MCQ
$10$ persons are seated around a round table. What is the probability that $4$ particular persons are always seated together?
  • A
    $\frac{1}{20}$
  • B
    $\frac{4}{10}$
  • $\frac{1}{21}$
  • D
    $\frac{3}{20}$

Answer

Correct option: C.
$\frac{1}{21}$
$10$ persons can sit around a table in $9!$ ways.
Consider the particular four persons as one unit.
Now, the entities are $6 + 1 = 7$
These $7$ entities can be arranged in $6!$ ways.
In the entities itself they can be arranged in $4!$ ways.
The required number of arrangements $= 6!4!$
Probability $= \text{nm}​ = \frac{!6!4}{9!} ​= \frac1{21}​$

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