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SVP
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one has 12 balls: 4 fluorescent green, 4 deep yellow, and 4 [#permalink]
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06 Mar 2003, 00:47
This topic is locked. If you want to discuss this question please repost 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.



SVP
Joined: 03 Feb 2003
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silence...
Who can crack it?



Intern
Joined: 18 Mar 2003
Posts: 4
Location: Munich

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:
46284= 378 possible groups with balls of all different colours



SVP
Joined: 03 Feb 2003
Posts: 1604

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.



CTO
Joined: 19 Dec 2002
Posts: 250
Location: Ukraine

don't agree [#permalink]
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18 May 2003, 05:19
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.



CTO
Joined: 19 Dec 2002
Posts: 250
Location: Ukraine

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 gyvgyv, where g, y, and v stand for the number of balls of that color in group 1 and group 2. We have such possibilities:
222222
321123
312132
132312
231213
123321
213231
It can be seen that each of them but 222222 is repeated twice: 321123 = 123321 for example. So, there're only 4 of them:
By dominant colors they are:
222222 : equidominant
321123 : green  violet
312132 : green  yellow
231213 : yellow  violet



Intern
Joined: 29 Aug 2003
Posts: 29
Location: Mumbai

i guess the answer should be 12.
whats the official answer?



GMAT Instructor
Joined: 07 Jul 2003
Posts: 770
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE

Re: combi 3 [#permalink]
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28 Nov 2003, 06:35
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



SVP
Joined: 30 Oct 2003
Posts: 1790
Location: NewJersey USA

I agree with AkamaiBrah. Answer should be 7



GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4288

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!



Intern
Joined: 27 Nov 2003
Posts: 33
Location: Moscow

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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CEO
Joined: 15 Aug 2003
Posts: 3454

Re: combi 3 [#permalink]
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18 Dec 2003, 04:57
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



Intern
Joined: 27 Nov 2003
Posts: 33
Location: Moscow

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 doublecounting
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Director
Joined: 28 Oct 2003
Posts: 501
Location: 55405

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.



Intern
Joined: 04 Oct 2003
Posts: 42
Location: Silicon Valley, CA

double counting is not requried.
consider
GYVgrp1 GYVgrp2
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










