[Math] Number of 3 digit numbers with distinct digits

Question

I need to find the number of 3 digit numbers without repetition (distinct digits).

MY ATTEMPT:
All 3-digit numbers:$100,101,102,103,…..,999$ (i.e. $1000$ numbers)

But we need to exclude following types of numbers

_11,_22,_33,…_99 type (8*9 numbers = 72 numbers because the first digit should not start with 0 or be similar to the other digits)

1_1,2_2,..,9_9 type (9*9 numbers =81 numbers)

11_,22_,…,99_ type (9*9 numbers =81 numbers)

111,222,333,…,999 type (9 numbers)

Required number of numbers = $1000-81-81-72-9=657$

But the actual answer is $648$. Where did I go wrong?

Answer 1

I think you just forgot to eliminate numbers ending with double $0$. Once you do this your method yields $900-81-81-81-9=648$.