Question

Ten persons , amongst whom are $A$ ,$B$ and $C$ to speak at a function. The number of ways in which it can be done if $A$ wants to speak before $B$ and $B$ wants to speak before $C$ is?

• A. $\frac{10!}{6}$
• B. $3!+7!$
• C. ${}^{10}{P}_3.7!$
• D. None of these

Answer 1

The correct option is A

$\frac{10!}{6}$

Explanation for the correct option:

To find the number of ways speakers to satisfy the given condition to speak the function:

Given, totally $10$persons.

For speakers $A$ ,$B$ and $C$ three places can be selected in ${}^{10}{C}_3$ ways.

Then $A$ ,$B$ and $C$ can get their priority in only one way${=}^{10}{C}_3\times 1$

Other seven speakers are selected in $7!$ ways.

Hence the required number of ways $=\left({}^{10}{C}_3\times 1\right)\times 7!$

$$\begin{gathered} =\frac{10!}{7!3!}\times 7! \\ =\frac{10!}{6}\left[\because 3!=3\times 2\times 1\right] \end{gathered}$$

Hence, option (A) is the correct answer.