A rectangular table seats 4 people on each of two sides, with every pe

We are given :

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TABLE
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Probability = # Favorable Outcomes / # Total Outcomes

Total Outcomes = 8 * 7 (Since a spot on the table for 2 people can be selected in 8 * 7 ways) = 56

Favorable Outcomes = 8 * 1 (One person can pick a seat in 8 ways, but to sit opposite to that person, there is only 1 option, thus 8*1) = 8

Probability = 8/56 = 1/7, thus the Answer is C.

Thanks for the Kudos!

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

Consider those 2 persson A and B so -

Fav condition:
A can sit in any of seat around the table no of ways - 8
B has to take seat opposite to A no of ways - 1
other member can take seat other 6 seats no of ways - 6!

All condition:
all possible combination without restriction no of ways - 8!

\(prob = fav/all => 8*1*6!/8!\)
\(prob = 1/7\)

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

Look at this:


If the first person comes and sits at A, it is exactly the same as sitting at D (table corner on his left and a person on his right, the shorter side of the rectangle on the left). Similarly, positions B and C are exactly the same.
Using same logic for the shorter sides of the rectangle too, we will get 4 unique places to sit for the first person, not 8 (with respect to the table)

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