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There are $16$ executives, including two brothers. In how many ways can they be arranged around a circular table if the two brothers can't be seated together?

I tried putting each of the brothers in between the $14$ executives by $13! \cdot 14C2 \cdot 2$.

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    $\begingroup$ I see that you are beginner. You must incorporate your thoughts/attempt with the question you ask $\endgroup$ – user481779 Feb 21 '18 at 9:23
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    $\begingroup$ Welcome to MathSE. Please edit your question to explain what you have attempted and where you are stuck so that you receive answers that address the specific difficulties you are encountering. Here is a tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Feb 21 '18 at 9:24
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    $\begingroup$ Are seatings which are equal up to a rotation considered equal? $\endgroup$ – Christian Blatter Feb 21 '18 at 9:32
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This can be easily calculated by substracting the number of ways in which they sit together from the total number of ways

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  • $\begingroup$ Is there any other method without subtracting them??.. $\endgroup$ – aneesh cool Feb 21 '18 at 9:23
  • $\begingroup$ I think this is the easiest $\endgroup$ – user481779 Feb 21 '18 at 9:25
  • $\begingroup$ @aneeshcool why do you want another method? This works fine. $\endgroup$ – drhab Feb 21 '18 at 9:28
  • $\begingroup$ Yes but i tried putting each of the brothers in between the 14 executives by 13!* 14C2 *2 $\endgroup$ – aneesh cool Feb 21 '18 at 9:31
  • $\begingroup$ @aneeshcool solve using the above method and you will get the same answer, I suppose $\endgroup$ – user481779 Feb 21 '18 at 9:36

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