## Algebra 2 Common Core

$120$ possible permutations.
Using the formula for permutations, $\frac{n!}{(n-r)!}$, $n = 6$ and $r = 3$. Substitute the values for $n$ and $r$: $\frac{6!}{(6-3)!}$ Simplify the denominator: $=\frac{6!}{3!}$ Expand the factorials: $=\frac{6\times 5\times 4 \times 3 \times3 \times 1}{3 \times 2 \times 1}$ Calculate: $\frac{720}{6} = 120$ There are $120$ possible permutations.