GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 22 Oct 2019, 07:16

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# From 6 positive numbers and 6 negative numbers, how many

Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 09 Jul 2005
Posts: 490
From 6 positive numbers and 6 negative numbers, how many  [#permalink]

### Show Tags

Updated on: 20 Apr 2014, 12:07
4
22
00:00

Difficulty:

45% (medium)

Question Stats:

67% (02:13) correct 33% (02:17) wrong based on 466 sessions

### HideShow timer Statistics

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.
Math Expert
Joined: 02 Sep 2009
Posts: 58432
Re: From 6 positive numbers and 6 negative numbers, how many  [#permalink]

### Show Tags

20 Apr 2014, 12:18
4
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.

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
Director
Joined: 13 Nov 2003
Posts: 647
Location: BULGARIA

### Show Tags

31 Oct 2005, 11:11
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
Director
Joined: 22 Aug 2005
Posts: 900
Location: CA

### Show Tags

31 Oct 2005, 11:24
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: 62
Location: California

### Show Tags

31 Oct 2005, 15:58
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
Joined: 17 Sep 2005
Posts: 62
Location: California

### Show Tags

31 Oct 2005, 16:22
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
Retired Moderator
Joined: 20 Dec 2013
Posts: 166
Location: United States (NY)
GMAT 1: 640 Q44 V34
GMAT 2: 710 Q48 V40
GMAT 3: 720 Q49 V40
GPA: 3.16
WE: Consulting (Venture Capital)
Re: From 6 positive numbers and 6 negative numbers  [#permalink]

### Show Tags

15 Apr 2014, 18:31
1
1
Potential combinations of 4 numbers that will yield positive product

++++ 4C6
---- 4C6
++-- 2C6*2C6

4C6 + 4C6 + 2C6*2C6
15 +15 + 15*15 = 255
_________________
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9701
Location: Pune, India
Re: From 6 positive numbers and 6 negative numbers  [#permalink]

### Show Tags

15 Apr 2014, 20:41
2
2
tauchmeister wrote:
Another question, from a set I found here at home.

I don't know from where the source is.

Thanks

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

When will the product of 4 numbers be positive? It is positive when the number of negative numbers in the product is even i.e. when we have 0 negative numbers or 2 negative numbers or 4 negative numbers.

3 cases:

- 0 negative numbers - All numbers are positive - Selected in 6C4 = 15 ways

- 2 negative numbers - 2 are positive and 2 are negative - Selected in 6C2*6C2 = 15*15 = 225 ways

- 4 negative numbers - All numbers are negative - Selected in 6C4 = 15 ways

Total number of ways = 15 + 225 + 15 = 255

_________________
Karishma
Veritas Prep GMAT Instructor

Director
Joined: 03 Aug 2012
Posts: 658
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]

### Show Tags

20 Apr 2014, 00:56
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
Intern
Joined: 13 Dec 2013
Posts: 35
GMAT 1: 620 Q42 V33
Re: From 6 positive numbers and 6 negative numbers, how many  [#permalink]

### Show Tags

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.
Intern
Joined: 13 Dec 2013
Posts: 35
GMAT 1: 620 Q42 V33
Re: From 6 positive numbers and 6 negative numbers, how many  [#permalink]

### Show Tags

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.

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: 228
Re: From 6 positive numbers and 6 negative numbers, how many  [#permalink]

### Show Tags

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.

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
Math Expert
Joined: 02 Sep 2009
Posts: 58432
Re: From 6 positive numbers and 6 negative numbers, how many  [#permalink]

### Show Tags

24 Apr 2014, 01:00
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.

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).

_________________
Intern
Joined: 01 Feb 2016
Posts: 9
Re: From 6 positive numbers and 6 negative numbers, how many  [#permalink]

### Show Tags

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
Intern
Joined: 13 Jul 2014
Posts: 3
Schools: Kellogg '18
Re: From 6 positive numbers and 6 negative numbers, how many gro  [#permalink]

### Show Tags

03 May 2019, 07:05
Thanks for the solution but I am not able to get this.

There are 6 +ve numbers, and if I need to choose 4, it is - 6*5*4*3 ways, why do you use 6C4 ? I know this is a fundamental concept, but I cant pin the difference.
Intern
Joined: 22 Oct 2018
Posts: 37
GMAT 1: 690 Q47 V38
GPA: 1
Re: From 6 positive numbers and 6 negative numbers, how many gro  [#permalink]

### Show Tags

03 May 2019, 11:49
1
vigneshrajendran wrote:
Thanks for the solution but I am not able to get this.

There are 6 +ve numbers, and if I need to choose 4, it is - 6*5*4*3 ways, why do you use 6C4 ? I know this is a fundamental concept, but I cant pin the difference.

You’re thinking of permutation which is the way to arrange 6 numbers however combination involves selection

6! Is permutation which demonstrates how many ways you can arrange 6 numbers = 6x5x4x3x2x1

6C4 however calculates how many ways you can select a group of 4 numbers from the 6 given which is calculated by:

6! / 4! (4-2)! = (6x5x4x3x2x1)/(4x3x2x1) x 2 = 15

The solutions above are pretty good so I have nothing else to add.

Hope this helps !

Posted from my mobile device
Non-Human User
Joined: 09 Sep 2013
Posts: 13392
Re: From 6 positive numbers and 6 negative numbers, how many  [#permalink]

### Show Tags

09 Jul 2019, 04:39
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: From 6 positive numbers and 6 negative numbers, how many   [#permalink] 09 Jul 2019, 04:39
Display posts from previous: Sort by