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# Here is one more that I got on my practice test: Out of

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Intern
Joined: 04 Oct 2004
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Here is one more that I got on my practice test: Out of [#permalink]  10 Oct 2004, 14:39
Here is one more that I got on my practice test:

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?

a) 6
b) 11
c) 17
d) 72
e) 210

-Irene
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Joined: 10 Oct 2004
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Re: PR Permutations [#permalink]  10 Oct 2004, 15:04
if u have to line them in increasing height and 4th and 6th tallest are there in the 5 selected from 7,u have to have them together

[quote="iostrovsky"]Here is one more that I got on my practice test:

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?

a) 6
b) 11
c) 17
d) 72
e) 210

-Irene[/quote]
Intern
Joined: 04 Oct 2004
Posts: 12
Followers: 0

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

Quote:
if u have to line them in increasing height and 4th and 6th tallest are there in the 5 selected from 7,u have to have them together

No you don't have to, just go to the next one.
Manager
Joined: 18 Sep 2004
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Location: Dallas, TX
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Ans C 17.

Find the possible combinations that would satisfy that 4th and 6th tallest not adjacent in line:

4th Tallest Not in Line = 6C5 = 6
6th Tallest Not in Line = 6C5 = 6
4th, 5th, 6th all in Line = 4C2 = 6

Subtract Duplicate
4th and 6th tallest not in line = 5 C 5 = 1

6+6+6-1 = 17

There's probably a easier way but this is what I got
Manager
Joined: 24 Aug 2004
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4 possible ways in which the condition can be met:

1. 4th and 6th tallest both are not selectied: 5C5 = 1
2. 4th is selected but not 6th : 5C4 = 5
3. 6th is selected but not 4th : 5C4 = 5
4. 4th, 5th and 6th all 3 are selected: 4C2 = 6

Summing up its 17.
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