Permutation HELP

[manihar.sidharth]

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?

1,008

1,296

1,512

2,016

2,268

I see many times these types of questions appear on GMAT mock tests.
So anybody can propose a generic approach to these type of questions.
OA after some discussion
Thanks
Sid

[GMATGuruNY]

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?

1,008

1,296

1,512

2,016

2,268

I see many times these types of questions appear on GMAT mock tests.
So anybody can propose a generic approach to these type of questions.
OA after some discussion
Thanks
Sid

Number of options for D = 8. (Any of the 8 positions.)
Number of options for E = 7. (Any of the 7 remaining positions.)

A, B and C must stand leftward of F, G and H.
Implication:
Of the 6 positions left, A, B and C must occupy the 3 most leftward.
Number of options for A = 3. (Any of the 3 most leftward positions.)
Number of options for B = 2. (Either of the 2 remaining most leftward positions.)
Number of options for C = 1. (Only 1 of the most leftward positions remains.)

Only 3 positions remain for F, G and H.
Number of options for F = 3. (Any of the 3 remaining positions.)
Number of options for G = 2. (Either of the 2 remaining positions.)
Number of options for H = 1. (Only 1 position remains.)

To combine the options above, we multiply:
8*7*3*2*1*3*2*1 = 2016.

The correct answer is D.

[eagleeye]

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?

1,008

1,296

1,512

2,016

2,268

I see many times these types of questions appear on GMAT mock tests.
So anybody can propose a generic approach to these type of questions.
OA after some discussion
Thanks
Sid

We know that ABC have to go before FGH. D and E can go unrestricted. Start off with D and E.

D can go in any of 8 places. No. of ways = 8
Once D is placed, E can go in any of 7 places. No. of ways = 7.

Now we have something like this
X X D X Y Y E Y.

Now ABC can go into any of X places and FGH can go into any of the Y places.
Among ABC, We have 3 options for the first X, 2 for the second X, and 1 for the last X.
Hence ABC between themselves can be arranged in 3*2*1 = 3! ways.
Similarly for the Y places, FGH can be filled in 3*2*! = 3! ways.

Total ways = 8*7*3!*3! = 56*36 = 2016 ways.

[Anurag@Gurome]

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?

1,008

1,296

1,512

2,016

2,268

I see many times these types of questions appear on GMAT mock tests.
So anybody can propose a generic approach to these type of questions.
OA after some discussion
Thanks
Sid

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 = 2,016

The correct answer is D.

[armand_h]

so if we have 8 letters A, B, C, D, E, F, G and H, how many permutations do we have, where ABC is before DEF.

Step 1: compute the total number of permutations: 8!
Step 2:
Now we need to know the number of permutations where ABC is before DEF.
There will be an identical number of permutations where ABD is before CEF.
There will be an identical number of permutations where ABE is before CDF and so on...
Hence step 2 is to compute how many groups of 3 letters we can make from ABCDEF, this is 6C3=6*5*4/3*2=20

Step 3:
So to summarise we have 8! permutations and 20 groups of 3. Each group of 3 is X times before the remaining 3 letters and we're after the value of X
Result is [spoiler]X=8!/20=8!/20=2,016[/spoiler]


[/spoiler]

[mehaksal]

m sorry m nt able to undrstnd how we are considering ABC before FGH case in eagleeye's solution???

[mcdesty]

m sorry m nt able to undrstnd how we are considering ABC before FGH case in eagleeye's solution???

We have 8 spots _ _ _ _ _ _ _ _
We have eight people ABCDEFG
We know ABC has(or they have(grammar nerds)) to come before FGH (We know this because the question says so)
D and E are free to sit wherever..
I don't know about you but I think it would be easier to seat D and E first..
We have 8 spots so whoever we choose first D or E,can sit in any of the 8 spots.
We have one person (D or E) left to seat and he/she can occupy any of the remaining 7 spots
Next we have to seat A B and C. We know they must come before FGH(Again the question told us so)
But FGH will occupy three spots eventually(at the back)
So ABC can occupy only three spots...
A can go into any of the 3
B can go into the 2 remaining
C can go into the last spot (Who you choose first, A B or C does not matter)
Follow the same logic for FGH then go back and reread Eagleeye's solution to fully understand...
Try 8 or 9 more problems like this one to whether you became an expert at arranging things(: