5*5*4*3*2*1*4 / 7*6*5*4*3*2*1 = 10/21

Even though there are many questions of this type, each is a scenario of its own. Hope one fine day (before the exam) would answer any of these types.

Similar question: https://www.beatthegmat.com/simple-proba ... tml#228317

I will give it a try in here. So please don't take it as my answer. I am trying to deduce a number that can be one of the answer choise.

7 people can be seated in 7 seats in 7! ways
Number of aisle seats=2
2 people can be arranged in 5 remaining seats= 5P2= 5!/3!=20
Remaining people =5 could be arranged in 5 seats= 5!

answer: 20 * 5!/ 7!=20/42= 10/21?

I approached it in a slightly different way:

2 people (who sat in the aisle previously) can only sit in the centre 5 seats = 5C2
Total combinations of where those same 2 people can sit = 7C2

Probability = 5C2 / 7C2

[spoiler]= 20/42 = 10/21[/spoiler]