1. ## different digits

how many integers between 100 and 1000 have different digits.?

im trying to solve this using permutations.

2. ## Re: different digits

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 $504+144=648$

1: 102
2: 103
3: 104
....
646: 985
647: 986
648: 987

4. ## Re: different digits

Originally Posted by DenisB
1: 102
2: 103
3: 104
....
646: 985
647: 986
648: 987
is that how many beers are on your wall?

5. ## Re: different digits

No, no! DenisB would never allow that many beers to remain on his wall!

6. ## Re: different digits

Why is every(hic!)body always a'pic(hic!)in on me...

7. ## Re: different digits

i dont understand your method. i dont see why using combinations is correct

8. ## Re: different digits

Hey markosheehan.

Do you understand what romsek said about fixing the first digit and then looking at the possibilities for the second and third?

9. ## Re: different digits

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.

10. ## Re: different digits

Originally Posted by romsek
We're talking a 3 digit number which obviously cannot start with the digit 0

you could look at it as the numbers w/no 0 + the numbers with 1 0

The first term is $3! \binom{9}{3}=504$
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.

The second term is $\binom{9}{1}2! \binom{8}{1} = 144$
This term is for numbers that contain one $0$

$\binom{9}{1}$ 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.