Out of 10 persons,A, B,and C are to speak at | vedclass

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(A) There are $10$ positions available for the $10$ speakers.First,we choose $3$ positions out of $10$ for $A, B$,and $C$,which can be done in $^{10}C

Vedclass · Accepted Answer

$10!/6$

Vedclass · Answer

(A) There are $10$ positions available for the $10$ speakers.First,we choose $3$ positions out of $10$ for $A, B$,and $C$,which can be done in $^{10}C_3$ ways.Once these $3$ positions are chosen,there is only $1$ way to arrange $A, B$,and $C$ such that $A$ speaks before $B$ and $B$ speaks before $C$ (i.e.,the order must be $A, B, C$).The remaining $7$ persons can be arranged in the remaining $7$ positions in $7!$ ways.Therefore,the total number of ways is $^{10}C_3 	imes 1 	imes 7!$.Total ways $= \frac{10!}{3! 	imes 7!} 	imes 7! = \frac{10!}{3!} = \frac{10!}{6}$.

Out of $10$ persons,$A, B$,and $C$ are to speak at a function. If $A$ wants to speak before $B$ and $B$ wants to speak before $C$,in how many ways can they speak?

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