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Combo Problem

by gmatusa2010 » Sun Oct 24, 2010 9:17 pm
I still don't have an intuitive understanding of this problem type, can someone answer these made up questions, thanks.

5 open slots, 2 red marbles 3 blue marbles, 4 green marbles

a) How many ways to arrange them in the five open slots?

9!/4!5!

b) How many ways to arrange them in the five open slots where order matters?

Is there anyway to restrict the order of the part that is in the slot?

c) How many ways to arrange them in the five open slots where order DOESN'T matter overall but order amongst the same-color marbles does matter?

Is there a way to do this besides doing different combo of 5, and solve them as mini-combo problem within the bigger problem.
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by Geva@EconomistGMAT » Mon Oct 25, 2010 1:29 am
I'll start by saying that you are digging too much into it - I don't think you'll encounter questions of this level in the GMAT.
gmatusa2010 wrote:I still don't have an intuitive understanding of this problem type, can someone answer these made up questions, thanks.

5 open slots, 2 red marbles 3 blue marbles, 4 green marbles

a) How many ways to arrange them in the five open slots?

9!/4!5! This is only true if you assume that marbles of the same color are still considered different from each other. I'll work under that assumption for (b) as well. In effect, this means that you have 9 marbles, and the colors do not really matter.

b) How many ways to arrange them in the five open slots where order matters?

Is there anyway to restrict the order of the part that is in the slot?
the difference between order matter and order doesn't matter is the division by k! in the (order doesn't matter) combinations formula. If the order matters, the formula is simply n!/(n-k)! - in this case, 9!/4!.
If the order doesn't matter, we divide by k! to get the familiar n!/(n-k)!*k! = 9!/4!5!.
Restricting the order is misleading - if the order of choice matters, then you actually have more options. For example, if the marbles are A, B, C, D, E, F, G...
Then the 9!/4! order matters formula means that you're counting such permutations as
ABCDE
BACDE
CABDE
etc.
and also all permutations of A, B, D, F, G (for example)
ABDFG
DABFG
GABFD
etc.
If the order doesn't matter, all of these permutations are one and the same: the first three options all sum up to {A, B, C, D, E}, the last three options (an others like it) all sum up to {A, B, D, F, G}. Thus, when the order doesn't matter, we discount by the number of permutations - the number of choice factorial, or k!.

c) How many ways to arrange them in the five open slots where order DOESN'T matter overall but order amongst the same-color marbles does matter?

Is there a way to do this besides doing different combo of 5, and solve them as mini-combo problem within the bigger problem.
Not really that I can see. But is is much higher in level than what the GMAT offers.
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by Geva@EconomistGMAT » Mon Oct 25, 2010 2:02 am
I just looked at this again, and realized that I didn't really answer your question, as my response basically ignores the fact that you have three colors. Presumably, balls of the same color are interchangeable, and this is not reflected in 9!/4!5! or 9!/4!.
If you were asking how many ways are there to arrange all 9 balls in a row, the answer would've been 9! divided by the number of internal arrangements of each group: 2!3!4!.
The reasoning behind this is that there are 9! ways to arrange 9 balls in a row, but withing these 9! permutations (think of them as strings of 9 letters each) there are duplicate permutations. If we had 9 balls A, B, C, D, E, F, G, H, I, then the 9! permutations will count all the following strings:

ABCDEFGHI
ACBDEFGHI
ACDBEFGHI etc.
as separate arrangements.

however, if we replace A,B, C with 3 identical blue marbles (all Bs), then the three strings above are actually the same string:
BBBDEFGHI.
In order to remove these duplicates, the 9! needs to be discounted by the number of internal arrangements of each group. The answer in this case would've been 9!/2!3!4! (2! for the internal arrangements involving the two reds, 3! for blues, 4! for greens).
Not sure how to factor in the idea of choosing only 5 balls out of these 9 - not even sure it can be done without looking at each and every scenario separately. In any case, too complicated for the GMAT.
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by gmatusa2010 » Mon Oct 25, 2010 4:13 am
Thanks Geva. I get most of what you are saying.

I was being confusing, so just to rehash. In A, the same colors are interchangeable (9!/5!4!, I get that). You might've alluded to this already at the end but just to clarify:

1) Is there a way to do A but restrict the repeats? (R1,R2,B,B,B is the same as R2,R1,B,B,B etc...). To put it another way, is there a way to combine concepts: the concept of selecting from a pool larger than the number of available seats AND the concept of putting the number of repeats in the denominator?

2) IS there a way to restrict the order so not only do repeats not count, the same combination (the order, this is common in team selection type problem) does not count (in the case above, 2 reds, 3 blues.) The manual way to do it would be work out all the combo that adds to 5 ex: 2 red+3 blue, 1 red+3 blue+1 green, 2 green etc....... (I actually ran into two problems that require both 1 and 2, both required cumbersome calculation so I was wondering if there's an easier way. [neither official gmat questions])

Thanks again.



Geva Stern wrote:I just looked at this again, and realized that I didn't really answer your question, as my response basically ignores the fact that you have three colors. Presumably, balls of the same color are interchangeable, and this is not reflected in 9!/4!5! or 9!/4!.
If you were asking how many ways are there to arrange all 9 balls in a row, the answer would've been 9! divided by the number of internal arrangements of each group: 2!3!4!.
The reasoning behind this is that there are 9! ways to arrange 9 balls in a row, but withing these 9! permutations (think of them as strings of 9 letters each) there are duplicate permutations. If we had 9 balls A, B, C, D, E, F, G, H, I, then the 9! permutations will count all the following strings:

ABCDEFGHI
ACBDEFGHI
ACDBEFGHI etc.
as separate arrangements.

however, if we replace A,B, C with 3 identical blue marbles (all Bs), then the three strings above are actually the same string:
BBBDEFGHI.
In order to remove these duplicates, the 9! needs to be discounted by the number of internal arrangements of each group. The answer in this case would've been 9!/2!3!4! (2! for the internal arrangements involving the two reds, 3! for blues, 4! for greens).
Not sure how to factor in the idea of choosing only 5 balls out of these 9 - not even sure it can be done without looking at each and every scenario separately. In any case, too complicated for the GMAT.
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by Geva@EconomistGMAT » Mon Oct 25, 2010 5:08 am
gmatusa2010 wrote:Thanks Geva. I get most of what you are saying.

I was being confusing, so just to rehash. In A, the same colors are interchangeable (9!/5!4!, I get that). You might've alluded to this already at the end but just to clarify:

1) Is there a way to do A but restrict the repeats? (R1,R2,B,B,B is the same as R2,R1,B,B,B etc...). To put it another way, is there a way to combine concepts: the concept of selecting from a pool larger than the number of available seats AND the concept of putting the number of repeats in the denominator?

No easy way to do this - that was the bottom line of my second post.

2) IS there a way to restrict the order so not only do repeats not count, the same combination (the order, this is common in team selection type problem) does not count (in the case above, 2 reds, 3 blues.) The manual way to do it would be work out all the combo that adds to 5 ex: 2 red+3 blue, 1 red+3 blue+1 green, 2 green etc....... (I actually ran into two problems that require both 1 and 2, both required cumbersome calculation so I was wondering if there's an easier way. [neither official gmat questions])
Again - no easy way. If you had an unlimited number of balls from each color, then we could attack it from the side of the slots, rather than the balls: There are five slots, and each slot has three options: red, green, blue. The answer in that case would've been 3*3*3*3*3=3^5. But with the limited number of balls fro each color (i.e. less than 5), I can find no other way than to go scenario by scenario.

Thanks again.
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