since, $1000000$ is a $7 -$ digit number and the number of digits to be used is also $7$. Therefore, the numbers to be counted will be $7 -$ digit only. Also, the numbers have to be greater than $1000000$ , so they can begin either with $1,2$ or $ 4.$

The number of numbers beginning with $1=\frac{6 !}{3 ! 2 !}=\frac{4 \times 5 \times 6}{2}=60,$ as when $1$ is fixed at the extreme left position, the remaining digits to be rearranged will be $0,2,2,2, $$4,4,$ in which there are $3,2 s$ and $2,4 s$

Total numbers begining with $2$

$=\frac{6 !}{2 ! 2 !}=\frac{3 \times 4 \times 5 \times 6}{2}=180$

and total numbers begining with $4=\frac{6 !}{3 !}=4 \times 5 \times 6=120$

Therefore, the required number of numbers $=60+180+120=360$