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by mehaksal » Fri Aug 24, 2012 9:42 am
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
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by neelgandham » Fri Aug 24, 2012 9:44 am
mehaksal,

Please find the solution here - https://www.beatthegmat.com/number-of-wa ... 56018.html
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by cypherskull » Sat Aug 25, 2012 4:28 am
I'll take a shot a this!

Since we have a condition that A, B and C must always stand before F, G and H, there can only be 1 way of placing ABC and FGH on the line as a group. However, each of ABC and FGH can be rearranged within themselves 6 ways:

ABC - 3! ways (=6)---(1)
FGH - 3! ways (=6)---(2)

Moving on, we don't have any restriction on the placement of D and E. So,

D can be placed on either of the 8 positions. ----(3)

Having selected D, E can be given any of the remaining 7 positions.----(4)

Total no. of ways ABCDEFGH can be arranged (given all the conditions)

= (1)*(2)*(3)*(4)
= 6*6*8*7
[spoiler]= 2016 ways (Ans: D).[/spoiler]

Please post the OA.
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by mehaksal » Sun Aug 26, 2012 8:45 pm
arrangement will be _ _ABC_ _FGH_ _
how about this :-

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
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by cypherskull » Sun Aug 26, 2012 9:42 pm
mehaksal wrote:arrangement will be _ _ABC_ _FGH_ _
how about this :-

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


Consider the scenario where D & E are placed on the line first.

D can take any of the 8 positions ----(1)
E can take any of the remaining 7 positions -----(2)

ABC can be placed in 6 ways ------------ (3)
FGH can be placed in 6 ways ------------ (4)

8*7*6*6 = 2016 ways
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
In your reasoning above, Once you've selected ABC and FGH, you'll have 2 positions remaining and not 6. So, the answer would become 6*6*2 = 72 which isn't correct. We have to consider the fact that D and E can take any of the 8 positions without restriction.

That's my take on this. Anyway, I know I'm going to struggle a lot in PnCs considering the fact that I have my gmat in 2 days :D

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by mehaksal » Wed Oct 03, 2012 4:21 am
i m still struggling! the day i gain full confidnc in PnC..i wl hv a grand party!
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by Brent@GMATPrepNow » Wed Oct 03, 2012 6:22 am
mehaksal wrote: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
Label the spaces #1, #2, .... #7, #8

Take the task of arranging all 8 people and break it into stages.

Stage 1: Place David
There are 8 spaces available, so this stage can be accomplished in 8 ways.

Stage 2: Place Elaine
There are 7 spaces remaining, so this stage can be accomplished in 7 ways.

Important: At this point, there are 6 spaces remaining, and Alice, Benjamin, Charlene must occupy the 3 spaces that are ahead of the last 3 spaces. We'll call these the "front spaces"

Stage 3: Place Alice
Alice must occupy one of the 3 front spaces. So this stage can be accomplished in 3 ways.

Stage 4: Place Benjamin
Benjamin must occupy one of the 2 remaining front spaces. So this stage can be accomplished in 2 ways.

Stage 5: Place Charlene
Charlene must occupy the last remaining front space. So this stage can be accomplished in 1 way.

At this point, there are 3 spaces remaining.

Stage 6: Place Frederick
There are 3 spaces remaining. So this stage can be accomplished in 3 ways.

Stage 7: Place Gale
There are 2 spaces remaining. So this stage can be accomplished in 2 ways.

Stage 8: Place Harold
There is 1 space remaining. So this stage can be accomplished in 1 way.

By the Fundamental Counting Principle (FCP) we can complete all 8 stages (and thus arrange all 8 people) in (8)(7)(3)(2)(1)(3)(2)(1) ways ([spoiler]= 2016 ways[/spoiler])

Answer = D

Cheers,
Brent

Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmat-counting?id=775
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by Brent@GMATPrepNow » Wed Oct 03, 2012 6:24 am
In my opinion, the majority of counting questions can be solved using the Fundamental Counting Principle (FCP). If you're interested in practicing this technique, here are a few questions to try:

https://www.beatthegmat.com/permutation-t122873.html
https://www.beatthegmat.com/a-certain-ship-t116519.html
https://www.beatthegmat.com/permutation-t112874.html
https://www.beatthegmat.com/permutation- ... 92779.html
https://www.beatthegmat.com/combinations ... 15538.html
https://www.beatthegmat.com/3-rings-t92252.html

Cheers,
Brent
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