Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 25 May 2017, 07:07

### 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
CEO
Joined: 15 Aug 2003
Posts: 3454
Followers: 67

Kudos [?]: 874 [0], given: 781

Out of seven models, all of different heights, five models [#permalink]

### Show Tags

23 Sep 2003, 14:55
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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?
Intern
Joined: 03 May 2003
Posts: 28
Followers: 0

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

### Show Tags

23 Sep 2003, 17:36
Ways to pick 5 models from 7
7C5 = 21

Ways in which 4th and 6th are adjacent = 4

21-4 = 17 ways
Intern
Joined: 16 Jul 2003
Posts: 32
Followers: 0

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

### Show Tags

23 Sep 2003, 19:31
I also got 17. But my approach was different.

Case I - ways when neither 4 nor 6 is selected = 5c5.
Case II - ways when either of the 2 ( #4or #6) is selected = 2c1*5c4.
Case III - ways when 4 and 6 are selected (here #5 is a must as 4 & 6
can't be adjacent and models are to arranged shortest to tallest. = 4c2 ( as #4,5,6 are to be selected )

Total ways, when 4 & 6 can't be adjacent and group is arranged from shortest to tallest = I+II+III = 1 + 10 + 6 = 17.

But

Can you pls. explain how did you get

Quote:
Ways in which 4th and 6th are adjacent = 4

Senior Manager
Joined: 22 Aug 2003
Posts: 257
Location: Bangalore
Followers: 1

Kudos [?]: 13 [0], given: 0

### Show Tags

24 Sep 2003, 01:17
First of all we are looking for arrangements not combinations/ways.

Total arrangements of 5 models = 7C5*5!------- a
But, the above arrangements include where 4th and 6the are together.
Number of ways when 4th & 6th are together = 5C3*4!------b
(Select 3 members from remaining 5 - 4th & 6th are already in group - and ways they can be arranged so that 4th & 6th remain togther = 4!)
Thus ways to arrange 5 models out of 7 such that 4 & 6 are not adjacent
= eqa - eqb
= 2280
Intern
Joined: 21 Jul 2003
Posts: 41
Location: India
Followers: 1

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

### Show Tags

24 Sep 2003, 05:17
Vicky,
How does your solution account for the arrangement of models from shortest to tallest ? I guess there is something we are missing here
CEO
Joined: 15 Aug 2003
Posts: 3454
Followers: 67

Kudos [?]: 874 [0], given: 781

### Show Tags

24 Sep 2003, 09:43
Exy has the best possible solution.

Vicks, there is no arrangement issue here..we know we have to arrrange from highest to lowest..
For every combo, we have only ONE way it will work

Thanks
praetorian
Intern
Joined: 03 May 2003
Posts: 28
Followers: 0

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

### Show Tags

24 Sep 2003, 16:08
The 4 ways in which 4 & 6 are adjacent are:
12346
23467
13467
12346
Senior Manager
Joined: 22 Aug 2003
Posts: 257
Location: Bangalore
Followers: 1

Kudos [?]: 13 [0], given: 0

### Show Tags

25 Sep 2003, 05:33
oh man.. my brain is working too much....
thanks praet...
Intern
Joined: 23 Aug 2003
Posts: 19
Location: ny
Followers: 0

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

### Show Tags

25 Sep 2003, 05:42
exy18 wrote:
The 4 ways in which 4 & 6 are adjacent are:
12346
23467
13467
12346

Your first and last orders are the same. What's the 4th way?
Intern
Joined: 16 Jul 2003
Posts: 32
Followers: 0

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

### Show Tags

25 Sep 2003, 10:20
Quote:
Your first and last orders are the same. What's the 4th way?

Exy mistyped the 4th, it should be 12467.

His method is much better & faster than My method !
25 Sep 2003, 10:20
Display posts from previous: Sort by