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Standing in the line

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by JeetGulia » Thu Jul 29, 2010 1:55 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
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by Rahul@gurome » Thu Jul 29, 2010 7:45 pm
Solution:
Let people Alice, Benjamin, Charlene, David, Elaine, Frederick, Gale, and Harold be A, B, C, D, E, F, G and H respectively.

Now A, B and C must be before F,G and H.
Two cases can arise.
Either (1) D and E are together
or
(2) D and E are not together.
Consider (1) first.
So if A B C F G H are arranged like this and D and E have to be together, there are 7 places in which they can be put in 7P1 *2 or 14 ways.
Also A. B and C can be arranged among themselves in 3! or 6 ways and F, G and H can be arranged among themselves in 3! or 6 ways.
Or If D and E have to be together, they can be arranged in 14*6*6 = 504 ways.
Consider (2) then.
If they are arranged like this A B C F G H, there are 7 places between them where D and E can be placed.
This can be done in 7P2 or 7*6 = 42 ways.
Also A. B and C can be arranged among themselves in 3! or 6 ways and F, G and H can be arranged among themselves in 3! or 6 ways.
So the possible arrangements we can have is 42*6*6 = 1,512.

Or total number of arrangements is 1512+504 = 2016 ways.

The correct answer is (D).
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by JeetGulia » Fri Jul 30, 2010 2:02 am
Rahul@gurome wrote:Solution:
Let people Alice, Benjamin, Charlene, David, Elaine, Frederick, Gale, and Harold be A, B, C, D, E, F, G and H respectively.

Now A, B and C must be before F,G and H.
Two cases can arise.
Either (1) D and E are together
or
(2) D and E are not together.
Consider (1) first.
So if A B C F G H are arranged like this and D and E have to be together, there are 7 places in which they can be put in 7P1 *2 or 14 ways.
Also A. B and C can be arranged among themselves in 3! or 6 ways and F, G and H can be arranged among themselves in 3! or 6 ways.
Or If D and E have to be together, they can be arranged in 14*6*6 = 504 ways.
Consider (2) then.
If they are arranged like this A B C F G H, there are 7 places between them where D and E can be placed.
This can be done in 7P2 or 7*6 = 42 ways.
Also A. B and C can be arranged among themselves in 3! or 6 ways and F, G and H can be arranged among themselves in 3! or 6 ways.
So the possible arrangements we can have is 42*6*6 = 1,512.

Or total number of arrangements is 1512+504 = 2016 ways.

The correct answer is (D).
Rahul..Could you please explain little bit how did you come up with

if A B C F G H are arranged like this and D and E have to be together, there are 7 places in which they can be put in 7P1 *2 or 14 ways.

and

If they are arranged like this A B C F G H, there are 7 places between them where D and E can be placed.
This can be done in 7P2 or 7*6 = 42 ways.
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by Rahul@gurome » Fri Jul 30, 2010 2:59 am
Take the case when D and E are together. This means the cases where they are standing next to each other.
The possibilities are
D E A B C F G H or E D A B C F G H
A D E B C F G H or A E D B C F G H
A B D E C F G H or A B E D C F G H
.
.
.
.
A B C F G H D E or A B C F G H E D

As seen from above we get 7 places for DE, 5 between A, B, C, F, G and H,1 before A and 1 after H. 7P1 is taken because D E are taken as one item and they have to be arranged in any of the seven places. This is multiplied by 2 because for every D E there is a possibility of E D.

In the second case where D and E are not together, we get 7 places again and in these 7 places D and E can be arranged in 7P2 ways. Since they are not standing together they are taken as 2 items.

I hope it is clearer now!
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