Probability of getting a number with two identical digits I have a fairly simply question which I am not sure about. A 3 digits number is being chosen by random (100-999). What is the probability of getting a number with two identical digits ? (like 101). Thank you !
 A: There are $900$ numbers. all equally likely. We now count the favourables, the ones with exactly two identical digits.
It is clear by symmetry that there is the same number from $100$ to $199$ as there is from $200$ to $299$, as there is from $300$ to $399$ and so on.
We count the number from $800$ to $899$, and multiply by $9$. 
Maybe the repeated digit is $8$, in which case the second occurrence can be at any of $2$ places. There are then $9$ choices for the third digit, for a total of $(2)(9)$.
Or maybe it  is the last $2$ digits that repeat. That gives $9$ possibilities.
Thus the total number of favourables is $9\left((2)(9)+9\right)$. 
A: The best way to approach these problems is to make lists of what could happen, and pick numbers one by one.
So the first number must be 1-9 The second number must be 0-9. (No restrictions so far.)
But we have to pick the third number carefully. If the first two numbers matched, we must pick a different number for the last one. If the first two did NOT match, we must copy one of these two for the last number.
So for this last step, we split into two cases (how many numbers can we pick the first two match, and how many numbers can we pick the first two not matching), and then add these together at the end. If you do so, you should get an answer similar to the one posted in comments.
A: There are ten numbers of the form aa*.  9 of them are not aaa. There are 9 possible chooses for a.  So there are 81 numbers where the first two digits are the same.
By exact same reasoning there are 81 numbers where where the first and last digit are the same.  
Counting the numbers where the last two digits are the same is a little different.  If we ignore *00 numbers for now there are 81 numbers by the same argument above. Except we can not count the number 0aa.  So that's only 72.  But we also have  9 *00 numbers. So there are 81 after all.  So there are 81 numbers where the last two digits are the same.
So there are 243 numbers with exactly 2 digits the same.   
