Question

The number of three-digit numbers with distinct digits that can be formed using the digits 1, 2, 3, 5, 8, and 9 is

Answer 1

Answer:

120

Step-by-step explanation:

Upon pondering the wording here, I am thinking this must be a permutation.  The words "distinct digits" tell me, for some reason, that order matters.  The formula for the permutations of this is

₆P₃ = $\frac{6!}{(6-3)!}$ which simplifies to

$\frac{6!}{3!}=\frac{6*5*4*3!}{3!}=120$