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From 6 positive numbers and 6 negative numbers, how many
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Updated on: 20 Apr 2014, 12:07
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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
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Originally posted by automan on 31 Oct 2005, 09:27.
Last edited by Bunuel on 20 Apr 2014, 12:07, edited 3 times in total.
Renamed the topic, edited the question and the OA.




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Re: From 6 positive numbers and 6 negative numbers, how many
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20 Apr 2014, 12:18
Enael wrote: I've been stuck in this problem.
I don't understand completely why we only take 6C2*6C2 when the group is composed of 2 +ves and 2 ves.
For me, there are 2 scenarios
Scenario 1: ++++ ++ 
Scenario 2: ++ ++ ++
Note: In these 2 scenarios, it is indifferent both: the order of the group and the order WITHIN the group.
Since every decision for a single group affects the other 2 groups, I don't understand how we can simply add up (1)+(2)+(3) as in the previous answer.
My reasoning:
Scenario 1: ++++ ++  6C4*2C2*6C2*4C4 = 225
Scenario 2: ++ ++ ++ 6C2*6C2*4C2*4C2*2C2*2C2
It get confused whenever we're supposed to add combinations vs multiply them.
Help is much appreciated. Not sure I understand your logic there but hope that the solution below will clear your doubts. 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 For the product of 4 numbers to be positive there must be: 0 positive numbers and 4 negative numbers > choosing all 4 numbers from 6 positive: \(C^4_6=15\); 2 positive numbers and 2 negative numbers > choosing 2 numbers from 6 positive numbers and 2 numbers from 6 negative numbers: \(C^2_6*C^2_6=225\); 4 positive numbers and 0 negative numbers > choosing all 4 numbers from 6 negative: \(C^4_6=15\). Total = 15 + 225 + 15 = 255. Answer: E. Also, note that the order of the chosen numbers is not important here. Meaning that if the positive numbers are 1, 2, 3, 4, 5, and 6 and negative numbers are 1, 2, 3, 4, 5, and 6, then {1, 2, 3, 4} selection is the same as {4, 3, 2, 1} selection, {1, 2, 3, 4} selection is the same as {4, 3, 2, 1} selection, {1, 2, 1, 2} selection is the same as {1, 1, 2, 2} selection... Hope it's clear.
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Joined: 22 Aug 2005
Posts: 1045
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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
Joined: 17 Sep 2005
Posts: 71
Location: California

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 rearranged 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)
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



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Re: From 6 positive numbers and 6 negative numbers, how many
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20 Apr 2014, 00:56
6 +ves 6 ves
Combinations can be taken for forming positive product.
(1).
 ++
6c2*6c2
(2).

6c4
(3).
++++
6c4
Total = (1) + (2) + (3)
900/4 + 30
255



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Posts: 55193

Re: From 6 positive numbers and 6 negative numbers, how many
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24 Apr 2014, 01:00
russ9 wrote: Bunuel wrote: Enael wrote: I've been stuck in this problem.
I don't understand completely why we only take 6C2*6C2 when the group is composed of 2 +ves and 2 ves.
For me, there are 2 scenarios
Scenario 1: ++++ ++ 
Scenario 2: ++ ++ ++
Note: In these 2 scenarios, it is indifferent both: the order of the group and the order WITHIN the group.
Since every decision for a single group affects the other 2 groups, I don't understand how we can simply add up (1)+(2)+(3) as in the previous answer.
My reasoning:
Scenario 1: ++++ ++  6C4*2C2*6C2*4C4 = 225
Scenario 2: ++ ++ ++ 6C2*6C2*4C2*4C2*2C2*2C2
It get confused whenever we're supposed to add combinations vs multiply them.
Help is much appreciated. Not sure I understand your logic there but hope that the solution below will clear your doubts. 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 For the product of 4 numbers to be positive there must be: 0 positive numbers and 4 negative numbers > choosing all 4 numbers from 6 positive: \(C^4_6=15\); 2 positive numbers and 2 negative numbers > choosing 2 numbers from 6 positive numbers and 2 numbers from 6 negative numbers: \(C^2_6*C^2_6=225\); 4 positive numbers and 0 negative numbers > choosing all 4 numbers from 6 negative: \(C^4_6=15\). Total = 15 + 225 + 15 = 255. Answer: E. Also, note that the order of the chosen numbers is not important here. Meaning that if the positive numbers are 1, 2, 3, 4, 5, and 6 and negative numbers are 1, 2, 3, 4, 5, and 6, then {1, 2, 3, 4} selection is the same as {4, 3, 2, 1} selection, {1, 2, 3, 4} selection is the same as {4, 3, 2, 1} selection, {1, 2, 1, 2} selection is the same as {1, 1, 2, 2} selection... Hope it's clear. Hi Bunuel, Your answer is very clear but i'm stuck on one part(and this is the same part that I get on stuck on with every problem). How do you know when the order is important vs. when the order isn't important? Meaning, I can almost never decipher if I need to use the P or C formula. I've read the GMAT club book but haven't had any luck differentiating between "ordered" vs. "unordered" pairs. Thanks, Russ This should come with practice. Not an universal rule, but below might help to distinguish: The words "Permutation" and "Arrangement" are synonymous and can be used interchangeably (order matters). The words "Combination", "Group" and "Selection" are synonymous and can be used interchangeably (order does not matters).
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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



Manager
Joined: 17 Sep 2005
Posts: 71
Location: California

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
Joined: 09 Jul 2005
Posts: 543

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
Joined: 20 Mar 2005
Posts: 196
Location: Colombia, South America

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
Joined: 09 Jul 2005
Posts: 543

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.



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Re: From 6 positive numbers and 6 negative numbers, how many
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20 Apr 2014, 11:43
I've been stuck in this problem.
I don't understand completely why we only take 6C2*6C2 when the group is composed of 2 +ves and 2 ves.
For me, there are 2 scenarios
Scenario 1: ++++ ++ 
Scenario 2: ++ ++ ++
Note: In these 2 scenarios, it is indifferent both: the order of the group and the order WITHIN the group.
Since every decision for a single group affects the other 2 groups, I don't understand how we can simply add up (1)+(2)+(3) as in the previous answer.
My reasoning:
Scenario 1: ++++ ++  6C4*2C2*6C2*4C4 = 225
Scenario 2: ++ ++ ++ 6C2*6C2*4C2*4C2*2C2*2C2
It get confused whenever we're supposed to add combinations vs multiply them.
Help is much appreciated.



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Joined: 13 Dec 2013
Posts: 36

Re: From 6 positive numbers and 6 negative numbers, how many
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20 Apr 2014, 17:29
Bunuel wrote: Enael wrote: I've been stuck in this problem.
I don't understand completely why we only take 6C2*6C2 when the group is composed of 2 +ves and 2 ves.
For me, there are 2 scenarios
Scenario 1: ++++ ++ 
Scenario 2: ++ ++ ++
Note: In these 2 scenarios, it is indifferent both: the order of the group and the order WITHIN the group.
Since every decision for a single group affects the other 2 groups, I don't understand how we can simply add up (1)+(2)+(3) as in the previous answer.
My reasoning:
Scenario 1: ++++ ++  6C4*2C2*6C2*4C4 = 225
Scenario 2: ++ ++ ++ 6C2*6C2*4C2*4C2*2C2*2C2
It get confused whenever we're supposed to add combinations vs multiply them.
Help is much appreciated. Not sure I understand your logic there but hope that the solution below will clear your doubts. 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 For the product of 4 numbers to be positive there must be: 0 positive numbers and 4 negative numbers > choosing all 4 numbers from 6 positive: \(C^4_6=15\); 2 positive numbers and 2 negative numbers > choosing 2 numbers from 6 positive numbers and 2 numbers from 6 negative numbers: \(C^2_6*C^2_6=225\); 4 positive numbers and 0 negative numbers > choosing all 4 numbers from 6 negative: \(C^4_6=15\). Total = 15 + 225 + 15 = 255. Answer: E. Also, note that the order of the chosen numbers is not important here. Meaning that if the positive numbers are 1, 2, 3, 4, 5, and 6 and negative numbers are 1, 2, 3, 4, 5, and 6, then {1, 2, 3, 4} selection is the same as {4, 3, 2, 1} selection, {1, 2, 3, 4} selection is the same as {4, 3, 2, 1} selection, {1, 2, 1, 2} selection is the same as {1, 1, 2, 2} selection... Hope it's clear. Thank you very much Bunuel, somehow I was reading that we needed to make 3 groups of 4 out of the 12 numbers. I have no idea why I interpreted the problem like that.



