Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 27 Mar 2015, 21:18

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:
Director
Joined: 05 Jan 2005
Posts: 561
Followers: 2

Kudos [?]: 12 [0], given: 0

Out of seven models, all of different heights, five models [#permalink]  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?
Director
Joined: 18 Feb 2005
Posts: 674
Followers: 1

Kudos [?]: 2 [0], given: 0

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
VP
Joined: 30 Sep 2004
Posts: 1490
Location: Germany
Followers: 4

Kudos [?]: 80 [0], given: 0

5c5 + 6c5 + 6c5 + 4c2 = 17
Director
Joined: 18 Feb 2005
Posts: 674
Followers: 1

Kudos [?]: 2 [0], given: 0

jpv, baner, anand,Paul and rest please take a plunge into this problem.
Intern
Joined: 19 Jul 2004
Posts: 43
Followers: 1

Kudos [?]: 3 [0], given: 0

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!)

ketan
Director
Joined: 18 Feb 2005
Posts: 674
Followers: 1

Kudos [?]: 2 [0], given: 0

Ketanm:

Shouldnt you be considering the 5 being selected from the 7 models?
VP
Joined: 30 Sep 2004
Posts: 1490
Location: Germany
Followers: 4

Kudos [?]: 80 [0], given: 0

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
Senior Manager
Joined: 25 Oct 2004
Posts: 250
Followers: 1

Kudos [?]: 6 [0], given: 0

7c5-3c3

Any inputs???

Whats the OA?
VP
Joined: 30 Sep 2004
Posts: 1490
Location: Germany
Followers: 4

Kudos [?]: 80 [0], given: 0

swath20 wrote:
7c5-3c3

Any inputs???

Whats the OA?

7c5 is clear ! but why do you chose 3c3 ?
Senior Manager
Joined: 25 Oct 2004
Posts: 250
Followers: 1

Kudos [?]: 6 [0], given: 0

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.
VP
Joined: 30 Sep 2004
Posts: 1490
Location: Germany
Followers: 4

Kudos [?]: 80 [0], given: 0

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.
Senior Manager
Joined: 25 Oct 2004
Posts: 250
Followers: 1

Kudos [?]: 6 [0], given: 0

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.
SVP
Joined: 03 Jan 2005
Posts: 2251
Followers: 13

Kudos [?]: 213 [0], given: 0

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)
Manager
Joined: 01 Jan 2005
Posts: 167
Location: NJ
Followers: 1

Kudos [?]: 0 [0], given: 0

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.
Manager
Joined: 15 Feb 2005
Posts: 247
Location: Rockville
Followers: 1

Kudos [?]: 6 [0], given: 0

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
Similar topics Replies Last post
Similar
Topics:
29 Out of seven models, all of different heights, five models 18 27 Oct 2007, 11:34
Out of seven models, all of different heights, five models 2 06 Apr 2006, 19:07
Out of seven models, all of different heights, five models 7 02 Apr 2006, 17:49
Out of seven models,all of different heights,five models 9 13 Mar 2006, 11:43
Out of seven models, all of different heights, five models 6 21 Apr 2005, 22:46
Display posts from previous: Sort by