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

 It is currently 20 Apr 2015, 22:40

### 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:
VP
Joined: 25 Nov 2004
Posts: 1495
Followers: 6

Kudos [?]: 38 [0], given: 0

Out of seven models, all of different heights, five models [#permalink]  21 Apr 2005, 22:46
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
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?
Manager
Joined: 05 Feb 2005
Posts: 116
Location: San Jose
Followers: 1

Kudos [?]: 1 [0], given: 0

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?

Total = A + B + C + D

where A = Number of arrangements that include 4th but not 6th
B = Number of arrangements that include 6th but not 4th
C = Number of arrangements that include both 4th and 6th
D = Number of arrangements that include neither

A = 5C4 x 1 = 5
B = 5C4 x 1 = 5
C = 5C3 x 1 = 10
D = 5C5 x 1 = 1

Total = 21

Lemme explain one term:
A = 5C4 x 1

5C4 = number of ways of choosing the members of the arrangement.
1 = Number of ways of arranging the 5 chosen models

In A, one member(4th) is already picked and one member(6th) is dropped. We just need to choose 4 more members from the remaining 5. Hence number of ways of choosing members for A is 5C4.

There is only 1 way to arrange the 5 chosen models.
_________________

Anyone who has never made a mistake has never tried anything new. -Albert Einstein.

Intern
Joined: 22 Apr 2005
Posts: 14
Followers: 0

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

Re: PS: Permutation and Combination [#permalink]  22 Apr 2005, 17:05
MA wrote:
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?

Is this an example of a question-type that could be on a real GMAT?

The GRE did not test for permutations and combinations...
VP
Joined: 25 Nov 2004
Posts: 1495
Followers: 6

Kudos [?]: 38 [0], given: 0

Re: PS: Permutation and Combination [#permalink]  22 Apr 2005, 17:16
kenmore3233 wrote:
Is this an example of a question-type that could be on a real GMAT? The GRE did not test for permutations and combinations...

yes, GRE math, i believe, is pretty easy but GMAT math is pretty difficult.
Manager
Joined: 13 Oct 2004
Posts: 239
Followers: 1

Kudos [?]: 7 [0], given: 0

Total possible arrangements = 7c5 = 21
Arrangements where 4 and 6 are adjacent implies that 5 is not picked and 4 and 6 are picked = 1. 4c3 = 4

Arrangements where 4 and 6 not adjacent = 21 - 4 = 17.

If the answer is 21, then all possible arrangements would have 4 and 6 non-adjacency?.
Manager
Joined: 05 Feb 2005
Posts: 116
Location: San Jose
Followers: 1

Kudos [?]: 1 [0], given: 0

Oops. You are right. My term C is incorrect.

C = 4C2 x 1 = 6

A = 5C4 x 1 = 5
B = 5C4 x 1 = 5
!!!! wrong C = 5C3 x 1 = 10
D = 5C5 x 1 = 1

!!!!wrong Total = 21

Total = 17.

Yes, prep_gmat you are right. And your method is elegant.
_________________

Anyone who has never made a mistake has never tried anything new. -Albert Einstein.

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

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

prep_gmat wrote:
Total possible arrangements = 7c5 = 21
Arrangements where 4 and 6 are adjacent implies that 5 is not picked and 4 and 6 are picked = 1. 4c3 = 4

Arrangements where 4 and 6 not adjacent = 21 - 4 = 17.

If the answer is 21, then all possible arrangements would have 4 and 6 non-adjacency?.

Can you explain this step please ?

Arrangements where 4 and 6 are adjacent implies that 5 is not picked and 4 and 6 are picked = 1. 4c3 = 4

Thanks
Similar topics Replies Last post
Similar
Topics:
Out of seven models, all of different heights, five models 5 11 Jun 2006, 12:23
Out of seven models, all of different heights, five models 3 16 Apr 2006, 06:00
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
Display posts from previous: Sort by