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In how many ways can $10$ people e seated on $24$ seats placed in a row, such that among each pair of seats equidistant from the beginning and end, at least one seat is empty?

This is how I attempted it:

We have $12$ pairs of seats. We first choose $10$ pairs for the $10$ people in $\binom{12}{10}$ ways. We can now say that each person has a choice of two seats. Hence each person can choose one of the seat among the two in $2$ ways. As there are $10$ people so $10$ people can choose $10$ seats in $2^{10}$ ways. Thus among each pair of seats , one seat will be empty and the remaining $4$ seats that were not included in the pairs of seats chosen will be empty anyways. So according to me, the total number of ways is

$\binom{12}{10}2^{10}$

However, the correct answer to the above problem is $12!2^{9}$

Could you please explain where did I go wrong and how can I get the correct answer?

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First consider the seats in pairs equidistant from the beginning and end, because we can seat at most one person in such pairs of seats (from the empty seat condition). Distributing people over each of the 12 pairs of seats, since we have 10 people and order does not matter, we have $^{12}P_{10}=\frac{12!}{2!}$ ways to do this.

Now within each pair of seats our person can be sat in the left seat or the right seat, 2 possibilities for each of 10 people gives $2^{10}$ ways to do this.

Finally we multiply our 2 numbers (since we need both of the above to occur) giving the required solution.

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  • $\begingroup$ Thanks for explaining ! I understood where I went wrong $\endgroup$ – Aditi Feb 23 '18 at 17:08
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You are missing a factor of $10!$ in your answer (for putting the people in the chosen chairs).

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  • $\begingroup$ Thank you very much for pointing it out ! I understood my mistake $\endgroup$ – Aditi Feb 23 '18 at 17:07

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