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The question is about one particular couple.
Choose any couple it doesnt matter who, there are 2 ways, left and right of him by which they sit together.
Now the ways they wont sit together is when you fix one of them and the remaining 11 slots are available.
In circular arrangement a person 3 seats away from the other person would be the same arrangement if you move the 1st person as you want or the 2nd person vice versa.

Thus 2/11.
Bunuel
6 men and 6 women are seated around a table. What is the probability that a particular couple are seated together?

A. 1/55
B. 3/55
C. 2/11
D. 36/55
E. 9/11
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we have 6 men and 6 women. so total 12 people.
find: P(particualr couple seated together)

two ways we can approach this.

Approach 1: using circular formula.

first find denominator, total num of arrangements.
we know formula for this, (n-1)!= (12-1)!= 11!

now for numerator, total favorable outcomes.
we will fix one couple and consider them as a single person.
so now we have total 11 people. 10 left out people and 1 single person in the form of 2 people.

so num of ways to arrange them, (11-1)! = 10!

now those 2 people can arranged in 2!= 2 ways.

so our numerator= 2* 10!

putting into formula, we have 2*10! / 11!

ans 2/11.

Approach 2.

lets fix one husband in any one chair. now we are left with 11 chair.
and there is only one particular wife for that husband.

but that wife has two seats to sit to make sure she is sitting besides him and that is one seat right to husband and one seat left to husband.

so she has two seat to choose from 11.

so probability = 2/11

either method gets the same ans. 2/11

choice C
Bunuel
6 men and 6 women are seated around a table. What is the probability that a particular couple are seated together?

A. 1/55
B. 3/55
C. 2/11
D. 36/55
E. 9/11
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