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In how many different orders can the people Alice, Benjamin,
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07 Jan 2014, 04:54
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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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In how many different orders can the people Alice, Benjamin,
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07 Jan 2014, 05:17
sanjoo 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?
A. 1,008 B. 1,296 C. 1,512 D. 2,016 E. 2,268 We want A, B, and C to be before F, G and H. Let's deal with this constraint first: consider A, B, and C as one unit {ABC} and F, G and H also as one unit {FGH}. According to the condition we need them to be arranged as {ABC}{FGH}. A, B, and C, within their unit can be arranged in 3! ways, similarly F, G and H , within their unit can also be arranged in 3! ways. So, we have a line with 6 people XXXXXX. D can take any place in the line so D has 7 options: *X*X*X*X*X*X*. Now, we have a line with 7 people and E can take any place there, so E has 8 options: *X*X*X*X*X*X*X*. Total = 3!*3!*7*8 = 2,016. Answer: D. Or: Place D and E first: 8 options for D, 7 options for E. We are left with 6 empty places. A, B and C must take first 3 places and F, G, H must take the remaining three. A, B and C can be arranged in their places in 3! ways. The same for F, G, and H. Total = 8*7*3!*3! = 2,016. Answer: D.
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Re: In how many different orders can the people Alice, Benjamin,
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07 Jan 2014, 05:54
sanjoo 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?
A. 1,008 B. 1,296 C. 1,512 D. 2,016 E. 2,268 first d,e can come at any point,=>8c2 * 2 next all three a,b,c should come before f,g,h so take all a,b,c as one unit and f,g,h as one unit. each of which can be written in 3! ways, hence total no of ways of arrangement are 8c2*2*3!*3!=2016



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Re: In how many different orders can the people Alice, Benjamin,
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07 Jan 2014, 07:43
Thanks bharat and Bunuel.. Bt m not geting that 3! for abc and 3! fgh .. how can we say that abc will come before fgh?? Above both solution shows that abc will sit togather and fgh will sit togather.. d and will sit anywhere.. I think above solution is also for this question too.. Q: In how many different ways can eight people a,b,c,d,e,f,g and h, can be standing on line that abc and fgh will always stand togather? If i m wrong correct me
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In how many different orders can the people Alice, Benjamin,
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07 Jan 2014, 07:51
sanjoo wrote: Thanks bharat and Bunuel.. Bt m not geting that 3! for abc and 3! fgh .. how can we say that abc will come before fgh?? Above both solution shows that abc will sit togather and fgh will sit togather.. d and will sit anywhere.. I think above solution is also for this question too.. Q: In how many different ways can eight people a,b,c,d,e,f,g and h, can be standing on line that abc and fgh will always stand togather? If i m wrong correct me In the first approach {ABC} is before {FGH} because we consider only {ABC}{FGH} ordering and not {FGH}{ABC}. In the second approach A, B and C take first 3 places and F, G, H take the remaining 3. So, here A, B and C are also before F, G, and H. Hope it's clear.
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Re: In how many different orders can the people Alice, Benjamin,
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07 Jan 2014, 09:42
sanjoo wrote: Thanks bharat and Bunuel.. Bt m not geting that 3! for abc and 3! fgh .. how can we say that abc will come before fgh?? Above both solution shows that abc will sit togather and fgh will sit togather.. d and will sit anywhere.. I think above solution is also for this question too.. Q: In how many different ways can eight people a,b,c,d,e,f,g and h, can be standing on line that abc and fgh will always stand togather? If i m wrong correct me suppose number the seats as below 1 2 3 4 5 6 7 8 it is given that all three abc are before all three fgh, we first choose 2 seats randomly for d,e if we choose 1,4 for d,e.. 2,3,5 should be filled by abc which are of 3!ways and 6,7,8 should be filled with fgh in 3! ways hence a,b,c and f,g,h are not together but seems together because of the formula



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Re: In how many different orders can the people Alice, Benjamin,
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07 Jan 2014, 11:07
Ok Bharat And Bunuel.. i got that.. Thanks alot
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Re: In how many different orders can the people Alice, Benjamin,
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22 Jul 2014, 16:57
What is wrong with my methodology below:
8 spots in line: _ _ _ _ _ _ _ _
ABC must be before DEF so possibilities are:
1) _ _ _ D E F _ _ For this option, ABC must occupy first three spots, and 3!*3!*2! = 72 ways 2) _ _ _ _ D E F _ For this option, ABC must occupy 3 of the first 4 slots, and D and E can be wherever, so 4!*2!*3! = 288 ways 3) _ _ _ _ _ D E F For this option, the remaining 5 people can be arranged in 5! ways = 720 ways
For a total of 1080 ways. Obviously this is wrong but I can't see why...



