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$\frac{3}{20}$
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 $= \text{nm} = \frac{!6!4}{9!} = \frac1{21}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Surface area times the rate of change of diameter.
Surface area times the rate of change of radius.
None of these.