The number of even numbers greater than 1000000 that can be formed using all the digits 1,2,0,2,4,2 and 4 is

Case (i): Number of 7 digit numbers using 0, 1, 2, 2, 2, 4, 4 ending with 0= $\frac{6!}{3! 2!}=60$

Number of 7 digit number using 0, 1, 2, 2, 2, 4, 4 ending with 4= $\frac{6!}{3!}-\frac{5!}{3!}=100$

                                             $-$ 

Number of 7 digit number using 0, 1, 2, 2, 2, 4, 4 ending with 2 $=\frac{6!}{2! 2!}-\frac{5!}{2! 2!}$

                                                                                                              $=180-30=150$

                                             $-$

Total number of permutations=60+100+150

                                                          =310