In how many different ways can five boys and five girls form a circle such that the boys and girls alternate?

 After fixing up one boy on the table, the remaining can be arranged in $4 !$ ways but boys and girls are to alternate. There will be $5$ places, one place each between two boys these five places can be filled by $5$ girls in $5 !$ ways.

Hence, by the principle of multiplication, the required number of ways $=4!\times 5!=2880.$