There are 10 persons who are to be seated around a circular table. Fin

1.	There are 10 persons who are to be seated around a circular table. Find the probability that two particular persons will always sit together.
Answer» Total number of ways in which 10 person can sit around a circular table = 9! (∵ We shall keep one place fixed and the rest of the 9 places will be filled in (9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1) ways asthere is no repetition in such case. The first variable place can be filled by any of the remaining 9 person, second place by any of the remaining 8 persons and so on.) If two particular persons (say A and B) sit together, then the number of ways in which 9 person (considering the pair of A and B as one unit) can sit round the table = 2! × 8!. (Here A and B can interchange places amongst themselves in 2! ways) ∴ Favourable number of cases = 2 × 8! ∴ Required probability = \(\frac{2\times8!}{9!}\) = \(\frac{2\times8!}{9\times8!}\) = \(\frac{2}{9}.\)

There are 10 persons who are to be seated around a circular table. Find the probability that two particular persons will always sit together.

Answer»

Total number of ways in which 10 person can sit around a circular table = 9!

(∵ We shall keep one place fixed and the rest of the 9 places will be filled in (9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1) ways asthere is no repetition in such case. The first variable place can be filled by any of the remaining 9 person, second place by any of the remaining 8 persons and so on.)

If two particular persons (say A and B) sit together, then the number of ways in which 9 person (considering the pair of A and B as one unit) can sit round the table = 2! × 8!. (Here A and B can interchange places amongst themselves in 2! ways)

∴ Favourable number of cases = 2 × 8!

∴ Required probability = \(\frac{2\times8!}{9!}\) = \(\frac{2\times8!}{9\times8!}\) = \(\frac{2}{9}.\)

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