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Re: From 6 positive numbers and 6 negative numbers, how many
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23 Apr 2014, 18:59
Bunuel wrote: Enael wrote: I've been stuck in this problem.
I don't understand completely why we only take 6C2*6C2 when the group is composed of 2 +ves and 2 ves.
For me, there are 2 scenarios
Scenario 1: ++++ ++ 
Scenario 2: ++ ++ ++
Note: In these 2 scenarios, it is indifferent both: the order of the group and the order WITHIN the group.
Since every decision for a single group affects the other 2 groups, I don't understand how we can simply add up (1)+(2)+(3) as in the previous answer.
My reasoning:
Scenario 1: ++++ ++  6C4*2C2*6C2*4C4 = 225
Scenario 2: ++ ++ ++ 6C2*6C2*4C2*4C2*2C2*2C2
It get confused whenever we're supposed to add combinations vs multiply them.
Help is much appreciated. Not sure I understand your logic there but hope that the solution below will clear your doubts. 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 For the product of 4 numbers to be positive there must be: 0 positive numbers and 4 negative numbers > choosing all 4 numbers from 6 positive: \(C^4_6=15\); 2 positive numbers and 2 negative numbers > choosing 2 numbers from 6 positive numbers and 2 numbers from 6 negative numbers: \(C^2_6*C^2_6=225\); 4 positive numbers and 0 negative numbers > choosing all 4 numbers from 6 negative: \(C^4_6=15\). Total = 15 + 225 + 15 = 255. Answer: E. Also, note that the order of the chosen numbers is not important here. Meaning that if the positive numbers are 1, 2, 3, 4, 5, and 6 and negative numbers are 1, 2, 3, 4, 5, and 6, then {1, 2, 3, 4} selection is the same as {4, 3, 2, 1} selection, {1, 2, 3, 4} selection is the same as {4, 3, 2, 1} selection, {1, 2, 1, 2} selection is the same as {1, 1, 2, 2} selection... Hope it's clear. Hi Bunuel, Your answer is very clear but i'm stuck on one part(and this is the same part that I get on stuck on with every problem). How do you know when the order is important vs. when the order isn't important? Meaning, I can almost never decipher if I need to use the P or C formula. I've read the GMAT club book but haven't had any luck differentiating between "ordered" vs. "unordered" pairs. Thanks, Russ



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Re: From 6 positive numbers and 6 negative numbers, how many
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24 Apr 2014, 19:23
Bunuel wrote: russ9 wrote: Hi Bunuel,
Your answer is very clear but i'm stuck on one part(and this is the same part that I get on stuck on with every problem). How do you know when the order is important vs. when the order isn't important? Meaning, I can almost never decipher if I need to use the P or C formula. I've read the GMAT club book but haven't had any luck differentiating between "ordered" vs. "unordered" pairs.
Thanks, Russ
This should come with practice. Not an universal rule, but below might help to distinguish: The words "Permutation" and "Arrangement" are synonymous and can be used interchangeably (order matters). The words "Combination", "Group" and "Selection" are synonymous and can be used interchangeably (order does not matters).Thanks a ton! I'll give this a shot.



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Re: From 6 positive numbers and 6 negative numbers, how many
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22 Apr 2016, 15:14
Hi Russ I share such confusion with you but just try 1 X 2 = 2 & 2 X 1 = 2 so both orders give the same positive 2 ,otherwise it is mentioned that repeat is allowed. Please Bunuel, correct me if I think wrong.



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Re: From 6 positive numbers and 6 negative numbers, how many
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16 Jun 2016, 12:09
For product of 4 nos. to be positive, the no. of + and  signed nos. should be even. So we can either pick two +ve nos and two ve nos or all four +ve or all four ve.
Total positive nos = 6 Total negative nos = 6
scenario1  2 positive and 2 negative = 6C2*6C2 =15*15 = 225(we multiple because each of the positive nos. could be combined with each of negative nos). scenario2 all four +ve. There are 6C4 ways to pick 4 positive nos. Thus result is 15 scenario3  all four ve. There are 6C4 ways to pick 4 negative nos. Thus result is 15
Total = 225+15+15=255



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Re: From 6 positive numbers and 6 negative numbers, how many
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16 Jun 2016, 12:09
For product of 4 nos. to be positive, the no. of + and  signed nos. should be even. So we can either pick two +ve nos and two ve nos or all four +ve or all four ve.
Total positive nos = 6 Total negative nos = 6
scenario1  2 positive and 2 negative = 6C2*6C2 =15*15 = 225(we multiple because each of the positive nos. could be combined with each of negative nos). scenario2 all four +ve. There are 6C4 ways to pick 4 positive nos. Thus result is 15 scenario3  all four ve. There are 6C4 ways to pick 4 negative nos. Thus result is 15
Total = 225+15+15=255



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Re: From 6 positive numbers and 6 negative numbers, how many
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Re: From 6 positive numbers and 6 negative numbers, how many
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