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In how many different orders can the people Alice, Benjamin,
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23 Jul 2014, 01:43
vaj18psu wrote: What is wrong with my methodology below:
8 spots in line: _ _ _ _ _ _ _ _
ABC must be before DEF so possibilities are:
1) _ _ _ D E F _ _ For this option, ABC must occupy first three spots, and 3!*3!*2! = 72 ways 2) _ _ _ _ D E F _ For this option, ABC must occupy 3 of the first 4 slots, and D and E can be wherever, so 4!*2!*3! = 288 ways 3) _ _ _ _ _ D E F For this option, the remaining 5 people can be arranged in 5! ways = 720 ways
For a total of 1080 ways. Obviously this is wrong but I can't see why... The problem with your solution is that D, E and F are not necessary to be adjacent. G and/or H could be between them. This will increase the number of arrangements for each case.
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In how many different orders can the people Alice, Benjamin,
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09 Jan 2016, 11:36
well, to be honest, I was lucky to pick the right answer.... i started this way: we have few options ABC D E FGH = 3!*5C1*4C1*3! = 6*5*4*6 = 720 D ABC E FGH = 5C1*3!*4C1*3! = 5*6*4*6 = 720 E ABC D FGH = 5C1*3!*4C1*3! = 5*6*4*6 = 720 D E ABC FGH = 5C1*4C1*3!*3! = 5*4*6*6 = 720. i get 2880. way to much. suppose we remove one of 720, we get 2160  still no match. where I did the mistake? ok, now I see. it might be ABC FGH D E or ABC D FGH E etc. or might be that ABC or FGH not one next to other... tricky question.



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Re: In how many different orders can the people Alice, Benjamin,
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26 Jan 2016, 00:02
Bunuel wrote: sanjoo 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?
A. 1,008 B. 1,296 C. 1,512 D. 2,016 E. 2,268 We want A, B, and C to be before F, G and H. Let's deal with this constraint first: consider A, B, and C as one unit {ABC} and F, G and H also as one unit {FGH}. According to the condition we need them to be arranged as {ABC}{FGH}. A, B, and C, within their unit can be arranged in 3! ways, similarly F, G and H , within their unit can also be arranged in 3! ways. So, we have a line with 6 people XXXXXX. D can take any place in the line so D has 7 options: *X*X*X*X*X*X*. Now, we have a line with 7 people and E can take any place there, so E has 8 options: *X*X*X*X*X*X*X*. Total = 3!*3!*7*8 = 2,016. Answer: D. Or: Place D and E first: 8 options for D, 7 options for E. We are left with 6 empty places. A, B and C must take first 3 places and F, G, H must take the remaining three. A, B and C can be arranged in their places in 3! ways. The same for F, G, and H. Total = 8*7*3!*3! = 2,016. Answer: D. Hi Brunel.. I am getting different ans.please tell me what is wrong with my approach 1)there are 8 spot.A,B,C,D,E AS ONE Subset and they are coming before F,G,H so can be arranged in 5! now considering F,G,H As one subset we can arrange them in 3! way..=so total is 5!*3! 2)another way 8 people can sit when ABC comes first and remaining 5 people can be in other subset so in total we can arrange them in 3!*5! so toal ways=A+B=720+720



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Re: In how many different orders can the people Alice, Benjamin,
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18 May 2016, 19:35
sanjoo 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?
A. 1,008 B. 1,296 C. 1,512 D. 2,016 E. 2,268 In response to a PM.... To answer these Qs, we have to find how many of the ways would be invalid...there are 6 people  A,B,C,F,H and G ... they can be arranged in 6! ways... BUT among them ONLY valid solutions are when B,C,A, are ahead of F,G,H..so they can be arranged in such scenario in 3!*3! ways.... so the fraction of TOTAL that will be VALID is 3!*3!/6! = 20....so answer = 8!/20 = 2016 D
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Re: In how many different orders can the people Alice, Benjamin,
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09 Mar 2018, 14:47
Hi All, This is a quirky permutation question. I think that the intent of the prompt is that we have a line of 8 people; from beginningtoend of that line, A, B and C must appear before we see F, G and H. This does NOT mean that A has to be first in line though. Here's how the math "works"… D and E are "floaters"  they can technically appear anywhere  so let's deal with them first: D can be in any of the 8 spots. Once we put D in a spot…. E can be in any of the 7 remaining spots… (8)(7) = 56 Now let's deal with the first group of 3: A, B and C. These three people can appear in any order, as long as they ALL appear before F, G and H. We can arrange A, B and C in 6 different ways: ABC ACB BAC BCA CAB CBA The same can be said for F, G and H: 6 different ways. (56)(6)(6) = 2016 If this were an official GMAT question, then the answers would likely be in ascending order. As you've listed them though… Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: In how many different orders can the people Alice, Benjamin,
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28 Mar 2018, 21:16
sanjoo 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?
A. 1,008 B. 1,296 C. 1,512 D. 2,016 E. 2,268 Leftmost arrangement of the non constraint element is DEABCFGH Using formula, the number of permutations is 2!*3!*3!*(7+6+5+4+3+2+1)=72*28=2016 For explanation of the formula see link below
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Re: In how many different orders can the people Alice, Benjamin, &nbs
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