Forming of numbers

This is equivalent to asking, ways to permute 3 objects from 6

$= \dfrac{6!}{(6-3)!} = \dfrac{6 \times 5 \times 4 \times \cancel{3!}}{\cancel{3!}} = \boxed{120}$

We can choose any one of the six digits for the units’ place. After placing one digit on the units' place, we can choose any one of the five remaining digits for the tens place. After placing a digit on the tens' place, we can choose anyone of the remaining four digits for the hundreds’ place. Hence, we can form $6\times5\times4=120$ different numbers.