Stirling's Formula for n! A group of $2N$ boys and $2N$ girls is divided into two equal groups. find the probability $p$ that each group will be equally divided into boys and girls. Estimate $p$ using Stirling's formula.
The author's solution is $$p=\frac{\binom{2N}{N}^2}{\binom{4N}{2N}}.$$ The author's estimate using Stirling's formula is $(\frac{2}{(N\pi)})^{0.5}$. I don't understand how it is calculated?
Stirling's formula is $n!\sim(2\pi)^{0.5}n^{n+0.5}e^{-n}$ where the sign $\sim$ is used to indicate the ratio of two sides tends to unity as $n\rightarrow \infty$.
 A: People have given hints, but sometimes it's good to see these things worked out explicitly.
Write $p$ in terms of factorials:
$$ p = \left( {(2n)! \over (n!)^2} \right)^2 \left( (4n)! \over (2n)!^2 \right)^{-1}  = {(2n)!^4 \over (4n)! (n!)^4 }$$
Then replace each of those factorials with its Stirling approximation:
$$ p \sim {\left(\sqrt{4\pi n} \left( {2n \over e} \right) ^{2n}\right)^4 \over\sqrt{8\pi n} \left( {4n \over e} \right)^{4n}  \left(\sqrt{2\pi n} \left( {n \over e} \right) ^{n}\right)^4}.  $$
Now a good general strategy for dealing with these quotients of Stirling approximations is to group all the constants together, all the factors of $\pi n$ to a constant power together, terms together, all the things like $(an)^{bn}$ together, and all the powers of $e$ together.  This gives
$$ p \sim {4^2 \over 8^{1/2} 2^2} {(\pi n)^2 \over (\pi n)^{1/2} (\pi n)^2} {(2n)^{8n} \over (4n)^{4n} n^{4n}} {e^{-8n} \over e^{-4n} e^{-4n}} $$
Then we can factor a bit more:
$$ p \sim {4^2 \over 8^{1/2} 2^2} {(\pi n)^2 \over (\pi n)^{1/2} (\pi n)^2}{2^8 \over 4^4} \left( {n^8 \over n^4 n^4} \right)^n \left( {e^{-8} \over e^{-4} e^{-4}} \right)^n $$
And you can easily see that this is
$$ p \sim 2^{1/2} \cdot (\pi n)^{-1/2} \cdot 1 \cdot 1^n \cdot 1^n $$
or, after dropping the trivial factors, $2/\sqrt{\pi n}$ like you're looking for.
