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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 13 Mar 2006, 11:43
Out of seven models,all of different heights,five models will be chosen to pose for photograph. If the five models are
to stand in a line from shortest to tallest, and the fourth-tallest and the sixth-tallest models cannot be adjacent,how many different
arrangements of five models are possible

6
11
17
72
210
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 [#permalink] New post 13 Mar 2006, 12:36
17.

Total arrangements =7C5 = 21.
Constraint is that if 4 and 6 are picked, 5 needs to be picked. Total number of such combinations are 1*1*4C3 = 4.

Total - Constraints = 21-4 = 17.
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 [#permalink] New post 13 Mar 2006, 13:02
lhotseface,
can you explain
"combinations are 1*1*4C3 = 4. "??
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 [#permalink] New post 13 Mar 2006, 14:58
lhotesface, can you please explain the solution? I think it should be 4c2 and not 4c3 in your solution.
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 [#permalink] New post 13 Mar 2006, 15:23
4C3 since if 4 and 6 are present , 5 CANNOT be present (so that 4 and 6 ARE lined up together).

Now we have to choose the 3 models out of the remaining 4 ( 4,6 selected and 5 not available for consideration ).
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 [#permalink] New post 13 Mar 2006, 17:32
Thanks lhotesface.
Total Number of possibilities = Number of ways of selecting any 5 models
- Number of ways of selecting 4th and 6th model such that they are together (i.e., not selecting 5)
Number of ways of selecting any 5 models = 7C5 = 21
Number of ways of selecting 4th and 6th model such that they are together (i.e., not selecting 5) = 4C3
In 4C3,
4 = We already selected 4,6 and not selected 5 => we are left with 4
3 = We already selected 4,6 and hence we should select 3 more.
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 [#permalink] New post 13 Mar 2006, 19:03
# of ways to select 5 models from 7 = 7C5 = 21

We must pick the 5th tallest if the 4th tallest and 6th tallest are picked so that 4th tallest person will not be standing adjacent to 6th tallest.

We want to subtract from total number of 21 groupings, so we need to calculate # of ways to pick 4th tallest and 6th tallest without picking 5 = 4C3 = 4

Total number of ways = 21 - 4 = 17
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 [#permalink] New post 14 Mar 2006, 06:16
The OA is 17 . Thanks a lot for the explanation .
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 [#permalink] New post 15 Mar 2006, 23:38
I am not very good at probabitliuy. Can someone explain to me in detail how they get the 4 exclusion?

Can someone explain in detail how the answer is arrived?
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 [#permalink] New post 16 Mar 2006, 08:50
elektraa wrote:
I am not very good at probabitliuy. Can someone explain to me in detail how they get the 4 exclusion?

Can someone explain in detail how the answer is arrived?


Models: (Shortest) 1,2,3,4,5,6,7 (Tallest)

First collect all possible combinations of 5 models from 7 models.
7C5 = 21. (Any 5 selected can be made to stand in order of height)

Now it is required that 4th & 6th tallest model should not be together.
So, find out combnation of 5 models in which "5" is not there, because only when "5" is not there, 4 & 6 will be together, got it?

Let's see, how many combinations of 5 models are there (in 7C5) such that 4 & 6 is present but 5 is not present.
Consider 4 & 6 selected (i.e. 2 models are selected & remaining 3 are to be selected), so we are left with 1,2,3,5,7. We don't want "5", so we are now left with 1,2,3,7.

Now, in how many ways can we select remaining 3 models from these 4 models? 4C3 = 4.

Hence, total = 7C5 -4 = 21-4 = 17
  [#permalink] 16 Mar 2006, 08:50
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