How many 6-digit numbers contain exactly 4 different digits? my solution is---->
 NUMBER can be 777210 this i denote by aaabcd
                ------   this can be done in ---> 10*1*1*9*8*7*[6!/3!]
                  {1 for a thrice}
 NUMBER can be 772210 this i denote by aabbcd
------  this can be done in ------------> 10*1*9*1*8*7*[6!/(2!*2!)]
           {1 for a and b }
and then multiply the answer by 0.9 to remove 0 case at unit place
but correct solution is 
C(10, 4)* [C(4, 2)* 6!/(2!)2 + C(4, 1)* 6!/3!] 
294840
 A: The trouble is that before permutations in  [ ], you are mixing up combinations and permutations.
A double check of the book answer by a slightly different formula is:
[Choose triple/double]$\times$ [Choose the remaining]$\times$ [permute]
which for patterns $AAABCD$ and $AABBCD$ yields 
$0.9[\binom{10}1\binom 93\cdot  \frac{6!}{3!} + \binom{10}2\binom 82\cdot\frac{6!}{2!2!}] = 294840$ 
A: In your aaabcd case, you have a factor of $\frac{6!}{3!}$ which I think is meant to account for "how many ways are there to choose 3 digit positions to be a?"
However, this number should not be $\frac{6!}{3!}$ but instead $\binom 63=\frac{6!}{3!\cdot 3!}$.
If your intention was to make this the multinomial coefficient $\binom{6}{3,1,1,1}$, then that will lead to overcounting, because for example 737527 will be counted both as abacda, acabda, aadabca, and so forth.
Similarly your second term has $\frac{6!}{2!2!}$ which overcounts for the same reason. You want something like $\binom{6}{2,2,1,1}\cdot\frac12\cdot\frac12$ to account for overcounting with the two pairs and the two singletons swapped, respectively.
This gives you the wrong numerical results for an approach that otherwise looks valid.
