The one question that decides which formula to use
Ask: "if I picked the same r items in a different order, would that count as a different outcome?" If yes, use permutation (nPr) — order matters, so ABC and BCA are counted separately. If no, use combination (nCr) — order doesn't matter, so ABC and BCA are the same outcome, counted once. This single question resolves the overwhelming majority of real confusion between the two.
The formulas, and how they relate to each other
nPr = n! ÷ (n − r)! counts every ordered arrangement of r items chosen from n. nCr = nPr ÷ r! counts unordered selections — dividing by r! specifically removes the r! different orderings that each unique group of r items could be arranged in, collapsing them down to one count per group. This is why nCr is always smaller than or equal to nPr for the same n and r: combination is permutation with the ordering information thrown away.
A worked example
Choosing 3 people from a group of 5 for three DIFFERENT roles (president, secretary, treasurer — order matters, since who gets which role is a different outcome): nPr(5,3) = 60 ways. Choosing the SAME 3 people from the same group of 5 to simply form a committee with no distinct roles (order doesn't matter — the same 3 people form the same committee regardless of selection order): nCr(5,3) = 10 ways. The 60-vs-10 gap is exactly a factor of 3! = 6, since each unique group of 3 people can be assigned to the 3 roles in 6 different ways.
Real-world cues that signal which one you need
Permutation situations: assigning distinct positions or ranks (1st/2nd/3rd place, specific job roles, a PIN code where digit position matters), arranging items in a sequence (a race finishing order, a password). Combination situations: forming an unordered group or team, choosing lottery numbers (the numbers drawn don't have a "1st drawn, 2nd drawn" significance to the prize outcome), selecting a subset where membership alone matters, not any internal order.
Why the gap between them grows with r
The permutation-to-combination ratio is always exactly r! — for r=3 it's 6×, but for r=5 it's 120×, and for r=10 it's already over 3.6 million times larger. This is worth remembering when sanity-checking a result: if a permutation count and its corresponding combination count for the same n and r don't differ by exactly r!, one of them was computed incorrectly.