Choosing a committee of $2$ members from a club of $10$ members
Terms
| Term | This example | Why |
|---|---|---|
| combination | $\binom{10}{2} = 45$ | Choosing a committee of $2$ from $10$ members has $\binom{10}{2} = (10 \cdot 9) / 2 = 45$ outcomes, which is half the $90$ ordered choices, because each committee can be ordered in $2$ ways. |
| permutation | ✗ | Choosing a committee of $2$ from $10$ members is not a permutation, because the order does not matter: Ana and Ben is the same committee as Ben and Ana. |