Question

Three persons A, B and C are to speak at a function with 5 other persons. If the persons speak in random order, the probability that A speaks before B and C speaks before A in that order is

• A. $\frac{3}{8!}$
• B. $\frac{{}^8{C}_2}{8!}$
• C. $\frac{1}{3!}$
• D. $\frac{1}{8!}$

Answer 1

Answer: (C)

$\frac{1}{3!}$

The $5$ people could arrange themselves in $5!=120$ ways. However, the relative positions of A, B, C are fixed in the order C, A, B. Hence, the number of ways of arranging the $5$ people keeping in mind the order = $\frac{5!}{3!}=\frac{120}{6}=20$
Therefore, required probability = $\frac{20}{5!}=\frac{1}{3!}$