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In how many different arrangements can 7 white suited persons and 5 black suited persons sit in a round table of 12 seats so that none of the black suited persons gets to sit next to each other?

The way I tried to approach this is by considering the possible arrangements of seating the black suited people and the white suited people as two groups.

three_possible_arrangements

And for each of these arrangements the white suited guys can rearrange their positions in 7! ways and for each of those rearrangements the black suited guys can rearrange their positions in 5! ways. So, the number of total possible seating arrangements we get are $3\times7!\times5!$

Is this approach correct? Are there other better ways to approach the problem?

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By convention, in a circular arrangement, only the relative order of the people matters. Suppose Andrew is one of the seven people in white suits. We will use him as a reference point.

Seat Andrew. The remaining six people in white suits can be arranged in $6!$ orders as we proceed clockwise around the table from Andrew. This creates seven spaces, one to the immediate left of each of the people in white suits. To separate the people in black suits, we must choose five of these spaces for the people in black suits. Once these spaces have been selected, the five people in black suits can be seated in those spaces in $5!$ orders as we proceed clockwise around the table from Andrew. Hence, there are $$6!\binom{7}{5}5!$$ distinguishable seating arrangements in which no two of the people in black suits sit in adjacent seats.

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    $\begingroup$ ... which agrees with the result the OP found with a different reasoning. $\endgroup$ – Henning Makholm Dec 15 '17 at 17:02
  • $\begingroup$ @HenningMakholm Good point. $\endgroup$ – N. F. Taussig Dec 15 '17 at 17:09
  • $\begingroup$ Thanks. This is a more efficient and elegant approach. $\endgroup$ – Tiash Dec 15 '17 at 17:52

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