Let the number of ways in which 9 children can be arranged around a circular table be $T$ .

Find the sum of the digits of $T$ .

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Details & Assumption - :
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1)Clock-wise & Anti-clockwise arrangements are considered the same.

2)Sum of digits of a number , say $2008$ , is $2+0+0+8 = 10$ .

The answer is 9.

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There are $9! = 362880$ ways to seat the nine people. For any seating, there are $9$ ways to rotate it around the table, so we divide by $9$ to get $\frac{362880}{9} = 40320$ (I have divided $362880$ by $9$ because two seatings are considered the same if one can be rotated to form the other).

Thus we get $T$ as $40320$ .

Sum of digits of $T = 9$ and that is the answer.