Out of the 6 digits 4 digits are fixed.

In the remaining two digits, both the digits could be same

or both the digits could be different.

**case (i): Both the digits are same.**

No. 6 digit numbers =$^4C_1\times \large\frac{6!}{3!}$

(Both the digits could be any of the 4 digits..so .$4C_1$)

**Case (ii) Both the digits are different**.

The two digits could be any 2 of the 4 digits. so $^4C_2$

$\therefore$ The no. of 6 digit numbers in this case = $^4C_2\times\large\frac{6!}{2!.2!}$

$\therefore$ The required no. of 6 digit numbers=$^4C_1\times \large\frac{6!}{3!}$$+^4C_2\times \large\frac{6!}{2!.2!}$+

$=6!(\large\frac{2}{3}+\frac{3}{2})$$=6!\times\large\frac{13}{6}$$=1560$