Inclusion exclusion — the series

All 24 arrangements of 4 objects, and the 9 that move every one. Every permutation of 4 objects drawn as a grid of cells, with the diagonal — where an object stays where it began — shaded, and the arrangements that avoid it entirely marked.

Nobody gets their own hat

Hand back a pile of hats at random and ask for the chance that not one person gets their own. The answer barely moves as the crowd grows — it is a third and a bit at four people, and a third and a bit at four thousand.

part 1 · probability

A 6-sided die rolled 6 to 40 times: some face missing, and the sum stopped early. Curves of the chance that some face of a die has not yet appeared against the number of rolls, with the inclusion–exclusion sum stopped after one, two and three terms drawn around the exact curve.

A sum stopped early still says something

Inclusion–exclusion corrects an overcount, then the correction's overcount, and so on to the end. Stop after any number of terms and the result is not merely an approximation: after an odd number it is too high and after an even number too low, always. So two or three terms bracket an answer whose full sum is out of reach — as long as the events being counted are rare.

part 2 · probability

How many get their own hat

The chance that nobody gets their own hat settles on 1/e. The chance that exactly one person does settles on 1/e too, exactly two on 1/(2e), exactly three on 1/(6e) — the Poisson distribution with mean 1. The reason is a set of averages that come out exactly 1 at every size, and the counts reach the limit so fast that eight hats are within six ten-thousandths of it.

part 3 · probability

Permutations that avoid 8 forbidden cells. A 5 by 5 grid with 8 forbidden cells shaded and one permutation that avoids them marked, beside the numbers of ways to place non-attacking rooks on the forbidden cells and the count of avoiding permutations they give.

The cells a permutation must miss

A derangement is a permutation that misses the diagonal of a square grid. Forbid any other set of cells instead and inclusion–exclusion still counts what is left — driven entirely by one list of numbers, the ways to place non-attacking rooks on the forbidden cells. Boards that look nothing alike can share that list, and rooks on a staircase turn out to count the ways to split a set.

part 4 · probability

5 couples seated so that no one sits beside a partner. A round table with 10 seats alternating women and men, labelled by couple, arranged so that no man sits next to his partner, with the number of such arrangements.

A round table with no couple together

Seat n couples round a table, men and women alternating, so that nobody sits beside their partner. Once the women are placed the men face a board of forbidden cells that bends round a corner — and that corner is the whole difficulty. The forbidden cells form a cycle, a count of non-adjacent points on a cycle finishes the problem, and the chance of a good seating creeps towards e^(−2) far more slowly than the hat problem reaches 1/e.

part 5 · probability

A single table for 9 guests over 4 nights. Small circles of 9 guests, one per night, each showing the night's seating as a closed zigzag path; every pair of guests is adjacent in exactly one of them.

Every pair side by side, once

Seat an odd number of guests at round tables for as many nights as it takes, the same table sizes every night, so that every two guests sit side by side on exactly one night. For a single table a zigzag turned a notch each night does it for any number of guests. For other table plans the answer is almost always yes — and for six guests at two tables of three, nine at tables of four and five, and eleven at three, three and five, an exhaustive search proves it is no.

part 6 · probability