Question

How many numbers greater than a million can be formed with the digits $2,3,0,3,4,2,3$?

Answer 1

Here one million $=10,00,000.$

Given numbers are $2,3,0,3,4,2,3.$
A number greater than $1000000$ has $7$ places, and all the digits are to be used. But $2$ is repeated twice and 3 is repeated thrice.
Total number of ways arrangeing $7$ digits $=\frac{7!}{(2!3!)}=420.$
But the numbers begin with 0 are no more seven-digit numbers then reject those numbers.
There fore, these can be arranged in $\frac{6!}{(2!3!)}=60$

Hence required number of arrangements $=420-60=360.$