[Math] 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

combinatorics

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

Best Answer

For an even no. last digit needs to be even = 4 choices

Total 4 digit even numbers = (Total even numbers) - (Total 3 digit even numbers)

Total even numbers: $4\cdot 6\cdot 5\cdot 4 = 480$

Total 3-digit even numbers: $3\cdot 5\cdot 4\cdot 1 = 60$

Total 4-digit even numbers: $480 - 60 = 420$