The learning goal
For two equal-role representatives, AB and BA describe one group, so divide 5×4 by 2!=2 to obtain 10. For chair and recorder, swapping roles changes the result, so all 20 assignments remain. A three-person equal-role group requires division by 3!=6 rather than 2.
Define the outcome before judging order
OpenStax’s counting principles distinguish multiplication, permutations and combinations. These labels, lists and variations are original editorial exercises, not a learning-effectiveness experiment.
Let A,B,C,D,E be five distinct people, each selectable at most once. One task chooses two equal-role representatives. Another chooses a chair and recorder, with no person doing both. The verb choose alone does not decide the counted object.
Three questions before counting
Identity
Are candidates distinct? Similar names do not establish one person.
Roles
Does swapping selected people change the result?
Repetition
May a candidate appear again? This example forbids it.
Enumerate all ten equal-role pairs
Pair A with B,C,D,E to obtain AB,AC,AD,AE. Pair B only with C,D,E, adding BC,BD,BE. C adds CD,CE, and D adds DE. Thus 4+3+2+1=10.
The enumeration convention prevents duplication rather than making A important. AB and BA contain the same people and, without distinct roles, represent one outcome. The complete list supports checking missing or repeated pairs directly.
| Start | New groups | Count |
|---|---|---|
| A | AB,AC,AD,AE | 4 |
| B | BC,BD,BE | 3 |
| C | CD,CE | 2 |
| D | DE | 1 |
| Total | All equal-role two-person groups | 10 |
When is twenty the correct answer?
There are 5 chair choices and then 4 recorder choices, giving 5×4=20. A-chair/B-recorder and B-chair/A-recorder are different assignments, so retain both.
When two numbered positions enumerate equal-role groups, each result appears in two orders, AB and BA. Dividing by 2 removes duplicate descriptions of one outcome. It does not halve the number of candidates.
Same people, different outcome objects
Merge role-swapped assignments because both roles use the same candidate list.
Test whether swapping matters: the two role assignments differ, so retain all 20.
| Task | AB versus BA | Result |
|---|---|---|
| Two equal-role representatives | One group | 5×4÷2=10 |
| One chair and one recorder | Different assignments | 5×4=20 |
Why not keep dividing by two for three people?
For three equal-role representatives, ordered positions give 5×4×3=60. ABC has six descriptions: ABC,ACB,BAC,BCA,CAB,CBA. Each group was counted 3!=6 times, so 60÷6=10.
That 10 equals the pair count for a different multiplicity reason. With chair, recorder and timekeeper, all six role assignments differ and all 60 remain. Do not copy an earlier answer because its number looks familiar.
Candidates: A, B, C, D, E; no repeated selection.
1. Choose two equal-role representatives.
2. Choose a chair and a recorder.
3. Choose three equal-role representatives.
4. Choose a chair, a recorder and a timekeeper.
For each, explain whether swapping people changes the outcome and how many ordered descriptions represent one result.| Object | Position count | Descriptions per outcome | Final count |
|---|---|---|---|
| Two people, equal roles | 20 | 2!=2 | 10 |
| Two people, distinct roles | 20 | 1 | 20 |
| Three people, equal roles | 60 | 3!=6 | 10 |
| Three people, distinct roles | 60 | 1 | 60 |
Record the reason instead of the formula alone
Locate the first step that merges different objects or retains repeated descriptions. Explain AB versus BA and ABC’s six orderings, then review identity, roles, repetition and restrictions. Changed conditions may require a new product and divisor.
Counting is not a probability model: the number of outcomes does not establish equal selection chances. Read independent versus mutually exclusive events, and use a first-wrong-step error record for your reasoning.
Completion conditions
- The object is a group or a role assignment.
- Identities differ and selection does not repeat.
- AB and BA can be explained as equal or different.
- All six ABC descriptions can be enumerated.
- Counting stays separate from equal-probability assumptions.
Common questions
Does arrange always imply permutation and choose always combination?
No. Test whether swapping changes the result and inspect roles, positions and restrictions; one verb does not define the outcome.
Why does choosing three also give 10?
A selected group of three corresponds to the two people left out. Still explain the three-person object and multiplicity rather than copying the pair answer.
Do ten groups automatically each have probability one tenth?
Counting alone does not establish the sampling mechanism. That conclusion needs an appropriate equal-probability selection process.
