Mixed A1–B2 Timed Sets + Finish-Clean Drills

Vandermonde: $\sum_k \binom{m}{k}\binom{n}{r-k} = \binom{m+n}{r}$ — choose $r$ people from $m$ men and $n$ women, split by how many men. Hockey stick: $\sum_{i=r}^{n} \binom{i}{r} = \binom{n+1}{r+1}$ — choose $r+1$ from $\{0,\dots,n\}$, split by the LARGEST element chosen. Committee-chair: $k\binom{n}{k} = n\binom{n-1}{k-1}$ — pick a committee with a chair, chair first or committee first. Each identity is one sentence naming the set and one sentence naming the two splits; that pair of sentences IS the proof, and it is the form you can reproduce at minute 80 of a session.

Compositions and partitions: compositions of $n$ (ordered) number $2^{n-1}$ — place bars in the $n-1$ gaps of a row of stones; with exactly $k$ parts, $\binom{n-1}{k-1}$ (stars and bars' cousin). Nonnegative solutions of $x_1 + \cdots + x_k = n$: $\binom{n+k-1}{k-1}$ — the workhorse. Partitions (unordered) have no simple finite closed form, which is exactly why they travel with generating functions: $\prod_{k \ge 1} \frac{1}{1-x^k}$, and Euler's trick of matching factor sets proves gems like 'partitions into odd parts = partitions into distinct parts' by pairing $\frac{1}{1-x^{2k-1}}$ against $(1+x^k)$ products.

The snake-oil method (Wilf's name, worth knowing) mechanizes hard sums: to evaluate $f(n) = \sum_k g(n, k)$, compute the generating function $\sum_n f(n) x^n$, SWAP the order of summation, recognize the inner sum in closed form, and read off coefficients. It converts cleverness-demanding identities into routine double-sum bookkeeping — the method to reach for when a binomial sum has no visible story.

Pitfall. Reaching for induction on identities that have a story. Induction on $\sum \binom{n}{k}^2$ works but costs fifteen minutes of algebra; the double count costs three sentences. On identities, always hunt the story first.

Catalan numbers $C_n = \frac{1}{n+1}\binom{2n}{n}$ count an absurd range of structures: balanced parenthesizations, monotone lattice paths that never cross the diagonal, triangulations of an $(n+2)$-gon, binary trees with $n$ internal nodes, non-crossing chord diagrams. The exam move is RECOGNITION: when a counting problem has a 'never go below zero' or 'non-crossing' or 'properly nested' constraint, conjecture Catalan, verify on $n = 1, 2, 3$ ($1, 2, 5$), then prove by bijection to a known Catalan family or by the reflection principle.

The reflection principle proves the path count directly: paths from $A$ to $B$ that TOUCH a forbidden line correspond bijectively to unrestricted paths from $A'$ (the reflection of $A$) to $B$ — reflect the prefix up to the first touch. Bad counts become unrestricted counts of a shifted problem, and subtraction finishes. This one bijection also settles the ballot problem: the probability that candidate $A$, winning $p$ votes to $q$ with $p > q$, stays STRICTLY ahead throughout the count is $\frac{p-q}{p+q}$.

The recurrence $C_{n+1} = \sum_{i} C_i C_{n-i}$ (split at the first return to zero / the root of the tree) is the other signature, and it is worth knowing that this recurrence plus generating functions yields the closed form: $C(x) = 1 + xC(x)^2$, solve the quadratic, expand the square root binomially. When a problem produces THAT recurrence, you are holding Catalan whether or not paths are visible.

Pitfall. Fuzzy 'first touch' bookkeeping. The reflection bijection needs a well-defined first crossing to reflect at — state it, and check the map is a bijection by describing its inverse (reflect back at the same first touch). A reflection argument without the inverse direction is half a proof.

This week's timed sets present A1-B2 difficulty MIXED in one sitting — the first time training matches the real December shape: three problems, ninety minutes, no labels telling you which is the gentle one. The skill being trained is ORDERING: five minutes reading all three, ranking by expected solve time for YOU (not by slot number), and committing to the plan. Putnam problem numbers estimate global difficulty; your personal ranking beats them often enough that the read-and-rank ritual pays for its five minutes.

Finish-clean discipline: a solved problem is not banked until the write-up would survive a hostile grader — every claim justified, edge cases closed ($n = 1$! empty sets! division by the thing that might be zero!), equality conditions stated. The finish-clean drill this week isolates that skill: take solved problems and harden the write-ups. Most solvers lose 2-4 points per session to write-up leaks on problems they SOLVED; this is the cheapest score improvement available to you.

Between-problem hygiene: when you park a problem at a checkpoint, write one line — current best idea, what you would try next — before switching. Re-entry costs three minutes when the parking note exists and ten when it does not; over a session with two switches, the notes buy you a quarter of a problem.

Pitfall. Reading only the first problem and diving. The five-minute read of all three is the highest-EV act in the session; the difference between attacking your best problem and the paper's first problem is routinely a full solve.