Count arrangements with product, sum and restriction rules

Multiply stages; add disjoint cases

Multiply the number of first-stage choices by the fixed number of permitted second choices for each first-stage result. Independence is not required; a constant number of permitted continuations is. Add counts only when cases are disjoint, so no outcome is counted twice

A label has three ordered positions, each allowing any of four symbols with repetition. How many labels exist?

A label has three ordered positions filled with different symbols chosen from four available symbols. How many labels exist?

Identical copies remove repeated descriptions

If n positions contain identical copies of each of several distinguishable types, with multiplicities summing to n, divide n! by the factorial of every multiplicity. Exchanging two identical copies does not create a new arrangement

Arrange seven tiles in a line: three identical red, two identical blue and two identical green. How many distinct colour orders exist?

State what counts as the same circle

For at least one distinct object around an unmarked circle, if rotations are equivalent but reflections are distinct, fix one object as an anchor and arrange the others. The count is the factorial of one fewer than the number of objects. Labelled seats instead give the factorial of the number of objects. If reflections are also equivalent, that is a different convention requiring separate justification

Use a block only for a required consecutive group

If three specified people must be consecutive in a line of six distinct people, treat them as one block with the other three: 24 outer orders and six internal orders, giving 144. To require two specified people not adjacent, subtract the adjacent-block count from all arrangements; the two cases must be complementary

Justify the repeated-object count, ungraded

Let n be the total number of objects and m₁, …, mₜ their type multiplicities, whose sum is n. Prove the displayed formula for distinct linear type-orders. Explain why every visible arrangement has the same number of artificially labelled versions

Model: remove artificial labels

Let n be the total number of objects, and let mᵢ be the number of copies of type i. Temporarily label every copy, giving the factorial of the total number of objects in labelled arrangements. For any fixed visible type-order, the labels of type i can be assigned to the positions for that type in the factorial of its multiplicity ways. The choices for the different types multiply, so every visible arrangement has the product of the multiplicity factorials in labelled versions. These equal groups partition all labelled arrangements, giving the quotient