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
    $\frac{4}{10}$
  • C
    $\frac{1}{21}$
  • D
    $\frac{3}{20}$

Answer

  1. $\frac{1}{21}$

Solution:

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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