P and Q are two friends standing in a circular arrangement with 10 mor...
Fix P at one point then number of places where B can be seated is 11.
Now, exactly three persons can be seated between P and Q, so only two places where Q can be seated. So, p = 2/11
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P and Q are two friends standing in a circular arrangement with 10 mor...
Solution:
To solve this problem, we need to consider two cases:
Case 1: P and Q are adjacent to each other
In this case, P and Q can be arranged in 2 ways (PQ or QP). The remaining 10 people can be arranged in 10! ways. However, we need to ensure that exactly 3 people are seated between P and Q. There are 7 people left after placing P and Q, so we need to choose 3 of them to sit between P and Q. This can be done in 7C3 ways. The remaining 4 people can be arranged in 4! ways. Therefore, the total number of arrangements in this case is:
2 x 10! x 7C3 x 4! = 20 x 10! x (7x6x5)/(3x2x1) x 4! = 20 x 10! x 35 x 24 = 20 x 35 x 10! x 24
Case 2: P and Q are not adjacent to each other
In this case, P and Q can be arranged in 2 ways (PQ or QP). The remaining 10 people can be arranged in 10! ways. However, we need to ensure that exactly 3 people are seated between P and Q. There are 5 people left after placing P and Q, so we need to choose 3 of them to sit between P and Q. This can be done in 5C3 ways. The remaining 7 people can be arranged in 7! ways. Therefore, the total number of arrangements in this case is:
2 x 10! x 5C3 x 7! = 20 x 10! x (5x4)/(2x1) x 7! = 20 x 10! x 10 x 7!
Total number of arrangements = 20 x 35 x 10! x 24 + 20 x 10! x 10 x 7!
= 20 x 10! x (35 x 24 + 10 x 7!)
= 20 x 10! x 2625
Now, we need to find the probability that exactly 3 persons are seated between P and Q. This is given by:
(total number of arrangements with exactly 3 persons between P and Q)/(total number of arrangements)
= (20 x 10! x 10 x 7!)/(20 x 10! x 2625)
= 10 x 7!/2625
= 2/11
Therefore, the probability that exactly 3 persons are seated between P and Q is 2/11.
Hence, option C is the correct answer.