Partitions and Their Counting I recently watched the movie on the Life of Ramanujan which had a scene about his work on partitions where he claims to have developed a formula and is challenged by a professor who takes up a big number and asks him to computer partitions by his formula while he does it manually. 
The number was very large and it must have been very difficult to keep track ot the partitions already made so how exactly did early mathematicians keep track of the one's they had already made?
Was there some technique or did they use some efficient method of sorting?
 A: The movie in question is The Man Who Knew Infinity based on Ramanujan's biography of the same name written by Robert Kanigel. 
The professor in question was P. A. MacMahon and the discussion was about evaluating $p(200)$.
MacMahon used the following recursive formula $$p(n) = p(n – 1) + p(n – 2) – p(n – 5) – p(n – 7) + p(n – 12) + p(n – 15) – \cdots=\sum_{j=1}^{\infty} (-1)^{j-1}\left\{p\left(n-\frac{3j^2-j}{2}\right)+p\left(n-\frac{3j^2+j}{2}\right)\right\}$$ to evaluate $p(200)$ and he got $$p(200)=3972999029388.$$ The use of the formula above requires one to evaluate all the values of the partition function upto $p(199)$ before evaluating $p(200) $. MacMahon was a great expert in numerical calculations and he manually made of a table of partition values till $p(200)$ using the above formula. In short it was not based on any magic but involved huge amount of labor. Most mathematicians (including Gauss and Ramanujan) are capable of performing such feat with reasonable effort. 
Ramanujan on the other hand used deep methods of complex analysis (with support from G H Hardy) and his own intuition about elliptic function theory to derive an asymptotic formula for evaluation of $p(n) $. Eight terms of this formula gave the value of $p(200)$ with an error of $0.004$. How Ramanujan got his intuition about these topics is mysterious and no one really knows.
