Permutations vs Combinations

The key question: does order matter?

• Permutation (order matters): Arranging 3 students in a line — ABC is different from BCA.
• Combination (order doesn't matter): Choosing 3 students for a committee — {A, B, C} is the same group regardless of order.

The formulas

Permutations: $P(n, r) = \dfrac{n!}{(n-r)!}$

Combinations: $C(n, r) = \dfrac{n!}{r!(n-r)!}$

Step-by-step: From 5 students (A, B, C, D, E), choose 3

Permutation: Arrange 3 in a line

$P(5, 3) = \frac{5!}{(5-3)!} = \frac{5!}{2!} = \frac{120}{2} = 60$

There are 60 different arrangements: ABC, ACB, BAC, BCA, CAB, CBA, ABD, ADB, ...

Combination: Choose 3 for a committee

$C(5, 3) = \frac{5!}{3! \cdot 2!} = \frac{120}{6 \cdot 2} = 10$

There are 10 different committees: {A,B,C}, {A,B,D}, {A,B,E}, {A,C,D}, {A,C,E}, {A,D,E}, {B,C,D}, {B,C,E}, {B,D,E}, {C,D,E}.

Why the relationship $P = C \times r!$

Each committee of 3 people can be arranged in $3! = 6$ ways. So $60 = 10 \times 6$. Permutations count each group multiple times (once per arrangement); combinations count each group once.

How to decide: permutation or combination?

Ask: "Would rearranging the same items create a different outcome?"

• Yes → permutation (e.g., passwords, race finishes, seating arrangements)
• No → combination (e.g., lottery numbers, teams, pizza toppings)

Try it in the visualization

Adjust $n$ and $r$. The comparison bar chart shows $P \gg C$ because permutations count more arrangements. The visual grouping shows how $r!$ arrangements of the same group collapse into one combination.