Answer
$10$ possible combinations.
Work Step by Step
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.
