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

120

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 =
`(6!)/((6-3)!)` which simplifies to

`(6!)/(3!)=(6*5*4*3!)/(3!)=120`