Answer
$15$ orders.
Work Step by Step
It is necessary to choose two of the six positions for boys.
Combination of 6 distinct childern taken 2 at a time:
$_6C_2=\frac{6!}{2!(6-2)!}=\frac{6!}{2!\times4!}=\frac{6\times5\times4\times3\times2\times1}{2\times1\times4\times3\times2\times1}=\frac{720}{48}=15$
Notice that, if you decide to choose four of the six positions for girls, the result remains the same:
Combination of 6 distinct childern taken 4 at a time:
$_6C_4=\frac{6!}{4!(6-4)!}=\frac{6!}{4!\times2!}=15$
