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(A) There are $10$ persons in total. The condition is that $A$ must speak before $B$,and $B$ must speak before $C$. This means the relative order of $

Vedclass · Accepted Answer

$\frac{10!}{6}$

Vedclass · Answer

(A) There are $10$ persons in total. The condition is that $A$ must speak before $B$,and $B$ must speak before $C$. This means the relative order of $A, B, C$ is fixed as $A, B, C$.First,we choose $3$ positions out of $10$ for $A, B, C$ in $^{10}C_3$ ways. Once these positions are chosen,there is only $1$ way to arrange $A, B, C$ such that $A$ is before $B$ and $B$ is before $C$.The remaining $7$ persons can be arranged in the remaining $7$ positions in $7!$ ways.Therefore,the total number of ways is $^{10}C_3 	imes 7! = \frac{10!}{3! 	imes 7!} 	imes 7! = \frac{10!}{3!} = \frac{10!}{6}$.

Ten persons,amongst whom are $A, B$ and $C$,are 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:

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