Count ordered arrangements using P(n,r) = n!/(n−r)! and understand when order matters.
P(n,r) = n! / (n−r)!
Order matters — ABC and BAC are different permutations. When you fill the first slot you have n choices, then n−1 remain for the second, and so on.
The falling factorial form makes this concrete:
P(n,r) = n × (n−1) × (n−2) × … × (n−r+1)
Special cases: P(n,0) = 1 (one way to choose nothing), P(n,n) = n! (arrange all items).
P(n,r) = r! × C(n,r)
Permutations count ordered arrangements — the order ABC ≠ BAC ≠ CAB.
Combinations count unordered subsets — {A,B,C} is the same group regardless of listing order.
Since each combination of r items can be arranged in r! ways, multiplying C(n,r) by r! recovers P(n,r).
One-on-one Algebra 2 tutoring builds strong intuition for when to use permutations vs combinations — we work through your actual homework and tests so the strategy clicks.