3.2 Permutations, Combinations, and Casework

Arrange or choose

A permutation counts ordered arrangements. A combination counts unordered selections. The formula sheet gives nPr, nCr, and factorial notation, but it does not decide which one applies. That decision comes from the context. If president, vice-president, and treasurer are chosen from the same group, order matters because the roles are different. If a three-person committee is selected and everyone has the same role, order does not matter.

Ask one question: if the same people or objects are listed in a different order, is the result considered different? If yes, use a permutation or slot method. If no, use a combination.

Factorials and formulas

Factorial notation means descending multiplication: 5! = 5 times 4 times 3 times 2 times 1. A permutation can be read as ordered choices: nPr = n!/(n-r)!. A combination can be read as the same ordered choices with the extra orderings divided out: nCr = n!/[r!(n-r)!].

This division is why 8C3 is smaller than 8P3. In 8P3, A-B-C and B-A-C are different arrangements. In 8C3, those are the same selected group.

Worked example: roles

A club has 9 members. It must choose a chair, a treasurer, and a secretary. No person can hold more than one office. How many executive slates are possible?

The offices are different, so order matters. Use 9P3 or slots: 9 choices for chair, 8 for treasurer, and 7 for secretary. The count is 9 times 8 times 7 = 504.

Using 9C3 would count only the group of three people and ignore which person has which office. That would undercount the slates by a factor of 3!, because every group of three can be assigned to the three offices in six ways.

Worked example: committee with a condition

A committee of 4 is chosen from 6 Grade 12 students and 5 Grade 11 students. How many committees have exactly 2 students from each grade?

This is a selection, not a ranking, so use combinations. Choose 2 of the 6 Grade 12 students and 2 of the 5 Grade 11 students. The count is 6C2 times 5C2 = 15 times 10 = 150.

Notice the multiplication: the final committee requires both decisions. If the question asked for either exactly 1 Grade 12 or exactly 2 Grade 12 students, those would be separate cases and you would add the case totals.

Worked example: repeated objects

How many distinct arrangements can be made from the letters in the word LEVEL?

There are 5 letters. The letter L repeats twice and the letter E repeats twice. If all letters were different, there would be 5! arrangements. Because swapping the two Ls does not create a visible new word and swapping the two Es does not create a visible new word, divide by 2! and 2!. The count is 5!/(2!2!) = 30.

Repeated-object questions are a common place to overcount. The calculator may happily return 120, but the question asks for distinct visible arrangements, not labelled physical letter tiles.

Casework without losing control

Casework means splitting a problem into non-overlapping scenarios. It is useful when there are restrictions such as at least, exactly, begins with, includes, or excludes. The rule is: multiply within a case, then add the separate cases.

For example, suppose a 3-person committee from 7 people must include Maya or Liam, but not both. Case 1 includes Maya and excludes Liam: choose the other 2 from the remaining 5, giving 5C2 = 10. Case 2 includes Liam and excludes Maya: again 5C2 = 10. Total: 20.

The cases must not overlap. If a case can contain both Maya and Liam, then the same committee may be counted twice. When cases are hard to make separate, try the complement. Count all committees and subtract committees with neither required person or with both, depending on the wording.

Diploma traps

Trap 1: using nPr because the answer feels larger. Larger is not a reason. The wording decides. A team, hand, sample, or committee usually points to nCr unless roles are assigned later.

Trap 2: missing hidden order. A password, seating row, race finish, or task schedule has positions. Even if the word permutation is absent, positions make order matter.

Trap 3: adding factors inside one case. Choosing two Grade 12 students and two Grade 11 students is not 6C2 + 5C2. A valid committee needs both choices, so multiply.

Trap 4: starting with the easy slot. In restricted arrangements, fill the restricted slots first. If a code must start with an even digit, count that first slot before the unrestricted slots.

A clean Math 30-2 response states the model in words before using calculator notation: order matters, so 9P3; order does not matter, so 6C2. That sentence prevents many setup errors.

Calculator note

Use the calculator to evaluate nPr, nCr, and factorials after the setup is chosen, not before. On a graphing calculator, 10P4 and 10C4 are both easy to enter, so the technology will not warn you that the context was misread. A reliable habit is to write a short reason beside the command: roles are different, order matters, or committee has no roles, order ignored. That written reason is often more valuable than the button sequence.