tarek99 wrote:
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?
A. 6
B. 11
C. 17
D. 72
E. 210
Okay..here's my take..
Let the models be(in ascending order of height)
A B C D E F G
Now..out of these 7..we have to select 5 and make them stand in ascending order of height.
According to the question, B and D cannot be standing together.
Imagine this, if we select any 5 of these, there will only be one way to make them stand.
Considering the situation, if both B and D were selected, and let's say that B and D were actually selected in the group; then they will stand together every time C is not the part of the group(as there will be no one to stand in the middle). So, our complement event is that B and D are selected, and C is not considered at all for selection. In this way, they will always stand together.
Number of ways to do this..
B and D are already in the group, so we have to select the remaining 3 out of 5. But wait, we also have to never consider C for selection. Finally then, we have to select the remaining 3 out of 4(where C is excluded)
\(4C3 = 4\)
This has to be subtracted from the total possible combinations.
\(7C5 - 4 = 17\)
Answer (C)