Is $\binom{52}{n}\cdot\binom{52-n}{n}$ maximised by $n=\frac{52}{3}$? If so why? I've been thinking a lot about cards, and recently about the combination of combination of hands....if you drew $n$ cards from a deck of $52$, and then drew another $n$ cards, how many combinations are there?
The number of combinations appears to peak at about $1\times10^{23}$ so I threw in my best guess at the peak number, 17, I realised that it would have to be the midpoint between the maximal of $\binom{52}{n}$ and $\binom{52-n}{n}$, which is going to be $\frac{26+13}{2}=17\frac13$ or $\frac{52}{3}$. Putting that in gives me $1.01446\times10^{23}$!
But all of this is from intuition, and I have no reason to believe this holds on all cases, nor do I know of the general case.
Is that number the maximising number? If so why, and is there a general case?
My guess at a general solution would be something like:
$$\prod_{i=0}^n\binom{m-i\cdot n}{n}\text{is maximised by }\begin{cases}\frac{m}{T_n(i+1)+1} & \text{if } i=0\\ \frac{m}{T_n(i)+1} & \text{if } i>0\end{cases}$$
But I've not formal Mathematics knowledge (just University level physics) so I'm not certain. Am I right? Why?
 A: Let $a_n$ be your product of binomial coefficients. It is clear that $a_n$ starts out small. 
We calculate $\dfrac{a_{n+1}}{a_n}$. Looks bad at first, but there is a tremendous amount of cancellation, and after a while we find that
$$\frac{a_{n+1}}{a_n}=\frac{(52-2n)(52-2n-1)}{(n+1)^2}.$$
We can now use a little algebra to find out where the  function 
$$f(t)=\frac{(52-2t)(52-2t-1)}{(t+1)^2}$$
is greater than $1$, and where it is less than $1$. This turns quickly into a problem about a quadratic. 
We get equality at roughly $16.8$. So for $n\le 16$, $a_{n+1}$ is "better" than $a_n$. The maximum value of $a_n$ is therefore at $n=17$. 
Details: We have 
$$a_n=\frac{52!}{n!(52-n)!}\frac{(52-n)!}{(n!)(52-2n)!}=\frac{52!}{n!n! (52-2n)!}.$$
Mechanically, it follows that 
$$a_{n+1}= \frac{52!}{(n+1)!(n+1)!(50-2n)!}.$$
Divide the second by the first. We get 
$$\frac{a_{n+1}}{a_n}=\frac{n!n!(52-2n)!}{(n+1)!(n+1)!(50-2n)!}.$$
Now $(n+1)!=n!(n+1)$ and $(52-2n)!=(52-2n)(52-2n-1)(50-2n)!$, and we have our simplified ration.
A: Let us define $\binom{n}{k}$ for non-integer $n$ and $k$ by:
$$ \binom{n}{k} = \frac{\Gamma(n+1)}{\Gamma(n-k+1)\Gamma(k+1)} $$
where $\Gamma(n) = \int_0^\infty x^{n-1} e^{-x} dx$ is the analytical continuation of the factorial (with $\Gamma(n+1) = n!$ for integer $n$). We must do this because the factorial function is not actually defined on the integers. One way to define it is the Gamma function, which matches up with the factorial for integer arguments. Now that we have the Gamma function, our problem reduces to calclus.
Then we have that:
$$ \binom{52}{n} \binom{52-n}{n} $$
$$ \frac{\Gamma(53)}{\Gamma(53-n)\Gamma(n+1)} \cdot \frac{\Gamma(53-n)}{\Gamma(53-2n)\Gamma(n+1)}$$
$$ \frac{\Gamma(53)^2}{\Gamma(53-2n)\Gamma(n+1)^2}$$
The numerator is a large number ($52!^2$). So let us worry about the denominator instead. The larger the denominator is, the smaller the fraction will be. To maximize the fraction, we should minimize the denominator. So:
$$ 0 = \frac{d}{dn} \Gamma(53-2n)\Gamma(n+1)^2 $$
$$ 0 = -2\Gamma^\prime(53-2n)\Gamma(n+1)^2 + 2\Gamma^\prime(n+1)\Gamma(n+1)\Gamma(53-2n) $$
$$ \Gamma^\prime(53-2n)\Gamma(n+1) = \Gamma^\prime(n+1)\Gamma(53-2n) $$
Which looks extremely ugly. One obvious way they could be equal is if $n+1 = 53-2n$, because then, by simple substitution, the two sides are equal. Solving for $n$, we get $n = \frac{52}{3}$. 
A: Hint Hope I'm not too late to the party. Let's consider logarithms since they are monotonically increasing and can do interesting things to products. Log of a product is a sum of logs: $$\log\left(\prod_{\forall i}a_i\right) = \sum_{\forall i}\log(a_i)$$ And the log of a factorial is a sum which can be approximated (the typical calculus rectangle area estimates) with an integral of the log function. Then we want to differentiate and set equal 0. The derivative of the integral of a log takes us back to log so using this approach we should be ending up with an equation involving a few logarithm terms instead which we can then solve for numerically or try and further use log-laws upon.
