# Number of 3 digit numbers with distinct digits

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?

• All 3-digit numbers are not 1000 to start with. – peter.petrov Sep 13 '16 at 13:56
• @peter.petrov Where did I say all 3 digit numbers are 1000? – user220382 Sep 13 '16 at 14:01
• You can also think it this way: You have 9 ways to choose the first digit (it cannot be zero), 9 ways to choose the second (it can be zero but it cannot be the same as the first one) and 8 ways to choose the third one (it can be zero but it cannot be the same as the first or second one). Total: $9 \times 9 \times 8 = 648$. – Giovanni De Gaetano Sep 13 '16 at 14:11
• @GiovanniDeGaetano Thanks.That's nice too :-) – user220382 Sep 13 '16 at 14:12
• @ZOZ Here: "MY ATTEMPT: All 3-digit numbers:100,101,102,103,.....,999 (i.e. 1000 numbers)" – peter.petrov Sep 13 '16 at 14:19

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