How many even numbers of four digits can be formed with the digits 0,1,2,3,4,5 and 6 no digit being used more? My attempt to solve this problem is:
First digit cannot be zero, so the number of choices only $6 (1,2,3,4,5,6)$
The last digit can be pick from $0,2,4,6$, so the number of choices only 4
Second digit can be only pick from the rest, so the number of choices only 5
Third digit can be only pick from the rest, so the number of choices only 4
The total number of choices is $6\cdot 4\cdot 5\cdot 4= 480$
So, is my solution true? Or I miss something?
Thanks

Consider two cases:
Case 1: first digit even
There are three ways to find the first digit, if the first digit is even.
There are 2,4, and 6.
There are three ways to find the last digit, because one is taken for the first digit.
There are 5 ways to find the second digit and 4 ways to find the third digit.
So, the total ways is: 3*5*4*3= 180 ways
Case 2: first digit odd
There are three ways to choose the first digit, there are: 1,3,5
There are 4 ways to choose the last digit, there are 0,2,4,6
There are 5 ways to choose the second digit and 4 ways to choose the third digit.
So, the total ways is: 3*5*4*4= 240 ways
Total ways: case 1+ case 2= 180+240= 420 ways
