Counting: a team or different roles

Choose two of four colleagues A,B,C,D. Why are there 6 teams but 12 coordinator–presenter assignments? The same people produce different outcomes when roles or order matter. Define an outcome before choosing a formula. Use powers and product identities.

Add or multiply

For one of two disjoint alternatives — one of 3 paper or 2 electronic formats — there are 3+2=5 choices. Overlap would count shared outcomes twice.

For staged choices with the same number of continuations after each first choice, multiply: 3 formats × 2 languages =6 packages. If continuation counts depend on the branch, count each branch separately and add. This counting rule does not assume probabilistic independence.

Roles: 12 ordered pairs

Teams: 6 unordered pairs

A,B,C,D are hypothetical participants. Tree: 4 choices for the first role, then 3 for the second; AA and other repeats are forbidden. Below, AB and BA merge. Diagram not to scale.

Order creates extra outcomes

The factorial n! is 1·2·…·n; 0!=1. Without repetition, for integers 0≤k≤n: an arrangement A selects k distinct objects in order, a permutation P orders all n, and a combination C selects without order:

$A_n^k=\frac{n!}{(n-k)!},\qquad P_n=n!,\qquad C_n^k=\frac{n!}{k!(n-k)!}.$

Choose the coordinator in 4 ways and a different presenter in 3. For a team AB and BA are identical, so divide by 2!:

$A_4^2=4\cdot3=12,\qquad C_4^2=\frac{12}{2!}=6.$

Generally each k-person team has k! orderings. Division by k! removes uniform overcounting; do not divide without checking that multiplicity is uniform. OpenStax

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Repetition, Pascal and the binomial theorem

With repetition and all n types available at each of k positions, there are nᵏ sequences: n choices at every position. If groups of identical objects have sizes m₁,…,mᵣ totaling n, use the second formula for their permutations:

$n^k;\qquad\frac{n!}{m_1!\cdots m_r!}\quad(m_1+\cdots+m_r=n).$

AABC gives 4!/2!=12 different strings. If order does not matter and k objects are selected from n types with unlimited supply:

$C_{n+k-1}^k\quad(n\ge1,\ k\ge0).$

Write k stars and n−1 separators. Star counts between separators specify each type’s count, including zero. This is a bijection. Two objects from three types give AA,AB,AC,BB,BC,CC — 6. MIT, combinations with repetition.

Pascal’s triangle has 1 at its edges; each interior number adds the two above:

$C_n^k=C_{n-1}^{k-1}+C_{n-1}^k\quad(1\le k<n).$

First rows: 1; 1,1; 1,2,1; 1,3,3,1; 1,4,6,4,1. A team either includes a specified person or excludes them, explaining the sum. For nonnegative integer n, the binomial theorem is:

${(a+b)}^n=\sum_{k=0}^n C_n^k a^{n-k}b^k.$

Cₙᵏ counts which k factors supply b while the rest supply a. Thus the cube has coefficients 1,3,3,1, not just two end terms. Binomial coefficients.

Before counting, decide whether swapping objects changes the outcome and whether an object may repeat. Teams and role assignments answer differently even with the same four people.