Multilingual hedge fund - Puzzle I'm having difficulty finding the solution for the following problem: A hedge fund has 70 employees. For any two employees $X$ and $Y$ there is a language that $X$ speaks but $Y$ does not, and there is a language that $Y$ speaks but $X$ does not. At least how many different languages are spoken by the employees of this hedge fund? My progress: From the given hint, I know there are 70 unique combinations, such that, for any two sets, there is at least one element that is present in one set and not in the other. From the formula of combination, I have: > $$x C y = 70.$$ I'm stuck at the above point, as there are two unknowns and one equation.

Without knowing the hint, I'm not entirely convinced that finding some minimal $x$ such that $\binom{x}{y} \geq 70$ for some $y$, will actually give you the smallest number of languages $x$. But after a bit of experimentation, this seems to be the right way to go. What you are missing is that for a given $x$, you want the $y$-value which gives you the maximum.

It is fairly well-known (from experience with the binomial distribution, anyway) that to maximize $\binom{x}{y}$ for $y$, pick $y = \lfloor x/2 \rfloor$. Also, the values $\binom{x}{y}$ will then grow roughly with $\sqrt{x!}$, so to get something as small as 70 it's easiest to just guess and check small numbers.

And indeed, $\binom{8}{4} = 70$.