From 6 positive numbers and 6 negative numbers, how many

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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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

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

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

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)

Potential combinations of 4 numbers that will yield positive product

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

4C6 + 4C6 + 2C6*2C6
15 +15 + 15*15 = 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

Answer (E)

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

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.

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.

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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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

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 !

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