6 digit number with digit(0-9) precisely 4 or 5 different digits I just want to be sure if i'm right so i have:
For 4 digits: 


*

*$C(10,4) = 210 $- to get number of combinations of 4 different digits

*1 digit repeat 2 more times  - $C(4,1) = 4 $

*number of other combinations -$ P*(3,1,1,1) = 120$

*2 digit repeats 1 more time - $C(4,2) = 6$

*number of other combinations - $P*(2,2,1,1) = 180$


Total number of 4 different digits :
$(210+4)*120+(210+6)*180 = 64560$

For 5 dif. digits


*

*$C(10,5) = 252$ - to get number of combinations of 5 different digits

*1 digit repeat 1 more times  -$ C(5,1) = 5 $

*number of other combinations -$ P*(2,1,1,1,1) = 360$
Total number of 4 different digits :
$(252+5)*360=92500$
at the and because of The OR i have to sum up for 4 and for 5 diff digits right ? :)

There is the complication that (perhaps) the first digit is not allowed to be $0$. The usual conventional meaning of $6$-digit number does not include
$001123$. 
Suppose that we count the number of choices where initial digit $0$ is allowed. If the count is $N$, then the count where initial digit $0$ is not allowed is $\frac{9}{10}N$.
Your basic procedure for finding $N$ is good, but there are mistakes at the end. For example, the number for exactly $4$ should be
$$\binom{10}{4}\left(\binom{4}{1}(120)+\binom{4}{2}(180)              \right).$$
