Use permutations and combinations to compute probabilities of compound events

Learning goal: Use permutations and combinations to count outcomes, compute probabilities of compound events in a uniform model by dividing favorable arrangements or selections by the total, and solve problems by choosing the correct counting tool.

Scope note. S.CP.9 is marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate a standard "for students who are planning on taking advanced courses". It is taught here in full, as extension content for students on that path.

Why counting matters for probability. In a uniform model every outcome is equally likely, so a probability is (favorable outcomes) / (total outcomes). When the outcomes are arrangements of several objects or selections of several objects from a larger set, the counts run into the thousands or millions, and listing is impossible. Two formulas do the counting. The question that decides which to use is: does the order of the chosen objects matter?

Permutations: order matters. A permutation is an ordered arrangement. The number of ways to arrange n distinct objects in a row is n! = n × (n − 1) × … × 2 × 1, because there are n choices for the first position, n − 1 for the second, and so on. The number of ways to choose and arrange r of n distinct objects is

nPr = n! / (n − r)! = n × (n − 1) × … × (n − r + 1), the product of r decreasing factors.

Example: 10P4 = 10 × 9 × 8 × 7 = 5,040, the number of four-digit codes using digits 0 to 9 with no digit repeated. Compare with 10⁴ = 10,000 codes when repetition is allowed; the difference is the codes with at least one repeated digit.

Combinations: order does not matter. A combination is an unordered selection. Choosing r objects from n when the order of selection is irrelevant gives

nCr = n! / (r! (n − r)!) = nPr / r!.

Dividing by r! removes the r! orderings of each chosen group, which the permutation count treats as different but the combination count treats as the same. Example: 10C3 = (10 × 9 × 8) / (3 × 2 × 1) = 720/6 = 120 committees of three from ten people. The committee {Ana, Ben, Cal} is one committee whether Ana or Cal was named first.

A useful check is symmetry: nCr = nC(n − r), because choosing which r to include is the same as choosing which n − r to leave out. So 5C2 = 5C3 = 10.

A fully worked probability with combinations. A club has 6 seniors and 4 juniors. Three members are chosen at random to attend a meeting; every group of three is equally likely. What is the probability that all three are seniors? The sample space is all 10C3 = 120 three-person groups. Favorable groups choose 3 of the 6 seniors: 6C3 = 20. So P(all seniors) = 20/120 = 1/6.

What is the probability of exactly two seniors? A favorable group has 2 of the 6 seniors and 1 of the 4 juniors. Those choices are made independently, so multiply: 6C2 × 4C1 = 15 × 4 = 60. P(exactly two seniors) = 60/120 = 1/2. Interpretation: half of the equally likely groups of three contain exactly two seniors. The product 15 × 4 is a small instance of the multiplication principle: for each of the 15 senior pairs, any of the 4 juniors completes the group, and different pairs or different juniors give different groups.

The same problem by the Multiplication Rule. "All seniors" can also be found stage by stage without replacement: (6/10)(5/9)(4/8) = 120/720 = 1/6, the same answer. The combination method counts unordered groups; the stage method counts ordered draws (10 × 9 × 8 = 720 of them, 6 × 5 × 4 = 120 favorable). Both are uniform models of the same experiment, and the ratio is the same because each unordered group corresponds to exactly 3! = 6 ordered draws in both numerator and denominator. When a problem asks about "exactly two seniors", the stage method needs to add three orders (SSJ, SJS, JSS), for example SJS is (6/10)(4/9)(5/8); each order has numerators 6, 5, 4 and denominators 10, 9, 8 in some arrangement, so each is 120/720 = 1/6, total 1/2; the combination method handled the orders automatically.

A worked probability with permutations. Five students, including Priya and Quinn, are seated at random in a row of five chairs. What is the probability that Priya and Quinn sit next to each other? Total arrangements: 5! = 120, all equally likely. Favorable: treat the pair as a block; there are 4 objects to arrange (the block and three others), 4! = 24 ways, and the pair can be ordered within the block in 2 ways, so 2 × 24 = 48 arrangements. P(together) = 48/120 = 2/5. Interpretation: in two fifths of the equally likely seatings, the two friends are adjacent. The complement, P(not together) = 72/120 = 3/5.

