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Out of seven models, all of different heights, five models

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Out of seven models, all of different heights, five models [#permalink] New post 04 Mar 2005, 14:59
Out of seven models, all of different heights, five models will be chosen to pose for a photograph. If the five models are to stand in a line from shortest to tallest, and the fourth-tallest and sixth-tallest models cannot be adjacent, how many different arrangements of five models are possible?


Can someone explain how we get the second part that will be subtracted from C(7,5)? Yes, i know the possible arrangements can be listed manually, but what is the shorter and more accurate path, using combinations?
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 [#permalink] New post 04 Mar 2005, 16:25
2394 ways

7C5*5! - 7C5*3!*2!

total - 4th and 6th together...


Assumtion is both 4th and 6th tallest are selected in the 5 models chosen
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 [#permalink] New post 05 Mar 2005, 05:56
5c5 + 6c5 + 6c5 + 4c2 = 17
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 [#permalink] New post 05 Mar 2005, 07:04
jpv, baner, anand,Paul and rest please take a plunge into this problem.
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 [#permalink] New post 05 Mar 2005, 09:14
My take on the problem. The five models out of seven can be choose such that

1) The 4th tallest and 6th tallest are present in the group of five
2) The 4th tallest is present but not the 6th tallest
3) The 6th tallest is present but not the 4th tallest
4) The 4th tallest and the 6th tallest are not present

Case 1:

The other three models can be choose out of remaining five models in 5C3 ways. Once the five models are choose they can be arranged as following

Consider 4th and 6th together as one group, hence total of 4 models (assuming 4th and 6th together as a single model) in 4! ways and these two models can internally be arranged in 2! ways.

Hence the total ways in which the models can be arranged with 4th and 6th always together is 4!*2!

Without considering any specific constraint total arrnagements are 5!. Hence required arrangements where the 4th and 6th models are never adjacent is

[5! - (2!*4!)]

Case 2:

Once the 4th model is choose, and 6th is not to be choosen, the remaining 4 models can be choosen out of 5 models in 5C4 = 5 ways. These five models can be arrnaged in 5! ways (no constraints on these). Hence total combinations = 5 * 5! ways

Case 3: This is similar to case 2 and hence total ways = 5*5!

Case 4: If 4th and 6th models are excluded then the remaining 5 have to be choosen and they can be arranged in 5! ways.

considering all cases the total number of arrangements are


[5! - (2!*4!)] + 5 * 5! + 5 * 5! + 5!

= (12 * 5!) - (2! * 4!)

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 [#permalink] New post 05 Mar 2005, 09:42
Ketanm:

Shouldnt you be considering the 5 being selected from the 7 models?
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 [#permalink] New post 06 Mar 2005, 13:26
christoph wrote:
5c5 + 6c5 + 6c5 + 4c2 = 17


when you consider the 4th and 6th tallest model to be in the groupe, then the 5th tallest has to be in the group as well, because the models has to stand from short to tallest:

_ _ 4 5 6 or _ 4 5 6 _ => 2 vacant positions

...that means you pick 2 from the remaining 4 => 4c2
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 [#permalink] New post 06 Mar 2005, 14:21
My answer would be
7c5-3c3

Any inputs???

Whats the OA?
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 [#permalink] New post 06 Mar 2005, 14:28
swath20 wrote:
My answer would be
7c5-3c3

Any inputs???

Whats the OA?


7c5 is clear ! but why do you chose 3c3 ?
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 [#permalink] New post 06 Mar 2005, 14:35
consider the case where 4 th and 6th models are always adjacent. We are given that the models are arranged as per their height and if we need the 4th and the 6th models to be adjacent that means that the 5th model cannot be chosen. Automatically the 5th model is ruled out. So what remains is a selection from models 1,2and 3 hence 3c3

Hope this explains....

Would love to know the OA though....

Last edited by swath20 on 06 Mar 2005, 15:04, edited 1 time in total.
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 [#permalink] New post 06 Mar 2005, 14:57
swath20 wrote:
consider the case where 4 th and gth models are always chosen. We are given that the models are arranged as per their height and if we need the 4th and the 6th models together that means that 5th cannot be chosen. Automatically the 5th model is ruled out. So what remains is a selection from models 1,2and 3 hence 3c3

Hope this explains....

Would love to know the OA though....


as i mentioned above: the order has to be from shortest to tallest and 4 and 6 cannot stand together. when we choose 4 and 6, 5 has to be between them. so 3 are already chosen and the remaining 2 spaces can be chose from the remaining 4.
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 [#permalink] New post 06 Mar 2005, 15:14
Christopher, I am sorry but I don't understand why u are saying the 5th model has to chosen. if the 4th and 6th are to be adjacent then 5th cannot be chosen because if the 5th model is chosen the order would be x,x,4th, 5th, 6th.

What I was trying to do was, from the total no of cases I was reducing the cases where 4th and 6th are adjacent, so it necessarily has to be 1,2,3,4 and 6

No of times when 4th and 6th are not together =Total no cases without restrictions - when models 4th and 6th are adjacent. ( And mind you this will happen only when the 5th model is not chosen) 1,2,3,4 and 6
=7c5-3c3

I hope I am not missing something.
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 [#permalink] New post 06 Mar 2005, 20:46
swath20 has the right approach, except he made a small typo. We need to take out the case where 4th and 6th model are adjacent from C(7,5). When these two people are adjacent, that means they both are selected, and 5th is not selected. We need to select three more people to make it five, from the remaining four people.

So it would be C(7,5)-C(4,3)
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 [#permalink] New post 07 Mar 2005, 02:06
The no of possibilities = total ways - when 4th n 6th are adjacent
c(7,5) - c(4,3)

as 4th and 6th are already there and 5th is not to be selected for them to be adjacent.
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 [#permalink] New post 07 Mar 2005, 09:25
agree with Honghu
7C5-1*4C3
unlike other problems in this one there would be only one way to place the 4th and 6th model as per the order required.
Hence, 4C3*1
Correct me if my reasoning seems wrong please
  [#permalink] 07 Mar 2005, 09:25
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Out of seven models, all of different heights, five models

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