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RandomGintoki
Krunaal
6 Clowns can be arranged in circular carousel in (6-1)! = 5! ways = 120

Now for the remaining 4 acrobats there are 6 places vacant between clowns, selecting 4 places = 6C4 = 15

4 acrobats can be arranged in 4! ways = 24

120*15*24 = 43,200

Answer B.
hey why did you take 6c4 ? Should the answer not be 4!*120 as after 6 clowns have been seated only 4 other chair remain
If we directly take 4! we are allowing arrangements where acrobats can sit together.

So when we sit 6 clowns in a circle, there are 6 distinct gaps around the circle => C1 _ C2 _ C3 _ C4 _ C5 _ C6 and one b/w C1 and C6 as they're sitting in a circle. Now acrobats need to be seated in these gaps, so that none of them sits together and there are 6 gaps, so we first choose 4 in 6C4 ways and then 4 acrobats can sit in 4! ways.
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