VeritasPrepKarishma wrote:

Bunuel wrote:

A rectangular table seats 4 people on each of two sides, with every person directly facing another person across the table. If eight people choose their seats at random, what is probability that any two of them directly face other?

(A) 1/56

(B) 1/8

(C) 1/7

(D) 15/56

(E) 4/7

The question asks for the probability of any two people (say A and B) facing each other.

Method 1:

A can take any seat. After that B should take the seat facing A which is 1 of the 7 remaining seats. So probability of A and B facing each other = 1/7

Method 2:

In how many ways can 8 people can sit on a rectangular table with 4 chairs on 2 opposite sides? The first person can sit in 4 ways (there are 4 distinct seats with respect to the table. Think circular arrangement). Now there are 7 distinct seats and 7 people so they can sit in 7! ways.

Total number of ways = 4*7!

In how many ways can 8 people sit such that A and B face each other? A goes and sits in 4 ways (as before). B sits directly opposite him in 1 way only. Now we have 6 distinct seats and 6 people. They can sit in 6! ways.

Probability = 4*6!/4*7! = 1/7

To solve this question we can do it by 2 methods....

Let the 2 persons facing each other be A & B

Method 1 : Place A at a position... Now there are 7 spaces for B to seat but in only one case it is opposite to B. So probability of B facing A = 1/7.

Method 2 : Counting total no. of arrangements

So, Total no. of ways of seating all the 8 people = 8!

(

VeritasPrepKarishma I disagree with you here for total no. of ways. Lets A is seated at corner seat then left and right seating of A is very different in nature. Hence it may not be considered as circular arrangement.)

Total no. of desired ways in which A & B faces each other = 4 (pair of opposite seats) * 2 (arranging A and B on the pair of opposite seat) * 6! (arranging other 6 people on the 6 seats) = 4*2*6!

So required probability = 4*2*6!/8! = 1/7

Answer C
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