Bijections, Homothety & Structured Counting

Study Sheet

Bijections, Homothety & Structured Counting

Reflection, rooks and derangements, Fibonacci diagonals, matchings by inclusion–exclusion, homothety in figures

Counting by Correspondence

Tip
The reflection bijection

Paths from (0,0)(0,0) to (n,n)(n,n) that never go above y=xy = x number (2nn)(2nn1)=Cn\binom{2n}{n} - \binom{2n}{n-1} = C_n: reflect each bad path across y=x+1y = x + 1 after its first touch to get a bijection with all paths to (n1,n+1)(n-1, n+1).

A bijection is a proof when both the map and its inverse are described and shown to land in the right set; "reflect" must come with "and the reflected path ends at (n1,n+1)(n-1, n+1), and reflecting again recovers the original". Catalan numbers count triangulations, parenthesizations, and non-crossing matchings by the same recurrence — an answer of 132132 or 429429 is a strong hint that Catalan is behind it.

Tip
Rooks, permutations, and forbidden positions

Non-attacking rook placements on an n×nn\times n board are permutations; forbidding the main diagonal gives derangements Dn=n!kn(1)k/k!D_n = n!\sum_{k\le n}(-1)^k/k!, and forbidding neighbors around a circle of 2n2n people gives a matching count by inclusion–exclusion over the cycle's edges: k(1)kmk(2n2k1)!!\sum_k(-1)^km_k(2n - 2k - 1)!! with mk=2n2nk(2nkk)m_k = \tfrac{2n}{2n-k}\binom{2n-k}{k}.

Inclusion–exclusion over a STRUCTURED family (the edges of a cycle, the diagonal cells) needs the count mkm_k of kk pairwise-compatible forbidden objects — for a cycle that is the number of kk-matchings, a Lucas-number-like quantity. Write the general term, compute mkm_k separately, and only then assemble the alternating sum.

Tip
Fibonacci diagonals and subset conditions

Subsets of {1..n}\{1..n\} whose size equals their minimum number FnF_n: sum (nkk1)\binom{n-k}{k-1} over kk, a shallow diagonal of Pascal's triangle. Subsets with no two consecutive elements number Fn+2F_{n+2}.

Two proofs to know: the diagonal identity j(njj)=Fn+1\sum_j\binom{n-j}{j} = F_{n+1} by Pascal's rule (both sides satisfy the Fibonacci recurrence), and the tiling bijection (squares and dominoes). When a count comes out Fibonacci, one of these is the intended proof.

Tip
Homothety in figures

The medial triangle is the image of ABCABC under the homothety at the centroid with ratio 12-\tfrac12; the anticomplementary triangle (lines through each vertex parallel to the opposite side) has ratio 2-2 and four times the area. Tangent circles are homothetic from their tangency point with ratio ±r2/r1\pm r_2/r_1.

A homothety sends lines to parallel lines, circles to circles, and scales every length by k|k| and every area by k2k^2. Locating its center — a centroid, a tangency point, an intersection of common tangents — is the whole problem; the ratio then reads off any pair of corresponding lengths.

Try it
Try it: Catalan by reflection

How many sequences of 55 H's and 55 T's have every prefix containing at least as many H's as T's?

(105)(104)=252210=42=C5\binom{10}{5} - \binom{10}{4} = 252 - 210 = 42 = C_5.