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26 Combinatorial AnalysisProperties

§26.15 Permutations: Matrix Notation

The set 𝔖n (§26.13) can be identified with the set of n×n matrices of 0’s and 1’s with exactly one 1 in each row and column. The permutation σ corresponds to the matrix in which there is a 1 at the intersection of row j with column σ⁡(j), and 0’s in all other positions. The permutation 35247816 corresponds to the matrix

26.15.1 [0010000000001000010000000001000000000010000000011000000000000100]

The sign of the permutation σ is the sign of the determinant of its matrix representation. The inversion number of σ is a sum of products of pairs of entries in the matrix representation of σ:

26.15.2 inv(σ)=∑ag⁢h⁢ak⁢ℓ,

where the sum is over 1≤g<k≤n and n≥h>ℓ≥1.

The matrix represents the placement of n nonattacking rooks on an n×n chessboard, that is, rooks that share neither a row nor a column with any other rook. A permutation with restricted position specifies a subset B⊆{1,2,…,n}×{1,2,…,n}. If (j,k)∈B, then σ⁡(j)≠k. The number of derangements of n is the number of permutations with forbidden positions B={(1,1),(2,2),…,(n,n)}.

Let rj⁡(B) be the number of ways of placing j nonattacking rooks on the squares of B. Define r0⁡(B)=1. For the problem of derangements, rj⁡(B)=(nj). The rook polynomial is the generating function for rj⁡(B):

26.15.3 R⁡(x,B)=∑j=0nrj⁡(B)⁢xj.

If B=B1∪B2, where no element of B1 is in the same row or column as any element of B2, then

26.15.4 R⁡(x,B)=R⁡(x,B1)⁢R⁡(x,B2).

For (j,k)∈B, B∖[j,k] denotes B after removal of all elements of the form (j,t) or (t,k), t=1,2,…,n. B∖(j,k) denotes B with the element (j,k) removed.

26.15.5 R⁡(x,B)=x⁢R⁡(x,B∖[j,k])+R⁡(x,B∖(j,k)).

Nk⁡(B) is the number of permutations in 𝔖n for which exactly k of the pairs (j,σ⁡(j)) are elements of B. N⁡(x,B) is the generating function:

26.15.6 N⁡(x,B)=∑k=0nNk⁡(B)⁢xk,

and

26.15.7 N⁡(x,B)=∑k=0nrk⁡(B)⁢(n−k)!⁢(x−1)k.

The number of permutations that avoid B is

26.15.8 N0⁡(B)≡N⁡(0,B)=∑k=0n(−1)k⁢rk⁡(B)⁢(n−k)!.

Example 1

The problème des ménages asks for the number of ways of seating n married couples around a circular table with labeled seats so that no men are adjacent, no women are adjacent, and no husband and wife are adjacent. There are 2⁢(n!) ways to place the wives. Let B={(j,j),(j,j+1)| 1≤j<n}∪{(n,n),(n,1)}. Then

26.15.9 rk⁡(B)=2⁢n2⁢n−k⁢(2⁢n−kk).

The solution is

26.15.10 2⁢(n!)⁢N0⁡(B)=2⁢(n!)⁢∑k=0n(−1)k⁢2⁢n2⁢n−k⁢(2⁢n−kk)⁢(n−k)!.

Example 2

The Ferrers board of shape (b1,b2,…,bn), 0≤b1≤b2≤⋯≤bn, is the set B={(j,k)| 1≤j≤n,1≤k≤bj}. For this set,

26.15.11 ∑k=0nrn−k⁡(B)⁢(x−k+1)k=∏j=1n(x+bj−j+1).

If B is the Ferrers board of shape (0,1,2,…,n−1), then

26.15.12 ∑k=0nrn−k⁡(B)⁢(x−k+1)k=xn,

and therefore by (26.8.10),