Question

In a conference 10 speakers are present. If S₁ wants to speak before S₂ and S₂ wants to speak after S₃, then the number of ways all the 10 speakers can give their speeches with the above restriction if the remaining seven speakers have no objection to speak at any number is-

• A. ¹⁰C₃
• B. ¹⁰P₈
• C. ¹⁰P₃
• D. $\frac{10!}{3}$

Answer 1

Answer: (D)

$\frac{10!}{3}$

Answer 2

If the order of speakers is S₁, S₃, S₂ (need not be consecutive) treat S₁, S₃, S₂ as identical and arrange. We obtain$\frac{10!}{3!}$. Similarly if the order of speakers is S₃, S₁, S₂ we obtain$\frac{10!}{3!}$.

Required = 2 $\frac{10!}{3!}$= $\frac{10!}{3}$