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?


Can someone explain how we get the second part that will be subtracted from C(7,5)? Yes, i know the possible arrangements can be listed manually, but what is the shorter and more accurate path, using combinations?

5c5 + 6c5 + 6c5 + 4c2 = 17

My take on the problem. The five models out of seven can be choose such that

1) The 4th tallest and 6th tallest are present in the group of five
2) The 4th tallest is present but not the 6th tallest
3) The 6th tallest is present but not the 4th tallest
4) The 4th tallest and the 6th tallest are not present

Case 1:

The other three models can be choose out of remaining five models in 5C3 ways. Once the five models are choose they can be arranged as following

Consider 4th and 6th together as one group, hence total of 4 models (assuming 4th and 6th together as a single model) in 4! ways and these two models can internally be arranged in 2! ways.

Hence the total ways in which the models can be arranged with 4th and 6th always together is 4!*2!

Without considering any specific constraint total arrnagements are 5!. Hence required arrangements where the 4th and 6th models are never adjacent is

[5! - (2!*4!)]

Case 2:

Once the 4th model is choose, and 6th is not to be choosen, the remaining 4 models can be choosen out of 5 models in 5C4 = 5 ways. These five models can be arrnaged in 5! ways (no constraints on these). Hence total combinations = 5 * 5! ways

Case 3: This is similar to case 2 and hence total ways = 5*5!

Case 4: If 4th and 6th models are excluded then the remaining 5 have to be choosen and they can be arranged in 5! ways.

considering all cases the total number of arrangements are


[5! - (2!*4!)] + 5 * 5! + 5 * 5! + 5!

= (12 * 5!) - (2! * 4!)

ketan

when you consider the 4th and 6th tallest model to be in the groupe, then the 5th tallest has to be in the group as well, because the models has to stand from short to tallest:

_ _ 4 5 6 or _ 4 5 6 _ => 2 vacant positions

...that means you pick 2 from the remaining 4 => 4c2

7c5 is clear ! but why do you chose 3c3 ?

as i mentioned above: the order has to be from shortest to tallest and 4 and 6 cannot stand together. when we choose 4 and 6, 5 has to be between them. so 3 are already chosen and the remaining 2 spaces can be chose from the remaining 4.

swath20 has the right approach, except he made a small typo. We need to take out the case where 4th and 6th model are adjacent from C(7,5). When these two people are adjacent, that means they both are selected, and 5th is not selected. We need to select three more people to make it five, from the remaining four people.

So it would be C(7,5)-C(4,3)

agree with Honghu
7C5-1*4C3
unlike other problems in this one there would be only one way to place the 4th and 6th model as per the order required.
Hence, 4C3*1
Correct me if my reasoning seems wrong please