how many integers between 100 and 1000 have different digits.?
im trying to solve this using permutations.
We're talking a 3 digit number which obviously cannot start with the digit 0
So we have 9 choices for the first digit, and then 9 for the second digit and 8 for the third
for a total of 648 numbers.
the fact that 0 cannot appear as the first digit is going to mess up your permutation calc.
you could look at it as the numbers w/no 0 + the numbers with 1 0
The first term is $3! \binom{9}{3}=504$
The second term is $\binom{9}{1}2! \binom{8}{1} = 144$
and these sum to $504+144=648$
i understand it now. so there are 9 different digits for the first number (0 isnt included) for the second digit there are 9 options(0 is included) for the third option there are 8
so 9*9*8=648
that^ seems to be the easiest way to explain it.
This is the term for the numbers with no $0$
There are $\binom{9}{3}$ ways to choose the 3 unique digits but since we do care about the ordering, i.e. $123 \neq 321$
we multiply by $3!$ to account for all the permutations of the $3$ digits.
This term is for numbers that contain one $0$The second term is $\binom{9}{1}2! \binom{8}{1} = 144$
$\binom{9}{1}$ is just the fact that there are 9 valid digits for the first digit in the number. $0$ is not valid.
$\binom{8}{1}$ is just the fact that since we know one of the remaining two digits is a $0$, there are only $8$ digits to choose from for the free digit.
But again ordering matters so we multiply by $2!$ to take into account the $2!$ permutations of these two remaining digits.