Question
P and Q sit in a ring arrangement with 10 persons. What is the probability that P and Q will sit together?
Answer: Option A
n(S)= number of ways of sitting 12 persons at round table:
= (12 - 1)! = 11!
Since two persons will be always together, then number of persons:
= 10 + 1 = 11
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
$$\eqalign{
& P\left( A \right) = \frac{{n\left( A \right)}}{{n\left( S \right)}} \cr
& = \frac{{10!\, \times 2!}}{{11!}} \cr
& = \frac{2}{{11}} \cr} $$
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n(S)= number of ways of sitting 12 persons at round table:
= (12 - 1)! = 11!
Since two persons will be always together, then number of persons:
= 10 + 1 = 11
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
$$\eqalign{
& P\left( A \right) = \frac{{n\left( A \right)}}{{n\left( S \right)}} \cr
& = \frac{{10!\, \times 2!}}{{11!}} \cr
& = \frac{2}{{11}} \cr} $$
Was this answer helpful ?
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