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Director  Joined: 09 Jul 2005
Posts: 543
From 6 positive numbers and 6 negative numbers, how many  [#permalink]

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3
19 00:00

Difficulty:   55% (hard)

Question Stats: 67% (02:14) correct 33% (02:17) wrong based on 653 sessions

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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

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.
##### Most Helpful Expert Reply
Math Expert V
Joined: 02 Sep 2009
Posts: 55193
Re: From 6 positive numbers and 6 negative numbers, how many  [#permalink]

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3
5
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.
_________________
##### General Discussion
VP  Joined: 22 Aug 2005
Posts: 1045
Location: CA

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1
1
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

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1
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) 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  Joined: 03 Aug 2012
Posts: 695
Concentration: General Management, General Management
GMAT 1: 630 Q47 V29 GMAT 2: 680 Q50 V32 GPA: 3.7
WE: Information Technology (Investment Banking)
Re: From 6 positive numbers and 6 negative numbers, how many  [#permalink]

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1
1
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
Math Expert V
Joined: 02 Sep 2009
Posts: 55193
Re: From 6 positive numbers and 6 negative numbers, how many  [#permalink]

### Show Tags

1
2
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).

_________________
Director  Joined: 13 Nov 2003
Posts: 756
Location: BULGARIA

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1
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

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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

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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

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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

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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.
Intern  Joined: 13 Dec 2013
Posts: 36
GMAT 1: 620 Q42 V33 Re: From 6 positive numbers and 6 negative numbers, how many  [#permalink]

### Show Tags

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.
Intern  Joined: 13 Dec 2013
Posts: 36
GMAT 1: 620 Q42 V33 Re: From 6 positive numbers and 6 negative numbers, how many  [#permalink]

### Show Tags

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. Manager  Joined: 15 Aug 2013
Posts: 243
Re: From 6 positive numbers and 6 negative numbers, how many  [#permalink]

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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
Manager  Joined: 15 Aug 2013
Posts: 243
Re: From 6 positive numbers and 6 negative numbers, how many  [#permalink]

### Show Tags

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.
Manager  B
Joined: 20 Apr 2014
Posts: 88
Re: From 6 positive numbers and 6 negative numbers, how many  [#permalink]

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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.
Intern  Joined: 01 Feb 2016
Posts: 10
Re: From 6 positive numbers and 6 negative numbers, how many  [#permalink]

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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
Intern  Joined: 01 Feb 2016
Posts: 10
Re: From 6 positive numbers and 6 negative numbers, how many  [#permalink]

### Show Tags

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
Non-Human User Joined: 09 Sep 2013
Posts: 10958
Re: From 6 positive numbers and 6 negative numbers, how many  [#permalink]

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_________________ Re: From 6 positive numbers and 6 negative numbers, how many   [#permalink] 05 Jul 2017, 00:34
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# From 6 positive numbers and 6 negative numbers, how many

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