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math Q..pls answer

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by Inspired » Mon Jan 18, 2010 12:27 pm
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?

6
11
17
72
210

Pls answer with explanation.
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Source: — Problem Solving |

by rohan_vus » Mon Jan 18, 2010 4:35 pm
Should be 17

7C5 - 4C3 = 17 --->Total Outcomes - Outcomes involving 4th and 6th adjacent to each other

7C5 --> is way of arranging from shortest to tallest with and without 4th and 6th tallest adjacent to each other
4C3 --> is way of arranging from shortest to tallest with 4th and 6th tallest adjacent to each other..Here's the tricky part--> Here Out of 7 , you aleady have 4th and 6th , so the remaining 3 should be selected from 1st tallest , 2nd tallest , 3rd tallest and 7th tallest only.

As 5th tallest cant be used because if 5th tallest is selected then 4th tallest and 6th tallest wont be adjacent so 5th tallest is not to be used .
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by Inspired » Mon Jan 18, 2010 11:49 pm
thank you rohan. that is the answer
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