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20 persons are to be seated around a circular table. Out of these 20 , 2 of them are brothers then number of arrangements in which there will be at least three persons between the brothers is.?

SO here's what I tried. First fix the postions of the two brothers then put the rest of the 18 people in the remaining places.

So there are 7 such cases.: 3,15 4,16 5,13 6,12 7,11 8,10 9,9.

So for example the first 3-15 case I first selected 3 out of 18 that is 18c3 and multiplied it by 3! and 15! to get the answer as 18! . The same results goes for all the other cases so the answer now is 7 * 18! . But the brothers themselves can be arranged in 2! ways .

So the answer im getting is 14*18!. The actual answer is 13*18!. SO what am I doing wrong?

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    $\begingroup$ What have you done? $\endgroup$ – Julian Rachman Jan 10 '15 at 5:42
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    $\begingroup$ I am adding that in a few minutes I tried editing with the app but didn't get through so. I'm doing it with a pc sorry for not writing that earlier. $\endgroup$ – Sudhanshu Jan 10 '15 at 5:55
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Every table arrangement can be expressed uniquely as a string of $20$ unique characters where the first character always represents brother $A$ (so that the circle is cut and streched into a line). There are $13$ positions into which the other brother can be placed. Once this has been selected there are $18$ seats for $18$ people which can be selected in $18!$ ways. Hence the answer is $13\cdot 18!$

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    $\begingroup$ not 16 positions for the second brother... $\endgroup$ – Joffan Jan 10 '15 at 5:48
  • $\begingroup$ Oh right, I forgot the three at the other end $\endgroup$ – Jorge Fernández Hidalgo Jan 10 '15 at 5:52
  • $\begingroup$ Thanks a lot. Yours is the best explaination I got. $\endgroup$ – Sudhanshu Jan 10 '15 at 6:10
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Take all the possibilities $(20-1)!$ and subtract the ones you do not like by putting the brothers close enough.
$$abx_3x_4\ldots x_{19}x_{20}$$ $$ax_2bx_4\ldots x_{19}x_{20}$$ $$ax_3x_4b\ldots x_{19}x_{20}$$ $$ax_2x_3\ldots x_{19}b$$ $$ax_2x_3\ldots bx_{20}$$ $$ax_2x_3\ldots bx_{19}x_{20}$$ So the total ways must be $19!-6*18!$

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Pick one brother, put him in any seat. Place three of the other people on his left and then on his right. With those 1+6 people chosen, place the remaining 13 people.

$20 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 13! = 260 \times 18!$

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  • $\begingroup$ They are in a circular permutation, cancel out the 20. $\endgroup$ – Phicar Jan 10 '15 at 5:56
  • $\begingroup$ No - nothing was said about rotations being equivalent. The circular table just means the seating of the two brothers is more restricted. $\endgroup$ – Joffan Jan 10 '15 at 6:11

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