Take the task of seating the 5 people and break it into stages.

We’ll begin with the most restrictive stage.

Stage 1: Select someone to sit in the MIDDLE seat
Since we're told Martha MUST sit in the middle seat, we must place her in that seat.
So, we can complete stage 1 in 1 way

Stage 2: Select someone to sit in seat #1
There are 4 people remaining to be seated, so we can complete this stage in 4 ways

Stage 3: Select someone to sit in seat #2
There are 3 people remaining to be seated, so we can complete this stage in 3 ways

Stage 4: Select someone to sit in seat #4
There are 2 people remaining to be seated, so we can complete this stage in 2 ways

Stage 5: Select someone to sit in seat #5
There is 1 person remaining to be seated, so we can complete this stage in 1 way

By the Fundamental Counting Principle (FCP), we can complete all 5 stages (and thus seat all 5 people) in (1)(4)(3)(2)(1) ways (= 24 ways)

Answer: 24

Note: the FCP can be used to solve the MAJORITY of counting questions on the GRE. So, be sure to learn

RELATED VIDEOS




_________________

How can we solve this question by permutation? \(\frac{n!}{(n-k)!}\)

I do not think you can use that formula.

This is basically a permutation wher the order does not matter.

You do have Martha in the middle but you can choose whatever you wanna of the other four people to sit in the other spots

I hope Brent could shed some more light on this.
_________________

There are very few "true" permutation questions on the GRE.

The Fundamental Counting Principal is a far more useful counting strategy.

Cheers,
Brent
_________________

You can use (n-1)! for the round table question , here n = no. of seats i.e. 5

So total number of ways = (5-1)! = 4!
_________________

You can use the permutations here: What you have to do is select Martha as the middle seat and we only want Martha to be on the middle seat so there is only one way of doing this.

Now there are four remaining people and there are four remaining seats, we can arrange them as we like so 4!=24.