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

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CEO
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Out of seven models, all of different heights, five models [#permalink] New post 23 Sep 2003, 13:55
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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?
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Answer [#permalink] New post 23 Sep 2003, 16:36
Ways to pick 5 models from 7
7C5 = 21

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

21-4 = 17 ways
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 [#permalink] New post 23 Sep 2003, 18:31
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


What's the official answer.
Senior Manager
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 [#permalink] New post 24 Sep 2003, 00:17
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
praet whats official answer...
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 [#permalink] New post 24 Sep 2003, 04:17
Vicky,
How does your solution account for the arrangement of models from shortest to tallest ? I guess there is something we are missing here :(
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 [#permalink] New post 24 Sep 2003, 08:43
Exy has the best possible solution.

17 is the correct answer

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
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Explination [#permalink] New post 24 Sep 2003, 15:08
The 4 ways in which 4 & 6 are adjacent are:
12346
23467
13467
12346
Senior Manager
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 [#permalink] New post 25 Sep 2003, 04:33
oh man.. my brain is working too much....
thanks praet...
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Re: Explination [#permalink] New post 25 Sep 2003, 04:42
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?
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 [#permalink] New post 25 Sep 2003, 09:20
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 !
  [#permalink] 25 Sep 2003, 09:20
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