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(a) In how many ways can a party of 4 boys and 4 girls be seated at a circular table so that no 2 boys are adjacent 

(b) In how many ways can 10 boys and 5 girls sit around a circular table, so that no two girls sit together?

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(a) Let the girls first take up their seats. They can sit in 3 ! ways. When they have been seated, then there remain 4 places for the boys each between two girls. Therefore the boys can sit in 4 ! ways. Therefore there are 3 ! × 4 !, i.e., 144 ways of seating the party. 

(b) Let B, denote the position of the boys around the table. 10 boys can be seated around the table in 9 ! ways. 

There are 10 spaces between the boys, which can be occupied by 5 girls in 10P5 ways. Hence, 

Total number of ways = 9! x 109!5 = \(\frac{9!10!}{5!}.\)

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