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# one has 12 balls: 4 fluorescent green, 4 deep yellow, and 4

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one has 12 balls: 4 fluorescent green, 4 deep yellow, and 4 [#permalink]

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05 Mar 2003, 23:47
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

one has 12 balls: 4 fluorescent green, 4 deep yellow, and 4 dark violet metallic. one want to divide them in 2 equal sets of 6 balls either. In how many ways can one do so, providing that either set should have balls of all given colors.

this is hard. I would say a real hard nut to crack.
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06 Mar 2003, 23:51
silence...

Who can crack it?
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19 Mar 2003, 05:29
Excellent website by the way!

(Nr. of groups without restrictions)-(Nr. of groups with only two colours)

so the nr. of groups without restrictions is: ┬╜ x 6C12= 462
( I multiplied by 1/2 because one group defines the other!)

nr. of groups with only two colours:
for example G G G G _ _ = 4C4 x 2C8= 1x 28
but there are three different colours so its 3x(1x28) = 84

so there are:
462-84= 378 possible groups with balls of all different colours
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19 Mar 2003, 05:44
The number of combinations with 2 colors is

GGGGYY
GGGGVV
GGGYYY
GGGVVV
GGYYYY
GGVVVV

so 6 for green, not 28
the same for Y and V -- total 18, but we count combinations twice (GGGVVV=VVVGGG)
Therefore, we have divide by 2 and get 9 prohibited combinations.
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18 May 2003, 04:19
I don't agree with your methods.

First, you should divide by 2 only AFTER you subtract the unfav combinations!

Second, stolyar, this are not combinations you're counting! I.e. GGGGYY makes 4C4*4C2 = 6 combinations, not one!

I used a different approach and i got another answer. Perhaps i just can't get your logic. Can you please check your solutions and mine so that we reconcile our answers? I'm trying to adapt this problem to quiz engine.

My solution:

Dividing in two groups is same as "taking" some balls while "leaving" others.

We can take 6 balls in 12C6 ways.

12C6 = 12! / 6! * 6! = 7 * 8 * 9 * 10 * 11 * 12 / 6 * 5 * 4 * 3 * 2 = 7 * 3 * 2 * 11 * 2 = 924.

The combinations that don't match:

No g (no v, no y) taken: 8C6. Here we count such takes as VVVYYY, VVVVYY, etc.
Total: no color taken - 3 * 8C6;

3 * 8C6 = 3 * 8! / 6! * 2! = 3 * 7 * 8 / 2 = 3 * 7 * 4 = 84

No g (no v, no y) left, but all colors taken: 4C1*4C1 = 16. Here we count such takes as GGGGYW, which would take all balls of a given color and leave none of them to the second group. Total: 3 * 16 = 48.

Answer: (924 - 48 - 84) / 2 = 792 / 2 = 396.

Why divide by 2? Because we "taken" and "left" groups are symmetrical, and we therefore have counted each combination twice.
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18 May 2003, 04:47
people, i'm shaken.

GGYYVV - GGYYVV
GGGYYV - GYYVVV
GGGVVY - GVVVYY
GGYYYV - GGYVVV

I bet all answers will fit into one of this classes...

It can be seen thus: let's denote a combination by as gyv-gyv, where g, y, and v stand for the number of balls of that color in group 1 and group 2. We have such possibilities:

222-222

321-123
312-132

132-312
231-213

123-321
213-231

It can be seen that each of them but 222-222 is repeated twice: 321-123 = 123-321 for example. So, there're only 4 of them:

By dominant colors they are:

222-222 : equidominant
321-123 : green - violet
312-132 : green - yellow
231-213 : yellow - violet
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27 Nov 2003, 09:06
i guess the answer should be 12.

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28 Nov 2003, 05:35
stolyar wrote:
one has 12 balls: 4 fluorescent green, 4 deep yellow, and 4 dark violet metallic. one want to divide them in 2 equal sets of 6 balls either. In how many ways can one do so, providing that either set should have balls of all given colors.

this is hard. I would say a real hard nut to crack.

Hmm. Let's see if I can crack this nut...

If we know we need one of each color in each bin, let's simply take out 2 balls of each color and simply assume one of each color is already in each bin. Now we have 6 balls left, two of each color.

Lets just look at Bin1 (since bin2 is completely dependend on how we fill bin1). Assume colors are A B and C. Since we have at most two balls left of any color, only possible combinations are:

ABC
AAB
AAC
BBA
BBC
CCA
CCB

Hence, IMO the answer in 7.

Here are the complete bin combinations (must have at least 1 and at most 3 of each color).

AABBCC
AAABBC
AAABCC
AABBBC
AABCCC
ABBBCC
ABBCCC
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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17 Dec 2003, 16:58
I agree with AkamaiBrah. Answer should be 7
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17 Dec 2003, 17:41
Guys, have you seriously found any thing this tough on the GMAT for anybody who got 700+?
Just curious. If yes, I think I would just choose randomly on a question like this one!
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18 Dec 2003, 01:39
Agree it's 7, I did same way as AkamaiBrah did. Just considered distribution of 3 pairs of balls cuase the other 6 can not be replaced.
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18 Dec 2003, 03:57
stolyar wrote:
one has 12 balls: 4 fluorescent green, 4 deep yellow, and 4 dark violet metallic. one want to divide them in 2 equal sets of 6 balls either. In how many ways can one do so, providing that either set should have balls of all given colors.

this is hard. I would say a real hard nut to crack.

anvar, stoolfi..why am i way off here?

1. Draw 6 making sure that you leave a minimum of 2 of each.
4C2 + 4C2 + 4C2 = 18 ways

2. Draw 6 making sure that you leave 2 of green and 1 yellow
Now it doesnt matter which colors you draw in the combination of 3,2,1
4C3 + 4C2 + 4C1 = 11 ways ( 3 + 2 +1 = 6 )

so 11 ways represent all those ways in which we can draw balls in set of 3,2,1

3. Draw 6 making sure that you leave 1 of green and 1 of yellow
Again, it doesnt matter which colors you draw in combination of 4,1,1
4C4 + 4C1 + 4C1 = 9 ways

total ways = 38 ways
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18 Dec 2003, 04:30
First set

Fixed Variations
------- -----------
GYV + GGY
GYV + GGV
GYV + GYY
GYV + GVV
GYV + GYV
GYV + YYV
GYV + YVV

It leaves us with a second set:

Fixed Variations
------- -----------
GYV + YVV
GYV + VYY
GYV + GVV
GYV + GYY
GYV + GYV
GYV + GVV
GYV + GGY

I believe in your variations there is a double-counting
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18 Dec 2003, 12:49
Quote:
anvar, stoolfi..why am i way off here?

I'm flattered that you would ask. I suspect double counting as well. If I have time this weekend, I'll take a closer look.
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21 Dec 2003, 16:08
double counting is not requried.
consider
GYV-grp1 GYV-grp2
321 123
312 132

and so on. So for each unique arrangement in group_1, group_2 has only 1 possible unique arrangement. And 222 has one possible way only. So ans is 3!+1 = 7
21 Dec 2003, 16:08
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