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They tried to make it trickier.

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sharadGmat
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They tried to make it trickier. [#permalink] New post   03 Jul 2006, 19:39
They tried to make it trickier..:)



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Re: They tried to make it trickier. [#permalink] New post   04 Jul 2006, 06:26
amansingla4, can you post an explanation as to how you got to your solution ?
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Re: They tried to make it trickier. [#permalink] New post   04 Jul 2006, 07:41
total combination: 9C3 = 84

@ least 2 Yellow:

1. comb of 2 yellow: 4C2 = 6
2. comb all yellow: 4C3 = 4
3. comb of 1 red: 2C1 = 2
4. comb of 1 brown: 3C1 = 3

6*2+6*3+4 = 34 (number of at least 2 yellow)

Exactly 1 yellow:
1. comb of 1 yellow: 4C1 = 4
2. comb of 1 red: 2C1 = 2
3. comb of 2 red: 2C2 =1
4. comb of 1 brown: 3C1 = 3
5. comb of 2 brown: 3c2=3

4*1+4*3+4*2*3=40 (number of exaclty 1 yellow)

therefore, (34+40)/84 = 74/84 = 37/42...hence B

is there an easier way? I know that this has taken me lot longer than 2 minutes...
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Re: They tried to make it trickier. [#permalink] New post   04 Jul 2006, 10:54
acfuture wrote:

therefore, (34+40)/84 = 74/84 = 37/42...hence B

is there an easier way? I know that this has taken me lot longer than 2 minutes...


B. An easier way is to note that ANY yellow in a mixture will lead to Jaune, the wording is meant to slightly confuse you.

Thus, take the (combos possible for all three colors minus the combos possible for no yellow colors) divided by total possibilities will give you your answer.

9C3 - 5C3 = 84 - 20 = 64

64/84 = 37/42
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Re: They tried to make it trickier. [#permalink] New post   04 Jul 2006, 13:08
amartin6165 wrote:
acfuture wrote:

therefore, (34+40)/84 = 74/84 = 37/42...hence B

is there an easier way? I know that this has taken me lot longer than 2 minutes...


B. An easier way is to note that ANY yellow in a mixture will lead to Jaune, the wording is meant to slightly confuse you.

Thus, take the (combos possible for all three colors minus the combos possible for no yellow colors) divided by total possibilities will give you your answer.

9C3 - 5C3 = 84 - 20 = 64

64/84 = 37/42


Agree with this but 5C3 is 10 not 20.
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Re: They tried to make it trickier. [#permalink] New post   04 Jul 2006, 17:06
amartin6165 wrote:
B. An easier way is to note that ANY yellow in a mixture will lead to Jaune, the wording is meant to slightly confuse you.

Thus, take the (combos possible for all three colors minus the combos possible for no yellow colors) divided by total possibilities will give you your answer.

9C3 - 5C3 = 84 - 20 = 64

64/84 = 37/42



Wow simplest possible answer :)
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Re: They tried to make it trickier. [#permalink] New post   05 Jul 2006, 01:31
amartin6165 wrote:
acfuture wrote:

therefore, (34+40)/84 = 74/84 = 37/42...hence B

is there an easier way? I know that this has taken me lot longer than 2 minutes...


B. An easier way is to note that ANY yellow in a mixture will lead to Jaune, the wording is meant to slightly confuse you.

Thus, take the (combos possible for all three colors minus the combos possible for no yellow colors) divided by total possibilities will give you your answer.

9C3 - 5C3 = 84 - 20 = 64

64/84 = 37/42


Very unclear,
as 9C3 - 5C3 = 84 -10 = 74, and 74/84 = 37/42

would you please elaborate 5C3.



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Re: They tried to make it trickier. [#permalink]
05 Jul 2006, 01:31
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