= Solution
Every allocation is determined by the set of $n$ children on the red bus, so there are $\binom{2n}{n}$ equally likely allocations. Exactly one places every boy on red and every girl on green. Hence
$$
\mathbb P(A_n)=\binom{2n}{n}^{-1}=\frac{(n!)^2}{(2n)!}.
$$
Applying <Stirling formula> to the numerator and denominator gives the <central binomial coefficient> asymptotic
$$
\binom{2n}{n}\sim\frac{4^n}{\sqrt{\pi n}},
$$
and therefore
$$
\boxed{\mathbb P(A_n)\sim\frac{\sqrt{\pi n}}{4^n}}.
$$
Back to article page
