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combination problem

by Taniuca » Mon Nov 01, 2010 6:29 pm
In how many different orders can the people Alice, Benjamin, Charlene, David, Elaine, Frederick, Gale, and Harold be standing on line if each of Alice, Benjamin, Charlene must be on the line before each of Frederick, Gale, and Harold?

a) 1,008

b) 1,296

c) 1,512

d) 2,016

e) 2,268


I have tried first to get the totals by grouping ABC and FGH as one unit so that I can have 4 elements to be 4!=24
Then, I have found the number of times DE can be which is 2!=2
ABC= 3!=6 and FGH=3!=6 Why the answer cannot be 24*6*6*2 =1728 that is not the right answer provided.
Why do I have to try to do it as a combination like the following
8!/ (2!)(6!) = 28
Then that value 28 multiplied with the options ABC, FGH, and DE = 28*2*6*6 will give me the right answer 2,010.

Could someone tell me what is that I don't have correct on my original concept?

Thanks in advance
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by Rahul@gurome » Mon Nov 01, 2010 8:12 pm
Solution:
A,B, C have to be before F, G, H.
Now D and E can occupy any space between each of A, B, C, F, G and H .
D and E can be case (1) together
Or case (2) separated by one other person.

Case (1) There are 7 places between A, B, C, F, G and H where they can be put in 7P1 * 2 ways. (multiplying by 2 because for every DE arrangement, there is an ED arrangement)
Besides A, B and C can be arranged among themselves in 3! Ways.
D, E and F can also be arranged among themselves in 3! ways.
So total ways of arranging A, B, C, D, E, F, G, and H such that A, B and C are always before F, G and H and D, E are together is 7P1*2*3!*3! = 504

Case (2) There are 7 places between A, B, C, F, G, and H where they can be put in 7P2 ways.
Besides A, B and C can be arranged among themselves in 3! Ways.
D, E and F can also be arranged among themselves in 3! ways.
So total ways of arranging A, B, C, D, E, F, G, and H such that A, B and C are always before F, G and H and D, E are not together is 7P2*3!*3! = 1512

So combining both cases correct answer is 1512 + 504 = 2016.
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by Ian Stewart » Mon Nov 01, 2010 8:17 pm
Taniuca wrote: I have tried first to get the totals by grouping ABC and FGH as one unit so that I can have 4 elements to be 4!=24
Then, I have found the number of times DE can be which is 2!=2
ABC= 3!=6 and FGH=3!=6 Why the answer cannot be 24*6*6*2 =1728
You are actually answering a slightly different question from the one that's asked. When you 'group ABC as one unit', you are then assuming that A, B and C are together in line, in some order. That is, you wouldn't be counting certain valid lineups, like ABDCEFGH, in which A, B and C do not appear in a row. So you haven't counted those lineups when either D or E is somewhere in the middle of the group ABC or the group FGH.

We can instead answer the question as follows: there are 8 places to put D, and then 7 places to put E. Once we've placed D and E, we know that A, B and C must go in the first three empty places, in some order; there are 3! orders we can put them in. Then in the last three empty spots, we must put F, G and H, and there are 3! orders to put them. So the answer is 8*7*3!*3! = 2016.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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