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there are 12 people in a room: 4 couples and 4 singles.

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Manager
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there are 12 people in a room: 4 couples and 4 singles. [#permalink] New post 27 Aug 2003, 01:43
there are 12 people in a room: 4 couples and 4 singles. picking 4 people at random, which is the probability of picking only one couple?

id say:

total no. of combinations 4c12=55*9

we pick a couple out of the 4: 1c4=4

now we can pick 2 singles: 2c4=6

or one person of one couple, one of another: 2c3*1c2*1c2=3*2*2=12

or one person of one couple, one single: 1c3*1c2*1c4=3*2*4=24

no. of ways of picking only one couple = 4*(6+12+24)=4*42

thus answer = 4*42/55*9

anyone? cheers, javi
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Probability [#permalink] New post 27 Aug 2003, 04:42
Total No of ways =12C4=495
NO. OF WAYS TO PICK 1 COUPLE =4C1
Now after choosing a couple rest 2 people can be picked up in 10C2 ways
Thus probability =4*45/495=4/11
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 [#permalink] New post 27 Aug 2003, 05:57
How about this approach:

Total # of ways: 12C4 = 495
Ways to pick a couple: 4C1 = 4 (makes sense as there are four couples)
Ways to pick two people after the couple = 10C2 = 45
BUT three of those groups will be the other three couples left, so it's really 45-3 = 42

so...(4 * (45-3))/495 = 168/495 reduces to 56/165
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Sept 3rd

Manager
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Re: Probability [#permalink] New post 27 Aug 2003, 07:09
aps_can wrote:
Total No of ways =12C4=495
NO. OF WAYS TO PICK 1 COUPLE =4C1
Now after choosing a couple rest 2 people can be picked up in 10C2 ways
Thus probability =4*45/495=4/11

aps_can, arent you including the possibility of having 2 couples in your counting?
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 [#permalink] New post 27 Aug 2003, 07:11
mciatto wrote:
How about this approach:

Total # of ways: 12C4 = 495
Ways to pick a couple: 4C1 = 4 (makes sense as there are four couples)
Ways to pick two people after the couple = 10C2 = 45
BUT three of those groups will be the other three couples left, so it's really 45-3 = 42

so...(4 * (45-3))/495 = 168/495 reduces to 56/165

mciatto, our solutions are equal! 4*42/55*9=56/165... yours makes sense to me... thx!
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Probability [#permalink] New post 27 Aug 2003, 10:17
Yes Mciatto is right! I missed that factor.
Probability   [#permalink] 27 Aug 2003, 10:17
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there are 12 people in a room: 4 couples and 4 singles.

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