In how many ways can 5 speakers S_1, S_2, S_3 | vedclass

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(A) There are $5$ speakers in total. The total number of ways to arrange $5$ speakers is $5! = 120$.In any arrangement,the relative order of $S_1, S_2

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(A) There are $5$ speakers in total. The total number of ways to arrange $5$ speakers is $5! = 120$.In any arrangement,the relative order of $S_1, S_2$,and $S_3$ can be in $3! = 6$ possible ways.These ways are: $(S_1, S_2, S_3), (S_1, S_3, S_2), (S_2, S_1, S_3), (S_2, S_3, S_1), (S_3, S_1, S_2), (S_3, S_2, S_1)$.Out of these $6$ ways,$S_3$ speaks after both $S_1$ and $S_2$ in only $2$ cases: $(S_1, S_2, S_3)$ and $(S_2, S_1, S_3)$.Thus,the probability that $S_3$ speaks after $S_1$ and $S_2$ is $\frac{2}{6} = \frac{1}{3}$.Therefore,the number of favorable ways is $\frac{1}{3} 	imes 5! = \frac{120}{3} = 40$.

In how many ways can $5$ speakers $S_1, S_2, S_3, S_4$,and $S_5$ give speeches one after the other if $S_3$ must speak after both $S_1$ and $S_2$?

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