count it

The desired numbers are of the following three mutually exclusive types.

(a) Numbers of six digits each.

(b) Numbers of five digits each.

(c) Numbers of four digits each.

For type (a) the left-most place can be filled in just $5$ ways (with $3,6,7,4$ or $2$ ). Then the next place in $5$ ways because $0$ is now eligible; etc.; the number of these numbers is $5 \times 5 \times 4 \times 3 \times 2 \times 1$ or $600$ numbers.

Similarly, there are $600$ numbers for type (b).

For type (c) the left-most place can be filled in $2$ ways (with $6$ or $7$ ). Then the next place in $5$ ways; etc.; the number of these numbers is $2 \times 5 \times 4 \times 3$ or $120$ numbers.

Finally, the number of numbers of all types is $600+600+120=$ $\color{#D61F06}\boxed{1320}$