<s>[INST] 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
 [/INST] Consider cases:


*

*The last digit is $0$.  Then you have six choices for the first digit, five choices for the second digit, and four choices for the third digit, giving $6 \cdot 5 \cdot 4 \cdot 1 = 120$ four digit even numbers that end in $0$.

*The last digit is $2$, $4$, or $6$.  Since the first digit cannot be zero, you have five choices for the first digit, five choices for the second digit, and four choices for the third digit, giving $5 \cdot 5 \cdot 4 \cdot 3 = 300$ four digit even numbers that end in $2$, $4$, or $6$.  
That exhausts the possibilities, so there are $120 + 300 = 420$ even four digit numbers that can be formed using the digits $0, 1, 2, 3, 4, 5, 6$.
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