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Manager
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Out of seven models, all of different heights, five models [#permalink]
 11 Jun 2006, 13:23
Out of seven models, all of different heights, five models will be chosen to pose for a photograph. If the five modesl are to stand in a line from shortest to tallest, and the fourth-tallets and the 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
pls explain....
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Re: Arrangements of cars [#permalink]
11 Jun 2006, 17:40
acfuture wrote: Out of seven models, all of different heights, five models will be chosen to pose for a photograph. If the five modesl are to stand in a line from shortest to tallest, and the fourth-tallets and the 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
pls explain....
is it C , 17
if its right , would post explanation.
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Manager
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C is right
If we only have to choose 5 from 7 so we got 7C5=21
But we have to exculde some ways in which 4 and 6 is adjacent
If we have 4&6 adjacent so we can't have 5. This way, along with 4 and 6, we must collect 3more models from the rest of 4 models. It means we got 4C3=4ways
And the answer must be 21-4=17
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C = 17
Number of ways 5 models can be chosen from 7 models = 7C5 = 21
Number of ways 4 and 6 are together
12346
13467
12467
23467
that is 4 ways
Hence total = 21-4 = 17
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C
Total number of ways 5 models can be chosen from 7 models = 7C5 = 21
4 and 6 are only adjacent when 5 is not included in the group. Therefore, to calculate the number of ways to choose five people when 5 is not included = 6C5 = 6.
However, we have to subtract 2 because there is a group that does not include 5, but also does not include both 4 and 6....(1, 2, 3, 4, 7) or (1, 2, 3, 6, 7).
Therefore 21 - (6 - 2) = 17
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OA is C
thanks everyone...
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