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

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Out of seven models, all of different heights, five models [#permalink] New post 21 Apr 2005, 22:46
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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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 [#permalink] New post 21 Apr 2005, 23:06
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.
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Re: PS: Permutation and Combination [#permalink] New post 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...
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Re: PS: Permutation and Combination [#permalink] New post 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.
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 [#permalink] New post 23 Apr 2005, 15:41
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?.
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 [#permalink] New post 23 Apr 2005, 23:05
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.
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 [#permalink] New post 24 Apr 2005, 06:25
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
  [#permalink] 24 Apr 2005, 06:25
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