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Out of seven models, all of different heights, five models [#permalink]
11 Jun 2006, 12:23

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?

Re: Arrangements of cars [#permalink]
11 Jun 2006, 16:40

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?

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

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

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).

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