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$\frac{3}{20}$
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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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