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'Number of ways...' question

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'Number of ways...' question

by kurt5828 » Mon Apr 12, 2010 7:11 pm
Hi everyone,

I'm stuck on this question:

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?

Can someone help? Thanks!
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by eaakbari » Mon Apr 12, 2010 7:59 pm
Is the answer 13824?
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by kurt5828 » Mon Apr 12, 2010 8:04 pm
13824 is incorrect; the problem I have is once I know the answer I can validate it in a bunch of presumably incorrect ways.

I'm hoping someone out there can break down the correct answer for me!
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by rockeyb » Mon Apr 12, 2010 8:15 pm
Let me give it a try .

This is an arrangement problem with a constraint that A, B , C should stand in front of F, G , H .

We have got 8 different people A, B, C,D,E,F,G,H .

Now D and E can take any position out of the 8 .

So Lets say 8 x 7 x_ x _x_ x _x_ x _

D can be arranged in 8 different positions and E can take any of the remaining 7 positions .

Now F , G and H can take any of the 6 th , 7 th or 8 th positions but can not take 3rd ,4th , 5 th positions as per the question.

So we can arrange F , G, H in 3! ways = 3x2x1

8 x 7 x_ x _x_ x 3 x 2 x 1

Similarly A , B , C can be arranged in 3! ways = 3 x 2 x1


8 x 7 x3 x 2 x1 x 3 x 2 x 1 = 2016 ways .

Is this the correct answer?
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by dxgamez » Mon Apr 12, 2010 8:35 pm
I'm gonna take a shot too.

DE can take up to 3 ways so that ABC can be in front of FGH.

(ABC) (FGH) (DE) or (DE) (ABC) (FGH) or (D) (ABC) (FGH) (E) , (E) (ABC) (FGH) (D)

1st case :- ABC can be arranged in 3! ways, so is FGH. DE can be arranged in 2! ways, Thus 3! x 3! x 2!

2nd case:- same like 1st case 2! x 3! x 3!

3rd case: same like 1st case...

no of ways = 3 x (3! x 3! x 2!) x (2! x 3! x 3!) x (2! x 3! x 3!) = 139968

looks so wrong though!
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by liferocks » Mon Apr 12, 2010 9:02 pm
we can divide the people in three groups
ABC,FGH,ED

now arrangement will be _ _ABC_ _FGH_ _

ABC can arrange among themselves in 3!=6 ways
FGH can arrange among themselves in 3!=6 ways
E can select any of the 6 blanks in 6C1 ways ie 6 ways
D can select any of the remaining 5 blanks in 5C1 ways ie 5ways

so ans is 6x6x6X5=216X5=1080ways
can we get the OA
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by kstv » Mon Apr 12, 2010 9:38 pm
liferocks wrote:we can divide the people in three groups
ABC,FGH,EDnow arrangement will be _ _ABC_ _FGH_ _
ABC can arrange among themselves in 3!=6 ways FGH can arrange among themselves in 3!=6 ways
E can select any of the 6 blanks in 6C1 ways ie 6 ways D can select any of the remaining 5 blanks in 5C1 ways ie 5ways
so ans is 6x6x6X5=216X5=1080ways
can we get the OA
see the bold part
why make ED a group , as your calc. shows they can stand anywhere. If we make ABC as Group I , FGH as Group II.Then arrange Group I, II , E and D in 4! possible ways . Of these, in half the arrangements Group II will be ahead of Group I and needs to be eliminated. For the rest I have same logic.

@ kurt5828 pl. you have to give us the answer choices maybe not the OA. You can make it up, embedding the right option. Even the experts like Stuart is a strong proponent of this . In GMAT no marks are given for steps involved in solving. Without giving the options you are robbing us of finding out fastest/cleverest method to solve the Q. Give it a thought. Thanks for a cracking Q.
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by kevincanspain » Mon Apr 12, 2010 10:24 pm
Rockeyb's answer looks great to me!

Here's a less elegant way that produces the same answer

Step One

Find number of permutations of ABC and that of FGH

3! X 3! = 36


Suppose we have ABCFGH

The two remaining letters can be placed consecutively in 7 x 2 = 14 ways
The two remaining letters can be placed non-consecutively in 7P2 = 42 ways


Thus the two remaining letters can be placed in 56 ways

The 8 letters can therefore be placed in 36 x 56 = 2016 ways
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by kstv » Mon Apr 12, 2010 11:13 pm
D and E can be arranged in any of the spaces numbered after ensuring that ABC is ahead of FGH

The two remaining letters can be placed together then they can be arranged in 7 x 2 = 14 ways
i.e in any of the 7 slots and between them they can be arranged in 2 ways

1 A 2 B 3 C 4 F 5 G 6 H 7

The two remaining letters can be placed non-consecutively in 7P2 = 42 ways
D or E is any of the 7 slots and then the other in the remaining 6 slots.

Yes!
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by kevincanspain » Tue Apr 13, 2010 1:25 am
Remember to look at rockeyb's solution- it's nicer
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by akhpad » Tue Apr 13, 2010 4:42 am
8P2 * 3! * 3! = 2016
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by kurt5828 » Tue Apr 13, 2010 5:49 pm
Thanks for the explanations! Very helpful. I apologize for not including answer choices - all I had was the answer! This was a review question and meant to teach principles (apparently...)
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