers : 12C4.15C4

Various ways to arrange this 8 (considering it is symetric) :

8C2.6C2.4C2.2C2/(4!.2)
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Re: There are 12 girls and 15 boys at a scool ball. In how many [#permalink] New post   09 Sep 2004, 11:45
Venksune's explanation is correct.

The answer is 12 C 4 * 12 P 4 = 16,216,200.

Thanks for all the explanations

Kedar
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Re: There are 12 girls and 15 boys at a scool ball. In how many [#permalink] New post   09 Sep 2004, 14:30
There is no hiden assumption that a pair = 1 boy + 1 girl ??



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Re: There are 12 girls and 15 boys at a scool ball. In how many [#permalink]
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