Models

Total combinations of picking 5 from 7 : 7C5 = 21
However, there're some cases that are not allowed to happen. Such cases are when the 4th and 6th tallest are sitting next to each other. In that case, of the 5 chosen models are the 4th and the 6th. So we need to pick the other three from the remaining 5 models. ( 7 models, the 4th and 6th tallest have been chosen). However, the 5th tallest model must be eliminated from the remaining pool, as the 5th tallest model when picked will prevent the 4th tallest and the 6th tallest from sitting next to each other.
So the remaining pool has only 4 models left : the 1st, the 2nd, the 3rd, the 7th tallest ones.
Total combinations to be eliminated: 4C3 = 4
Total combinations that satisfy the requirement: 21 - 4 = 17
So pick C.

singhsa wrote:Out of 7 models, 5 will be selected for a photo. if the 5 models are to stand in a line from shortest to tallest and if all are of different heights, and if the fourth and sixth tallest models cannot sit side by side, how many different arrangements of 5 models are possible?

6
11
17
72
210

OA - C

For the moment, let's ignore the stipulation that the 4th and 6th models cannot be placed adjacent to each other.

If we choose any 5 models, there will be only 1 way to arrange them: from shortest to tallest. So any combination of 5 gives us 1 possible arrangement.

Number of possible combinations = 7C5 = 21. So there are 21 ways to arrange the models from shortest to tallest.

Since a few of these arrangements will be bad -- those that place the 4th and 6th models next to each other -- we need an answer choice just a bit less than 21.

The correct answer is C.

Wow.....all the three explanations are different and easy to understand.

Thanks a lot guys!!!