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From 6 positive numbers and 6 negative numbers, how many

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Director
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From 6 positive numbers and 6 negative numbers, how many [#permalink] New post 31 Oct 2005, 08:27
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

0% (00:00) correct 100% (00:42) wrong based on 1 sessions
From 6 positive numbers and 6 negative numbers, how many groups of 4 numbers, yielding a positive product, can be formed?

A. 720
B 625
C. 30
D. 960
E. 255
[Reveal] Spoiler: OA

Last edited by Bunuel on 16 Apr 2014, 00:42, edited 2 times in total.
Renamed the topic, edited the question and the OA.
Director
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 [#permalink] New post 31 Oct 2005, 10:11
selecting 4C6 of each group would give 30 as an ans. But if we can select 2C6 pos and 2C6 neg then it would give 225. Finally we can have a combination of both which gives 255.Think it is a bit unclear
VP
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 [#permalink] New post 31 Oct 2005, 10:24
Agree with automan. Also, not clear if 6 numbers are all distinct.

Believing numbers are all different:
If we think order in the group is important, it would be:

all 4 positive OR 2 -ve 2 +ve OR all 4 -ve
6P4 + 6P2 * 6P2 + 6P4 = 1620

if order in group not important:
6C4 + 6C2 * 6C2 + 6C4 = 255
Manager
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 [#permalink] New post 31 Oct 2005, 14:58
I am getting F!

We got 2 sets of 6 elements each.
In order for the product of 4 numbers to be positive we need even number (0,2,4,..) of -ve numbers. The following permutations of 4 numbers will yield a +ve product

# of -ve numbers, # of +ve numbers
0, 4
2, 2
4, 0

Assuming different elements in each set and given that we are trying to identify "groups of numbers" we have:
6P0*6P4 + 6P2*6P2 + 6P4*6P0
= 1*360 + 30*30 + 360*1
= 360 + 900 + 360
= 1620
Manager
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 [#permalink] New post 31 Oct 2005, 15:22
On a second thought, I think going with combinations Vs permutations makes sense because:
# of -ve numbers, # of +ve numbers
0, 4 could be {}, {1,2,3,4}
2, 2 could be {-1,-2}, {1/2,100}
4, 0 could be {-1,-4,-6,-9.9},{}

For example, in the 2nd case above the group of 4 numbers formed by taking {-1,-2}, {1/2,100} together is {-1,-2,1/2,100}, which is no different from the re-arranged group {100,-1,1/2,-2} as far as the product of those 4 elements is concerned.

Since order isn't an issue, go combinations!
6C0*6C4 + 6C2*6C2 + 6C4*6C0
= 1*15 + 15*15 + 15*1
= 15 + 225 + 15
= 255
Still getting an F (none of the above) :wink:


mbaqst wrote:
I am getting F!

We got 2 sets of 6 elements each.
In order for the product of 4 numbers to be positive we need even number (0,2,4,..) of -ve numbers. The following permutations of 4 numbers will yield a +ve product

# of -ve numbers, # of +ve numbers
0, 4
2, 2
4, 0

Assuming different elements in each set and given that we are trying to identify "groups of numbers" we have:
6P0*6P4 + 6P2*6P2 + 6P4*6P0
= 1*360 + 30*30 + 360*1
= 360 + 900 + 360
= 1620
Director
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 [#permalink] New post 31 Oct 2005, 15:57
The solution is C. I don't have the OE and my reasoning was the same that you all used to reach to 255. I will be working on this....

Any suggestion??
Manager
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 [#permalink] New post 31 Oct 2005, 18:45
automan wrote:
The solution is C. I don't have the OE and my reasoning was the same that you all used to reach to 255. I will be working on this....

Any suggestion??


I think there is something wrong with your OA, as I think this is a combination problem rather than a permutation as what it matters is the sign of the number and not its order.

so in that case 12c4 would be the maximum amount of groups that could be made with that set, ignoring the sign of its multiplication and that is equal to 495 that is less than C

I think mbaquest has the right answer
Director
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 [#permalink] New post 01 Nov 2005, 02:08
Ups, sorry. The OA is 225. I'm very sorry for the inconvenience. Anyway, I can not explain this solution. There must be a typo.
  [#permalink] 01 Nov 2005, 02:08
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