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There are 10 speakers in a conference $S_1, S_2..S_{10}$ and $S_1$ speaks only after $S_2$ has spoken, then find number of ways in which speakers speak.

Using string method, let $S_1$ and $S_2$ be one block. Also there is only one way to arrange $S_1$ and $S_2$ inside the block.

So total permutations are $9!$

But given answer is $\frac{10!}{2}$. Where am I going wrong?

Aditya
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  • To illustrate the ideas already expressed by the answers, your algorithm would overlook the sequence ${S_2, S_3, S_1, S_4, \cdots }.$ – user2661923 Mar 08 '21 at 04:31
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    Further, as others have stated, in the $(10)!$ sequences possible, by symmetry you know that in (1/2) of them $S_1$ precedes $S_2$ and in the other (1/2), vice versa. – user2661923 Mar 08 '21 at 04:33

2 Answers2

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If you consider $S_1$ and $S_2$ are one block, you are counting the ways in which $S_1$ speaks immediately after $S_2$.

Hint There are $10!$ arrangements of 10 speakers, show that $S_1$ speaks after $S_2$ in half of them.

N. S.
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  • That’s what I also thought initially, and that is where my problem is. I don’t know how to show that $S_2$ speaks after $S_1$ in half the cases – Aditya Mar 08 '21 at 04:21
  • @Aditya Total number of cases= Cases in which S1 speaks before S2+Cases in which S1 speaks after S2, now since there is no distinction made between S1 and S2, by symmetry both of them will be equal. – V.G Mar 08 '21 at 04:23
  • @Aditya Let $A$ denote the ways when $S_1$ speaks before $S_2$ and $B$ the ways when $S_1$ speaks after $S_2$. Prove that interchanging $S_1$ with $S_2$ gives a bijection between $A$ and $B$. – N. S. Mar 08 '21 at 05:33
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If you make $S1$ and $S2$ one block, you make their relative speaking positions also fixed. As per question, they can speak in any position as long as $S1$ speaks after $S2$. It is easy to see that $S1$ will speak after $S2$ in half of all permutations and before $S2$ in rest half.

So the answer is $\displaystyle \frac{10!}{2}$.

Edit: on your point about not being convinced on the same. Let's see this way. If $S2$ is in the first position, there are $9!$ ways for arranging rest of the speakers and all are permissible. If $S2$ is in second position, $8$ permissible positions for $S1$ and $8!$ ways for rest of the speakers. Continue with same logic and total number of permutations come to

$9! + 8! \cdot (8 + 7 + 6 + 5 + 4 + 3 + 2 + 1) = 1814400$.

Math Lover
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