SOLUTION: Determine how many numbers between 10 and 1000 inclusive contain a repeated digit and the rest of the digits(if any) are distinct.(eg:101,229). My solution : For two digits

Question 1024465: Determine how many numbers between 10 and 1000 inclusive
contain a repeated digit and the rest of the digits(if any)
are distinct.(eg:101,229).
My solution :
For two digits number : [P(9,1)*P(9,1)-1]+[[P(9,1)*P(9,1)]/9]=89.
For three digits number : [9*9*8]+[9*9]+[9*9]+[9*9]=891.
891 + 89 = 980 numbers.
Is this the correct answer?
Thanks in advance
Answer by Edwin McCravy(20054) (Show Source): You can put this solution on YOUR website!

No, that's way too many!  Think of the number of ways
there are to choose the digits.  Don't depend on
permutation formulas.  Think it through.

Case 1:  2 digit numbers  
There are 9 2-digit numbers, 11,22,33,...,99

Case 2:  3 digit numbers 

There are three types of these numbers: AAB, ABA, and ABB 
(For example 773, 505, and 911)

For each of these 3 types,
There are 9 choices for the A, 1 through 9 (can't use 0) 
There are 9 remaining choices for the B, (can use 0)

That's 3*9*9 = 243

Grand total: 9+243 = 252.

Edwin

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