Three children,each accompanied by a guardian | vedclass

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(B) Let the children be $C_1, C_2, C_3$ and their respective guardians be $G_1, G_2, G_3$.There are $6$ persons in total.The condition is that for eac

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(B) Let the children be $C_1, C_2, C_3$ and their respective guardians be $G_1, G_2, G_3$.There are $6$ persons in total.The condition is that for each pair $(G_i, C_i)$,the guardian $G_i$ must be interviewed before the child $C_i$.In any arrangement of $6$ persons,there are $2!$ ways to arrange the pair $(G_i, C_i)$ such that $G_i$ comes before $C_i$ or $C_i$ comes before $G_i$.Since there are $3$ such pairs,the total number of unrestricted arrangements is $6!$.For each pair,only $1$ out of $2!$ arrangements is valid (i.e.,$G_i$ before $C_i$).Thus,the number of valid ways is $\frac{6!}{2! 	imes 2! 	imes 2!} = \frac{720}{8} = 90$.

Three children,each accompanied by a guardian,seek admission in a school. The principal wants to interview all the $6$ persons one after the other subject to the condition that no child is interviewed before its guardian. In how many ways can this be done?

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