# Finding number of cases , arranging people around circular table

Suppose we have a circular table and it contains 10 seats how many way we can arrange 15 people in this table ?

$\dbinom {15}{10} \cdot \dfrac {15!}{5!}$

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Are the seats distinct? – Sarastro Dec 3 '12 at 7:41
Yes, I am pretty sure – Node.JS Dec 3 '12 at 7:43
Do you consider two arrangements as equal if one arrangement is generated by rotation or reflection from the other arangement? – miracle173 Dec 3 '12 at 7:46
The convention for circular tables is that arrangements that differ by a rotation are counted as being the same. – André Nicolas Dec 3 '12 at 7:46

It rather depends on what you regard as identical seating arrangements.

Start with ${15 \choose 10}$ for choosing ten out of fifteen, as you presumably did.

Multiply by $10!$ for the ways of seating these ten in identified seats.

Perhaps divide by $10$ if each of the rotations count as the same.

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if those seats are not identical,what would the answer be ? – Node.JS Dec 3 '12 at 7:47
If the rotations do not count as the same then do not divide by $10$. But see André Nicolas's comment on the question. – Henry Dec 3 '12 at 7:59

The number of ways of choosing $k$ objects from $n$ objects with order mattering is given by

$$_n P_k = n(n-1)(n-2)\cdots(n-k+1) = \frac{n!}{(n-k)!} = k! \left({n \atop k}\right)$$

(read "n permute k"). Explanation: Take $n$ objects. Choose one for the first position, leaving $n-1$ left. Then choose from those $n-1$ for the second. And so on, making $k$ choices in all. Note that the precise nature of the choices does not affect the number of ways of making the choices, and so by the multiplication principle, you have the given product.

So in your case, you want to choose 10 from 15 with order mattering, thus you have $_{15} P_{10}$. If rotations are not to be distinguished, then divide this by 10 to get $\frac{_{15} P_{10}}{10} = 9! \left({15 \atop 10}\right)$, which is 1,089,728,640.

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