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From 6 positive numbers and 6 negative numbers, how many

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From 6 positive numbers and 6 negative numbers, how many [#permalink] New post 31 Oct 2005, 08:27
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 20 Apr 2014, 11:07, 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 [#permalink] New post 19 Apr 2014, 23:56
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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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Re: From 6 positive numbers and 6 negative numbers, how many [#permalink] New post 20 Apr 2014, 11:18
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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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Re: From 6 positive numbers and 6 negative numbers, how many [#permalink] New post 24 Apr 2014, 00:00
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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).

_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

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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
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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
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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
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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
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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??
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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
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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.
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Re: From 6 positive numbers and 6 negative numbers, how many [#permalink] New post 20 Apr 2014, 10:43
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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Re: From 6 positive numbers and 6 negative numbers, how many [#permalink] New post 20 Apr 2014, 16:29
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. :shock:
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Re: From 6 positive numbers and 6 negative numbers, how many [#permalink] New post 23 Apr 2014, 17:59
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 [#permalink] New post 24 Apr 2014, 18:23
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.
Re: From 6 positive numbers and 6 negative numbers, how many   [#permalink] 24 Apr 2014, 18:23
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