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In a conference 10 speakers are present. $S(1)$ wants to speak before $S(2)$ and $S(2)$ wants to speak before $S(3)$, then the number of ways all the 10 speakers can give their speeches with the above restriction if the remaining speakers have no objection to speak at any number?

exilednick
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3 Answers3

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Outline

Unrestriced, there are $10!$ permutations. In all these arrangements, the S(1), S(2), S(3) appear in 6 different ways. We are interested in only one of them. So the final answer is $\displaystyle \color{blue}{\frac16 10!}$

Edit

Also look here for an uncluttered answer of a similar problem.

Community
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Shailesh
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Hint: In order to specify a working ordering, we need to specify the following pieces of information:

  • What 3 positions (out of 10) are occupied by the speakers $S(1), S(2),$ and $S(3)$. How many ways are there to do this?
  • The order of the 7 other speakers. How many ways are there to do this?

Then, we must combine the two pieces of information. However, since an ordering is uniquely determined by that info, the total number will be the product of the two numbers above.

Aaron
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  • While I do not understand the reason for the downvote, you could clarify your answer by asking which $3$ of the $10$ positions are occupied by speakers $S(1)$, $S(2)$, and $S(3)$. – N. F. Taussig Oct 27 '15 at 16:51
  • @N.F.Taussig You are right. Saying "the first 3 speakers" is a bit confusing, although I think context makes it clear what is meant. I will change it. – Aaron Oct 27 '15 at 16:54
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the time slots can be represented as a group of 10 objects wherein the rightmost is the 1st to speak the next the 2nd and so on. e.g., $|*||*|||*|$, with * representing one selected and | one not selected. wherein the rightmost/first selected is always occupied by $S(1)$, the next selected by $S(2)$ and the leftmost/last selected by $S(3)$. then all orderings of remaining $7$ so answer is ${10\choose 3}7!$

miniparser
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