Find out the number six digit natural numbers such that each digit in number appears at least twice? 

In each of the six digit numbers $333333,225522,118818,707099$ each digit in number appears at least twice.Find out the number of such six digit natural numbers.


My Attempt
We observe that three possible situation may occur.  


*

*There is only one digit. 

*There are two digits.  


*

*One digit occurs twices and second digit occurs four times.  

*both the digits occur three times.  


*there are three digits each occurring two times.  


Now we count each of the cases  


*

*There are $9$ such numbers.  

*$ $


*

*First digit can be chosen in $9 \choose 1$ way. If first digit is repeated twice then the second occurrence may be placed in $5 \choose 1$ positions. The second digit can be chosen in $9 \choose 1$ ( now zero can be chosen) and they be placed in the remaining four places in one way. Thus number of ways are $9*5*9=405 $
If first digit is repeated four times then number of ways are $9*10*9=810$ Thus total count in 2a is $405+810=1215$  

*Total count is $9*10*9=810$  


*Total count =$9*5*9*6*8=19440$  


Thus grand total = $9+1215+810+19440=21474$
But answer supplied is $11754$. Can anyone help to to get right answer and spot mistake in my argument? Thanks in advance.
 A: You counted all the cases except the case in which there are three digits that each appear twice.
There are $9$ ways to select the leading digit, $5$ ways to choose the other position for the leading digit, $\binom{9}{2}$ ways to select the other two digits, and $\binom{4}{2}$ ways to choose the positions of the smaller of those digits.  Therefore, the number of six-digit positive integers in which there are three digits that each appear twice is 
$$\binom{9}{1}\binom{5}{1}\binom{9}{2}\binom{4}{2} = 9720$$
You counted each such case twice.  For instance, take the number $837783$.  You counted it in two ways:


*

*You selected $8$ as the leading digit, then chose to place it in the fifth position.  You selected $3$ as the second digit, then chose to place it in the second and sixth positions.  You chose $7$ as the third digit and placed it in the third and fourth positions.

*You selected $8$ as the leading digit, then chose to place it in the fifth position.  You selected $7$ as the second digit, then chose to place it in the third and fourth positions.  You chose $3$ as the third digit and placed it in the second and sixth positions.

