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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] New post 05 Mar 2003, 23:47
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. :wall
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 [#permalink] New post 06 Mar 2003, 23:51
silence...

Who can crack it?
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 [#permalink] New post 19 Mar 2003, 05:29
Excellent website by the way!

:idea: (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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 [#permalink] New post 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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don't agree [#permalink] New post 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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 [#permalink] New post 18 May 2003, 04:47
people, i'm shaken.

Is the answer actually 4???

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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 [#permalink] New post 27 Nov 2003, 09:06
i guess the answer should be 12.

whats the official answer?
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Re: combi 3 [#permalink] New post 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. :wall


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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 [#permalink] New post 17 Dec 2003, 16:58
I agree with AkamaiBrah. Answer should be 7
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 [#permalink] New post 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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 [#permalink] New post 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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Re: combi 3 [#permalink] New post 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. :wall


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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 [#permalink] New post 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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 [#permalink] New post 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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 [#permalink] New post 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
  [#permalink] 21 Dec 2003, 16:08
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