In how many different orders can the people Alice, Benjamin, Charlene, David, Elaine, Frederick, Gale, and Harold be standing in line if each of Alice, Benjamin, and Charlene must be in line before each of Frederick, Gale, and Harold?
A)1,008
B)1,296
C)1,512
D)2,016
E)2,268
OA D
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Probability Problem
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aditiniyer
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Label the spaces #1, #2, .... #7, #8mehaksal 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
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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Hi aditiniyer,
If you find these types of questions challenging, then it sometimes helps to 'work your way up' to the situation that's defined in the prompt (so that you can understand the logic involved and not make any math mistakes).
Assuming that certain people (A/B/C) had to be in line ahead of certain other people (F/G/H)....
IF... you only had two people (A and H), then you would have (1!)(1!) = 1 possible order: AH
IF... you only had four people (A/B and G/H), then you would have (2!)(2!) = 4 possible orders: ABGH, ABHG, BAGH, BAHG.
IF... you only had six people (A/B/C and F/G/H), then you would have (3!)(3!) = 36 possible orders. While I won't list them all out, here is WHY that occurs...
A, B and C don't have to be arranged in that order - there are 6 possibilities:
ABC
ACB
BAC
BCA
CAB
CBA
Similarly, there are 6 possibilities for F/G/H. Thus, any of the 6 ABC options can be 'paired' with any of the 6 FGH options - (6)(6) = 36 options
Now, if we include a 7th person who can be placed ANYWHERE, then we have to multiply that prior total by 7... (36)(7) = 252 options.
And if we include an 8th person who can be placed ANYWHERE, then we have to multiply that preceding total by 8... (252)(8) = 2016 options.
Final Answer: D
GMAT assassins aren't born, they're made,
Rich
If you find these types of questions challenging, then it sometimes helps to 'work your way up' to the situation that's defined in the prompt (so that you can understand the logic involved and not make any math mistakes).
Assuming that certain people (A/B/C) had to be in line ahead of certain other people (F/G/H)....
IF... you only had two people (A and H), then you would have (1!)(1!) = 1 possible order: AH
IF... you only had four people (A/B and G/H), then you would have (2!)(2!) = 4 possible orders: ABGH, ABHG, BAGH, BAHG.
IF... you only had six people (A/B/C and F/G/H), then you would have (3!)(3!) = 36 possible orders. While I won't list them all out, here is WHY that occurs...
A, B and C don't have to be arranged in that order - there are 6 possibilities:
ABC
ACB
BAC
BCA
CAB
CBA
Similarly, there are 6 possibilities for F/G/H. Thus, any of the 6 ABC options can be 'paired' with any of the 6 FGH options - (6)(6) = 36 options
Now, if we include a 7th person who can be placed ANYWHERE, then we have to multiply that prior total by 7... (36)(7) = 252 options.
And if we include an 8th person who can be placed ANYWHERE, then we have to multiply that preceding total by 8... (252)(8) = 2016 options.
Final Answer: D
GMAT assassins aren't born, they're made,
Rich
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The line has eight spaces. We know that A, B, and C must be somewhere in the first five spaces, and that F, G, and H must be somewhere in the last five.
From there, one approach is using casework. Let's say that X is the member of the FGH group that's furthest ahead in line. We've got three possibilities here:
#1: X is 4th
#2: X is 5th
#3: X is 6th
Now let's compute each.
#1:
If X is 4th, then A,B,C are #1, #2, and #3. We can arrange them in 3! ways. One of FGH is fourth, so there are 3 options there. The other four people come at the end: 4!. In all, this is 3! * 3 * 4!, or 432.
#2:
If X is 5th, then A,B,C are somewhere from one to four, along with either D or E. So we arrange those four people in 4! ways, and have two options (D or E), giving us 2 * 4!. Then we've got one of FGH (3 options), then our other three people (3!). So this is 2 * 4! * 3 * 3!, or 864.
#3:
If X is 6th, we've got 5! for the first five, then 3! for the last three, for 720 arrangements.
