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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

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

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

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

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

Show Tags

20 Apr 2014, 00:56

1

This post received KUDOS

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 _________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

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...

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.

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.

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). _________________

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

Show Tags

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).

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

Show Tags

01 Mar 2016, 01:10

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]

Show Tags

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.

gmatclubot

Re: From 6 positive numbers and 6 negative numbers, how many
[#permalink]
22 Apr 2016, 15:14

So, my final tally is in. I applied to three b schools in total this season: INSEAD – admitted MIT Sloan – admitted Wharton – waitlisted and dinged No...

HBS alum talks about effective altruism and founding and ultimately closing MBAs Across America at TED: Casey Gerald speaks at TED2016 – Dream, February 15-19, 2016, Vancouver Convention Center...

By Libby Koerbel Engaging a room of more than 100 people for two straight hours is no easy task, but the Women’s Business Association (WBA), Professor Victoria Medvec...