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In how many ways can $k$ people out of $n$ be seated at one round table and the remaining $n-k$ people at another table?

In my specific case, there are 14 people and I want to know the way to seat 8 of them at one round table and the others at another round table?

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closed as off-topic by Davide Giraudo, mlc, kingW3, Chris Brooks, Arnaldo May 23 '17 at 17:19

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Davide Giraudo, mlc, kingW3, Chris Brooks, Arnaldo
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Minor English lesson (assuming the OP is not a native speaker): An individual sits "at" a table; a group of people can sit either "at" or "around" a table. But sitting "on" a table means you're actually on top of the table, along with the plates and the silverware. $\endgroup$ – Barry Cipra May 23 '17 at 14:43
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    $\begingroup$ 1. Are the two tables distinct ? 2. Do both tables have $8$ seats ? $\endgroup$ – true blue anil May 23 '17 at 14:48
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Choose 8 people out of 14, and put them at the first table. The number of possible combinations equals:

$${14 \choose 8} = 3003$$

Then, there are $8!$ ways for these 8 people to be seated, and $6!$ for the other 6. However, since it is a round table, we must divide by 8 and 6 respectively. As such, we get:

$${14 \choose 8} \cdot 7! \cdot 5! = 1,816,214,400$$

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