PEAZIT Aha card
Go straight to resources ↓A combination selects a group without ordering it
When selecting two different people, Anna–Boris is the same pair as Boris–Anna. Divide the ordered selection count by two.
| Relation | Meaning |
|---|---|
| 5 × 4 = 20 | There are 5 choices for the first person and 4 for the second. |
| 2! = 2 | Each pair can be written in two orders. |
| C(5,2) = 20 / 2 = 10 | Ignoring order leaves 10 teams. |
One worked example
- 5 × 4 = 20There are 5 choices for the first person and 4 for the second.
- 2! = 2Each pair can be written in two orders.
- C(5,2) = 20 / 2 = 10Ignoring order leaves 10 teams.
Watch out: Elements are not repeated here. In general C(n,k) = n!/[k!(n−k)!] for integers 0 ≤ k ≤ n.
Goal, prerequisites and card origin
What you will learn: Remove duplicate counting of the same group.
What you need first: The multiplication rule and distinct people.
An original PEAZIT working card. Independent expert review and effectiveness testing have not yet taken place. · v1
Show connections
When selecting two different people, Anna–Boris is the same pair as Boris–Anna. Divide the ordered selection count by two.