Question 883540
I get a different number.
I checked by figuring it different ways, but I might be missing something.
 
If you use 4, 5, or 6 as the first digit, your number will surely be greater than 3,255.
 
The first digit cannot be 2, and that greatly limits the choices.
 
Choosing 3 as the first digit limits your choices for the other digits, because
3,222, 3,223, 3,224, 3,225, 3,226,
3,232, 3,233, 3,234, 3,235, 3,236,
3,242, 3,243, 3,244, 3,245, 3,246,
3,252, 3,523, 3,254, and 3,255 are not greater than 3,255,
so you have {{{19}}} less choices.
 
Other than the limitations above,
you have 5 choices (2, 3, 4, 5, or 6) for each digit.
That gives you {{{5*5*5=125}}} choices for the last 3 digits.
 
Using 3, 4, 5, or 6 as a first digit (4 choices),
combined with those {{{125}}} choices for the last 3 digits gives you
{{{4*125=500}}} numbers that
use only 2, 3, 4, 5, and 6 as digits, with 3, 4, 5, or 6 as first digit.
 
If we subtract from those {{{500}}} numbers
the {{{19}}} that start with 3 and are not greater than 3,255,
we get {{{500-19=highlight(481)}}}.
 
NOTE: Listing the numbers is not a good way to count them.
I tried listing the numbers that started with 3,
and my count did not add up, because I was missing some numbers.
I know I make lots of distracted mistakes, so I always try to check my results.
I check using different ways, because when figuring things the same way, I usually make the same mistake.