## Algebra 2 Common Core

$10$ possible combinations.
Using the formula for combinations, $\frac{n!}{r!(n-r)!}$, where $n =$ the size of the set, and $r=$ the size of the combinations. $n = 5$, and $r = 2$ Substitute these values for $n$ and $r$ in the formula: $\frac{5!}{2!(5-2)!}$ Simplify the dominator: $=\frac{5!}{2! \times 3!}$ Expand the factorials: $=\frac{5\times 4 \times 3 \times2 \times 1}{2 \times 1 \times 3 \times2 \times 1}$ Calculate: $\frac{120}{12} = 120 \div 12 = 10$ There are $10$ possible combinations.