S_1, S_2, ldots, S_{10} are the speakers in a | vedclass

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(D) There are $10$ speakers in total. The condition is that $S_1$ must speak after $S_2$. In any arrangement of $10$ speakers,there are only two possi

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$\frac{10!}{2}$

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(D) There are $10$ speakers in total. The condition is that $S_1$ must speak after $S_2$. In any arrangement of $10$ speakers,there are only two possibilities for the relative order of $S_1$ and $S_2$: either $S_1$ speaks before $S_2$ or $S_2$ speaks before $S_1$. Since these two cases are equally likely,exactly half of the total permutations $(10!)$ will satisfy the condition that $S_2$ speaks before $S_1$ (which is equivalent to $S_1$ speaking after $S_2$). Therefore,the required number of ways is $\frac{10!}{2}$.

$S_1, S_2, \ldots, S_{10}$ are the speakers in a conference. If $S_1$ addresses only after $S_2$,then the number of ways the speakers address is

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