Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

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.

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.

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.

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

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).

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

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!

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. _________________

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

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