Question

15 persons among whom are A and B, sit down at random at round table. The probability that there are exactly four persons between A and B is

(a) \(\frac{9!}{14!}\)

(b) \(\frac{10!}{14!}\)

(c) \(\frac{2.9!}{14!}\)

(d) None of these

Answer 1

(d) None of these

15 persons can be seated round a table in 14! ways. If 4 persons are to be included between A and B, they can be arranged in ¹³C₄ × 4! ways, now total number of arrangements is ¹³C₄ × 4! × 9! × 2! = 2.13!

Hence the probability is \(\frac{2.13!}{14!} = \frac 2{14} = \frac 17\).