The Empty Arrangement

At a glance

• Part 1: 2! =2, 1! =1, and the untouched zero-seal display is exactly one complete arrangement.
• Part 2: Choosing the first case gives 𝑛! =𝑛(𝑛 −1)!. At 𝑛 =1, this forces 0! =1.
• Part 3: Every collection has one identity permutation. For the empty collection, it is the unique empty instruction list.

Key idea: no objects is not the same as no valid way

An arrangement is a complete assignment, not a pile of marks. When there are no seals and no cases, the blank assignment already meets every requirement: no seal is omitted, no case is unfilled, and no case receives two seals.

There is one such blank assignment. A second one would need to differ somewhere, but there is no seal or case where a difference could occur.

1. Prove the empty count

For two seals 𝐴,𝐵, the complete arrangements are:

1. case 1 contains 𝐴, and case 2 contains 𝐵;
2. case 1 contains 𝐵, and case 2 contains 𝐴.

Seal 𝐴 must occupy either case 1 or case 2. Once that choice is made, the location of 𝐵 is forced. These two cases are disjoint and exhaustive, so

For one seal 𝐴 and one case, the only complete arrangement places 𝐴 in that case. Thus

For zero seals and zero cases, the untouched display is valid:

• no seal is unused, because there are no seals;
• no case is empty, because there are no cases; and
• no case contains more than one seal.

It is also the only possible arrangement. Any different arrangement would have to contain at least one seal-case assignment, but neither a seal nor a case exists. Therefore

2. Derive the factorial recurrence

For 𝑛 ≥1, build a complete arrangement by first choosing the seal for case 1.

• There are 𝑛 choices for case 1.
• After one seal is chosen, 𝑛 −1 distinct seals remain.
• Those remaining seals can be arranged in the remaining cases in (𝑛 −1)! ways.

Each finished arrangement has exactly one seal in case 1, so it appears in exactly one of these 𝑛 groups. The groups are disjoint and cover every arrangement. By the multiplication principle,

Substitute 𝑛 =1:

Since 1! =1,

and therefore

This agrees with the one empty arrangement from Part 1. It also agrees with the product form

When 𝑛 =0, that list contains no factors, and the empty-product rule assigns it the value 1.

3. Find the identity permutation

A permutation records one destination for every object, with every destination used exactly once. The identity permutation sends each object 𝑥 to itself.

For any collection 𝑆, every arrow in an identity permutation is forced:

for each 𝑥 in 𝑆. Therefore there can be at most one identity permutation. The collection of all these self-arrows is a valid reversible permutation, so an identity permutation also exists. Hence it is unique.

If 𝑆 is empty, there are no inputs and no destinations. The identity permutation is the empty instruction list: it contains no arrows. It is valid because no input or destination is omitted, and it is unique because there is no possible arrow by which another instruction list could differ.

Check

The three arguments agree:

• direct counting finds one complete zero-object arrangement;
• the factorial recurrence forces 0! =1; and
• the empty collection has one identity permutation, represented by the same empty assignment.

Each argument supplies both existence and exclusion, so “exactly one” is proved rather than inferred from a pattern.

Teaching note

The proof uses a general logical pattern sometimes called vacuous truth: a requirement such as “every seal is used” cannot fail when there are no seals. Students do not need that term. A complete response should say why the blank arrangement satisfies every rule and why a second arrangement would require an impossible seal-case pair.