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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 implies the relative order mu

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 implies the relative order must be $A, B, C$.In any arrangement of $10$ persons,there are $3! = 6$ possible relative orders for the three specific persons $A, B$,and $C$ (i.e.,$ABC, ACB, BAC, BCA, CAB, CBA$).Out of these $6$ possible relative orders,only $1$ order ($A$ before $B$ and $B$ before $C$) satisfies the given condition.The total number of ways to arrange $10$ persons is $10!$.Since only $1/6$ of the total arrangements satisfy the condition,the required number of ways is $\frac{10!}{6}$.

Ten persons,among whom are $A, B$,and $C$,are to speak at a function. The number of ways in which they can speak if $A$ wants to speak before $B$ and $B$ wants to speak before $C$ is:

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