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Probability Level 2

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 $T$ .

Details & Assumption - :

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.

3 solutions

Tushar Malik
Mar 7, 2015

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.

Great solution @Tushar Malik !

Phillip Han - 4 years, 3 months ago

Oh , I wrote the answer as 40320 and got it wrong ... forgot that the question asked is the sum of the digits ...

Ashwin Hebbar - 6 years, 2 months ago

You still had 2 more tries. What even

Vasudev Chandna - 6 years, 2 months ago

Noel Lo
May 20, 2015

I would prefer to use the method (9-1)! = 8! = 40320 (whose digits sum up to 9). For circular permutations, the number of ways n people can sit is (n-1)!. Try it for any positive integral value for n. This general formula should work.

Fox To-ong
Feb 12, 2015

its a cyclical permutation = ( n - 1 )! = 8! = 1,2x3x4x5x6x7x8 = 40320 T = 4 + 0 + 3 + 2 + 0 = 9

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The answer is 9.

3 solutions

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