Problem Solving

Seven people are sitting in a theater watching a show. The row they are
in contains seven seats. After intermission, they return to the same row
but choose seats randomly. What is the probability that neither of the
people sitting in the two aisle seats was previously sitting in an aisle seat?

(a) 3/7
(b) 10/21
(c) 11/21
(d) 4/7
(e) 13/21

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]