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

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Manager
Joined: 21 Dec 2005
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Out of seven models, all of different heights, five models [#permalink]  02 Apr 2006, 17:49
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?

Thanks
Senior Manager
Joined: 22 Nov 2005
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Here is the explanation.
Total number of ways models can be arranged = 7C5

Total number of ways 4th and 6th models can be chosen together = 4C3
Here we have to exclude 5th modle so only 4 models are left

hence total number of ways = 7C5 - 4C3

=17
Manager
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Thanks

Could you please explain why after choosing five models, we have the sixth model and why is it that its 4C3. What does 4C3 mean in this case..are we choosing 4 out of 3 why so?

Thanks
VP
Joined: 20 Sep 2005
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The constraint is "when we choose 4 and 6, we can't choose 5"...that leaves with choosing 3 models from the remaining 4.

jodeci wrote:
Thanks

Could you please explain why after choosing five models, we have the sixth model and why is it that its 4C3. What does 4C3 mean in this case..are we choosing 4 out of 3 why so?

Thanks
GMAT Club Legend
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Total number of ways to pick 5 models = 7C5 = 7!/5!2! = 21 ways

Constraint: 6th talles and 4th tallest cannot be adjacent to each other:

Number of ways 64 are adjacent to each other:

Can be 64321 --> 1 way
Can be 764XX --> 3C2 ways = 3!/2!1! = 3 ways

Total = 1 + 3 = 4

# of different arrangements = 21-4 = 17
Manager
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Can be 64321 --> 1 way

How can we 3C2 yet it is an arrangement.

I only find 76432, 76421 as the only viable options. I do not know about 764231 Please explain why we have this

Can be 764XX --> 3C2 ways = 3!/2!1! = 3 ways

Thanks
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5068
Location: Singapore
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Kudos [?]: 194 [0], given: 0

jodeci wrote:
Can be 64321 --> 1 way

How can we 3C2 yet it is an arrangement.

I only find 76432, 76421 as the only viable options. I do not know about 764231 Please explain why we have this

Can be 764XX --> 3C2 ways = 3!/2!1! = 3 ways

Thanks

Because the last two places can be taken by any of the 3 remaining models who are shorter than 4. We cannot take 5 as that would mean model 6 and model 4 would not be standing side by side.

Since we have 3 models and only 2 places, the number of ways to pick a pair of 2 models from a pool of 3, is 3C2.

In case you were interested, the 3 ways to do it:

76432
76431
76421
Senior Manager
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I had very long way of doing this thanks for quick explanations ywilfred.
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