Question

In a conference 10 speakers are present . if $S_1$ wants to speak before $S_2$ & $S_2$ wants to speak after $S_3$ 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. ${}^{10}{C}_3$
• B. ${}^{10}{P}_8$
• C. ${}^{10}{P}_3$
• D. $\frac{10!}{3{}^{10}}$

Answer 1

The correct option is C ${}^{10}{P}_3$

This is a question on permutation and combination.

There are $7$ speakers where the restrictions do not apply.

For the first three speakers we have $=S_1,S_3,S_2$

$S_3$ can also speak before $S_1$

There are two possibilities for $S_1$ and $S_3$ to take place and there is no third possibility.

For $S_2$ it has to remain in the same position.

We have,

$\frac{10!}{(10-7)!}=\frac{10!}{3!}={}^{10}{P}_3$