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$P$ and $Q$ sit in a ring arrangement with 10 persons. What is the probability that $P$ and $Q$ will sit together?









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Option(A) is correct

$n(S)=$ number of ways of sitting 12 persons at round table: 


Since two persons will be always together, then number of persons: 


So, 11 persons will be seated in $(11−1)!=10!$ ways at round table and 2 particular persons will be seated in $2!$ ways.

n(A)= The number of ways in which two persons always sit together $=10!×2$

\(P(A)=\dfrac{n(A)}{n(S) }\)



(4) Comment(s)


No compex maths needed here:

wherever the first person sits in the ring, there are only two sits next to him (right and left).

So the solution is always 2/(N-1) where N is the total number of people in the ring.

Question is a bit misleading, as it actually means 'with 10 OTHER people"

Poonam Pipaliya

i don't understand this.

How is it $(12-1)!$?

If seats are 12 then total arrangement would be $12!$, isn't it?


When round tables are considered, we take $(n-1)!$ arrangements for $n$ number of people sitting there. Since on round table there is no start or end point like in a row (where arrangements would be $n!$)

To take an simple example, consider only 2 people sitting around round table. There is only ONE $(=(2-1)!=1!)$ way of sitting there and not $2!=2$ ways.

You may do the math for higher number of people if you wish to.


There is flaw in the question. As in the 1st statement it says $P$ & $Q$ and in the second statement it says $X$ & $Y$'s probability to sit together.

Kindly correct the question.