<s>[INST] Counting clarification In the text I am reading there's a question:
From the digits $0, 1, 2, 3, 4, 5, 6$, how many four-digit numbers with
distinct digits can be constructed? How many of these are even numbers?
I get the first part correctly by multiplying $6*6*5*4 = 720$.
To calculate the second part, I follow this process.
The unit's place can be chosen 4 different ways $(0,2,4,6)$.
The ten's place can be chosen 6 ways ($7-1$ ways)
The hundred's place can be chosen 5 ways($7 - 2$ already chosen)
The thousand's place can be chosen 3 ways(6 (digits excluding 0) - 3 already chosen).
So, this becomes $4*6*5*3 = 360$. But the answer in the text is $420$. 
Where am I going wrong?
 [/INST] To count four-digit even numbers composed of only the digits $0, 1, 2, 3, 4, 5, 6$, we treat those numbers that end in zero as a special case.
We consider cases:
Case 1:  The units digit is zero.  
There are six choices for the thousands digit (since we must exclude zero), five choices for the hundreds digit (since we must exclude the thousands digit and zero), and four choices for the tens digit (since we must exclude the thousands digit, the hundreds digit, and zero).  Therefore, there are $6 \cdot 5 \cdot 4 = 120$ four-digit even numbers composed only of the digits $0, 1, 2, 3, 4, 5, 6$ that end in zero.
Case 2:  The units digit is nonzero.
We have three choices for the units digit ($2$, $4$, and $6$) since the last digit must be even.  We have five choices for the thousands digit (since both the units digit and zero are excluded), five choices for the hundreds digit (since the thousands digit and the units digit are excluded), and four choices for the tens digit (since the thousands digit, hundreds digit, and the units digit are excluded).  Thus, there are $3 \cdot 5 \cdot 5 \cdot 4 = 300$ four-digit even numbers composed only of the digits $0, 1, 2, 3, 4, 5, 6$ that do not end in zero.  
Since these cases are mutually exclusive, the number of four-digit even numbers composed only of the digits $0, 1, 2, 3, 4, 5, 6$ is $6 \cdot 5 \cdot 4 + 3 \cdot 5 \cdot 5 \cdot 4 = 120 + 300 = 420$.
You undercounted the number of choices for the thousands digit when the units digit was zero.   
Andre Nicolas has provided an elegant alternative solution.
</s>