Large sample spaces. A hypothetical lottery asks a player to choose 5 different numbers from 1 to 40, with the order of the numbers irrelevant, and pays the top prize if all 5 match the drawn set. There are 40C5 = (40 × 39 × 38 × 37 × 36) / 120 = 658,008 possible sets, each equally likely if the draw is fair, so the probability of matching all five with one ticket is 1/658,008. The number is small not because anything is rigged but because a great many sets exist. This is a hypothetical game invented for the arithmetic; real games have their own rules and their odds must be read from those rules.

Choosing the tool. Ask the order question first. Codes, rankings, seatings, and "first, second, third" prizes are ordered; when no item can be used twice they are permutations, counted by nPr, but a code that allows repeated digits is counted by nʳ instead (10⁴ = 10,000 four-digit codes). By convention 0! = 1, so nC0 = nCn = 1: there is exactly one way to choose nothing, or everything. Committees, hands of cards, sets of lottery numbers and groups of items sampled together are combinations. If the problem has stages with different conditions, such as "two seniors and one junior", count each part with combinations and multiply. If the problem says "at least", either add the cases or use the complement: P(at least one junior in the group of three) = 1 − P(no juniors) = 1 − 1/6 = 5/6.

The misconception. Using a permutation count for an unordered selection inflates both numerator and denominator by the same factor r! only if you do it consistently; mixing the two, for instance counting favorable groups with 6C3 = 20 but total outcomes with 10P3 = 720, gives 20/720 = 1/36, which is wrong. Numerator and denominator must count the same kind of object, both ordered or both unordered.

Worked permutation count: How many four-digit codes use the digits 0 to 9 with no repeats, and what is the probability that a random four-digit code (any digits, repeats allowed, 10,000 equally likely codes) has all digits different? 10P4 = 10 × 9 × 8 × 7 = 5,040 codes have no repeats. P(all different) = 5,040/10,000 = 63/125 = 0.504. Just over half of random codes have four distinct digits.

Worked combination count with two groups: From 6 seniors and 4 juniors, three are chosen at random. P(exactly one senior) needs 1 senior from 6 and 2 juniors from 4: 6C1 × 4C2 = 6 × 6 = 36 favorable of 10C3 = 120 groups, so 36/120 = 3/10. P(no seniors) = 4C3/120 = 4/120 = 1/30. Check: P(0 seniors) + P(1) + P(2) + P(3) = 4/120 + 36/120 + 60/120 + 20/120 = 120/120 = 1.

Worked card hand: Five cards are dealt from a standard deck; every 5-card hand is equally likely, 52C5 = 2,598,960 hands. The number of hands that are all hearts is 13C5 = 1,287, so P(all hearts) = 1,287/2,598,960 ≈ 0.000495. P(all one suit, any suit) is four times that, 5,148/2,598,960 ≈ 0.00198, since the four suits give mutually exclusive cases.

Represent and record: Make a two-column chart headed "order matters: permutation" and "order does not matter: combination" and sort ten short situations into it (seating five people, choosing a three-person committee, a top-three finish in a race, a hand of cards, a four-digit code, a set of lottery numbers, and four of your own). For each, write the count with its formula and its value. Then write the club problem (6 seniors, 4 juniors, choose 3) as a probability distribution over the number of seniors, 0 to 3, and verify the four probabilities add to 1.

Independent performance: Pose and solve three probability problems in uniform models: one requiring permutations, one requiring combinations from two subgroups, and one that uses "at least" and the complement. For the combination problem also solve it by the stage-by-stage Multiplication Rule and show that the answers agree. Success criteria: each problem states why order does or does not matter; the total and favorable counts are computed with the same kind of object; every probability is a reduced fraction with a decimal approximation; the complement calculation is written as 1 minus a computed probability, not guessed.