Adding all this up, we've got 432 + 864 + 720 = 2016. (What a great problem this would've been last year
)
From there, one approach is using casework. Let's say that X is the member of the FGH group that's furthest ahead in line. We've got three possibilities here:
#1: X is 4th
#2: X is 5th
#3: X is 6th
Now let's compute each.
#1:
If X is 4th, then A,B,C are #1, #2, and #3. We can arrange them in 3! ways. One of FGH is fourth, so there are 3 options there. The other four people come at the end: 4!. In all, this is 3! * 3 * 4!, or 432.
#2:
If X is 5th, then A,B,C are somewhere from one to four, along with either D or E. So we arrange those four people in 4! ways, and have two options (D or E), giving us 2 * 4!. Then we've got one of FGH (3 options), then our other three people (3!). So this is 2 * 4! * 3 * 3!, or 864.
#3:
If X is 6th, we've got 5! for the first five, then 3! for the last three, for 720 arrangements.
Adding all this up, we've got 432 + 864 + 720 = 2016. (What a great problem this would've been last year
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Might be a good idea to change the subject to this thread - this isn't really a probability problem, though combinatorics and probability do often go hand in hand.
Can you explain the part in red. I don't see why we multiply by 7 and 8[email protected] wrote:Hi aditiniyer,
If you find these types of questions challenging, then it sometimes helps to 'work your way up' to the situation that's defined in the prompt (so that you can understand the logic involved and not make any math mistakes).
Assuming that certain people (A/B/C) had to be in line ahead of certain other people (F/G/H)....
IF... you only had two people (A and H), then you would have (1!)(1!) = 1 possible order: AH
IF... you only had four people (A/B and G/H), then you would have (2!)(2!) = 4 possible orders: ABGH, ABHG, BAGH, BAHG.
IF... you only had six people (A/B/C and F/G/H), then you would have (3!)(3!) = 36 possible orders. While I won't list them all out, here is WHY that occurs...
A, B and C don't have to be arranged in that order - there are 6 possibilities:
ABC
ACB
BAC
BCA
CAB
CBA
Similarly, there are 6 possibilities for F/G/H. Thus, any of the 6 ABC options can be 'paired' with any of the 6 FGH options - (6)(6) = 36 options
Now, if we include a 7th person who can be placed ANYWHERE, then we have to multiply that prior total by 7... (36)(7) = 252 options.
And if we include an 8th person who can be placed ANYWHERE, then we have to multiply that preceding total by 8... (252)(8) = 2016 options.
Final Answer: D
GMAT assassins aren't born, they're made,
Rich
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Matt@VeritasPrep
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Suppose you've arranged your six people: ABCDEF. When you go to place G, you have the following empty spaces:hoppycat wrote: Can you explain the part in red. I don't see why we multiply by 7 and 8
_ A _ B _ C _ D _ E _ F _
There are seven _'s, so we've got 7 ways to place that person. If we chose the second one, for instance, our order would become AGBCDEF.
Once you've got seven people, of course, the same logic will give you 8 places to place the eighth person.
Thanks Matt!Matt@VeritasPrep wrote:Suppose you've arranged your six people: ABCDEF. When you go to place G, you have the following empty spaces:hoppycat wrote: Can you explain the part in red. I don't see why we multiply by 7 and 8
_ A _ B _ C _ D _ E _ F _
There are seven _'s, so we've got 7 ways to place that person. If we chose the second one, for instance, our order would become AGBCDEF.
Once you've got seven people, of course, the same logic will give you 8 places to place the eighth person.
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Matt@VeritasPrep
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No probhoppycat wrote:Thanks Matt!Matt@VeritasPrep wrote:Suppose you've arranged your six people: ABCDEF. When you go to place G, you have the following empty spaces:hoppycat wrote: Can you explain the part in red. I don't see why we multiply by 7 and 8
_ A _ B _ C _ D _ E _ F _
There are seven _'s, so we've got 7 ways to place that person. If we chose the second one, for instance, our order would become AGBCDEF.
Once you've got seven people, of course, the same logic will give you 8 places to place the eighth person.
