Permutations and combinations.
Permutations count ordered arrangements; combinations count unordered selections. The first question to ask: does swapping two picks give a different answer?
Fifty athletes are running a race; medals go to the gold, silver, and bronze finishers. How many ways can the three medals be distributed? L2's counting principle handles this almost directly — 50 choices for gold, but once that's set there are only 49 left for silver, then 48 for bronze. The medals come out to 50 × 49 × 48 = 117,600.
Now flip the question. Same 50 athletes, but instead of medals you're choosing a three-person committee. Three seats, all equal in rank, all interchangeable. The committee {Alice, Bob, Carol} is the same committee no matter which order you pick the members in. How many committees? Definitely fewer than 117,600 — each committee got counted six times by the medal count (once for each of the 3! = 6 orderings). So 117,600 / 6 = 19,600 committees.
The difference between these two problems is the difference between a permutation (order matters, like medals) and a combination (order doesn't, like a committee). Same 50 athletes, same 3 picks, but two very different answers. The first question to ask on any "how many ways" problem is — does swapping two picks change the answer?
Order matters? Permutation. Order doesn't? Combination.
Permutation — order matters. The number of ordered arrangements is P(n, r) = n × (n−1) × ··· × (n−r+1) = n! / (n−r)!.
Combination — order doesn't. The number of unordered selections is C(n, r) = P(n, r) / r! = n! / [(n−r)! · r!].
A combination is always smaller than the corresponding permutation by exactly a factor of r! — the number of ways to reorder r picks. When order doesn't matter, all those reorderings collapse into one selection, so we divide them out.
Both formulas reduce to the counting principle with shrinking choices: 50, 49, 48 for r=3 picks from n=50, and so on.
Left: medals are distinguishable (gold vs silver vs bronze), so a different award assignment is a different arrangement — permutation. Right: committee seats are identical, so swapping two members gives the same committee — combination. The combination count is the permutation count divided by r! = 3! = 6.
A permutation is an ordered arrangement: P(n, r) = n! / (n−r)!. A combination is an unordered selection: C(n, r) = n! / [(n−r)! · r!]. They differ by the factor r!, which counts the ways to reorder any single selection of r items.
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- Factorial (n!) n! = n × (n−1) × (n−2) × ··· × 1. So 3! = 6, 4! = 24, 5! = 120, 10! = 3,628,800. By convention, 0! = 1. The factorial counts the number of ways to arrange n distinct items in a row.
- P(n, r) — permutation Number of ordered ways to pick r items from n: P(n, r) = n × (n−1) × ··· × (n−r+1), which is r terms total. Equivalently, n!/(n−r)!. Use when the position or identity of each pick matters.
- C(n, r) — combination Number of unordered ways to pick r items from n: C(n, r) = P(n, r) / r!. Use when the order or position doesn't matter and every selection of the same r items is the same answer.
- The litmus test "Does swapping two of my picks give me a different answer?" If yes (medals, schedules, podium positions), it's a permutation. If no (committees, card hands, pizza toppings), it's a combination. Ask this question first, before reaching for any formula.
- Shrinking product The descending multiplication n × (n−1) × (n−2) × … that shows up in every permutation. P(10,3) = 10 × 9 × 8 = 720. Stop after r factors. The factorial is just the shrinking product all the way down to 1.
Distributing three medals to 50 athletes.
"50 athletes are running a race. A gold medal is to be given to the winner, a silver medal to the second-place finisher, and a bronze medal to the third-place finisher. Assume that there are no ties. In how many possible ways can the 3 medals be distributed?"
Apply the litmus test.
If athletes Alice, Bob, and Carol finish 1st-2nd-3rd, that's a different outcome than Carol-Alice-Bob finishing 1st-2nd-3rd. The order changes who gets which medal. Order matters — this is a permutation.
→ permutation, not combinationIdentify n and r.
Picking 3 medalists out of 50 athletes: n = 50, r = 3.
Compute the permutation.
P(50, 3) is the product of r = 3 shrinking choices starting from n = 50:
Compute in two steps:
2,450 × 48 = 117,600
Just over a hundred thousand possible medal distributions for a single 50-athlete race.
→ multiply 3 shrinking choicesThree problems. Litmus test first.
Aldo has 15 European cities he'd like to visit. On his next vacation, he has time for only 3: one on Monday, one on Tuesday, one on Wednesday. He won't visit the same city twice. How many different schedules are possible?
A club has 12 members and wants to select a 3-person subcommittee (no titles — all three roles are equal). How many different subcommittees are possible?
A pizzeria has 10 toppings. How many different 3-topping pizzas can be made? (All toppings are added at once; the order doesn't matter.)
Three fast questions before you move on.
Q1. Which scenario is a permutation (order matters)?
Q2. Compute P(8, 2).
Q3. Why is C(n, r) always ≤ P(n, r) for r ≥ 1?
Order is the question; everything else follows.
Most counting questions in life are one of these two flavors. Sports lineups (permutation: batting order matters), poker hands (combination: 5 cards is 5 cards regardless of dealing order), lottery picks (combination: 6 numbers is 6 numbers), passwords (permutation: ABCD ≠ DCBA), committee assignments (combination: three interchangeable seats), seating arrangements (permutation: chair 1 ≠ chair 2).
Once you train your eye for the litmus test, the formulas become rote: pick r out of n, multiply the shrinking choices, and divide by r! if order doesn't matter.
Next: Lesson 5 introduces expected value — turning the probabilities and counts we've built so far into an average payout. The math behind fair games, insurance premiums, and the casino's edge.
Continue to Lesson 05Different angle? Need another rep? These are optional — tap any that look helpful.