Identity, Maths Club of IISER Kolkata

Gourav Banerjee, a 21MS student, goes to the main canteen of IISER Kolkata for dinner at some arbitrarily scheduled time between 8 and 9 pm. He frequently meets an anonymous, beautiful girl in the mess and begins to wonder whether the girl is stalking him or if their meeting is just a coincidence. So he tries to compute the probability of meeting that girl in the mess during dinner time given the following constraints: Both Gourav and the girl go to mess for having dinner at some random time between 8 - 9 pm. Because of the Queue at the mess, both stay in the mess for minimum of 30 min. What do you think? Solution Let $x$ denote the time when Gourav enters the mess and let y denote the time when girl enters the mess. Here we take origin to be the 8 pm mark and a distance of 1 unit represents 1 hour on both $x$ and $y$ axis so all possible coordinates within the unit square $ABCD$ represents an event where Gourav and the girl both visit the canteen. Now the favourable coordinates which ...

Here's a curious polynomial identity I stumbled upon: for $r, n \in \mathbb{N}$ and $r > n$, \[ f(n, r) = r^n - \binom{r}{1}(r - 1)^n + \binom{r}{2}(r - 2)^n + \dots + (-1)^{r - 1}r = 0. \] This can be proved by brute force expansion, but there is a really nice combinatorial argument in connection with the following problem. How many surjective functions of the form $f\colon \{1, 2, \dots, n\} \to \{1, 2, \dots, r\}$ are there? Consider such a function $f$; for each $i \in \{1, 2, \dots, n\}$, we have $r$ choices for $f(i)$, giving us at most $r^n$ functions. However, $f$ must be surjective, so we must get rid of those functions which fail to hit some $j \in \{1, 2, \dots, r\}$ in the codomain. There are $(r - 1)^n$ functions whose codomain misses a particular $j$, and $\binom{r}{1}$ ways to choose this missing element, hence we take away their product. This isn't the end, since now we've removed those functions which miss at least two points ,...