Out of seven models, all of different heights, five models : Quant Question Archive [LOCKED]
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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]

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25 Nov 2004, 14:18
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?
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25 Nov 2004, 18:08
Hmmm, well, maybe this is the 2nd tough combination problem from PR.

Let's arrange people from shortest to tallest:
A-B-C-D-E-F-G
4th and 6th tallest persons are D and F in red
I proceeded with favorable outcome approach

1- If D and F are in the group of 5, then E has to be there so that D and F be separate as well as the group be arranged from shortest to tallest. Hence, if we have D, E and F in the group of 5, we have 4C2 ways of selecting the 2 remaining people{A,B,C,G} to form the group of 5. 4C2 = 6

2- If only D is in the group of 5 and F is to be excluded, we have 5C4 ways of selecting 4 people from the group made up of A,B,C,E or G. Remember that F is already removed and should not be among the sample group. 5C4 = 5

3- If only F is in the group of 5 and D is to be excluded, it is the same process as in step 2 above. 5C4 = 5

Answer is sum of all those: 6+5+5 = 16
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Paul

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PR test problem:permuts & combins. [#permalink]

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26 Nov 2004, 02:46
I get 17.

Total possibilities: 7C5 = 21

Forbidden possibilities: those where D and F are in the selected group with the exclusion of E: we take D and F we have A, B, C and G to choose 3: 4C3=4

21-4=17

The problem is that Iâ€™ve read Paulâ€™s post and his reasoning seems fine. I canâ€™t see the flaw in any of the two approaches, please help.
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26 Nov 2004, 05:08
artabro, you are right. I forgot to count possibility when D and F are not part of the group of 5. 5C5 = 1 way of choosing 5 people among {A,B,C,E,G} to form a group where D and F are not together.
Answer is then 16+1 = 17

I think when I did it originally, I proceeded with unfavorable outcome approach like you did. My method is just too long and more prone to mistakes
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Paul

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26 Nov 2004, 15:06
Thanks for those great explanations.
26 Nov 2004, 15:06
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