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?
 A: If we already chose $0$ earlier, then the thousands place can be chosen in $4$ ways, not $3$. So your method undercounts. We give one method of calculating correctly. There are others.
There are two possible cases.
Case 1: The thousands place is odd. There are $3$ choices. For every such choice, the units place can be chosen in $4$ ways, and the remaining $2$ in $(5)(4)$ ways.
Case (2): The thousands place is even. There are $3$ choices. For every such choice there are $3$ ways to select the units place, and $(5)(4)$ ways to select the rest.
Add. We get $(3)(4)(5)(4)+(3)(3)(5)(4)$.
A: 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.
A: There's probably a nicer way to write this, but for me (I've never done combinatorics) it helped to put it in this way:
For the 4-digit number to be even it needs to end in $0, 2, 4, 6$. If it ends in $0$, then there are $(6)(5)(4)$ possible numbers, otherwise we can have again two possibilities: the ten's being $0$ or non-zero. If it's zero then we have $(5)(4)$ numbers, otherwise, we can have another two possibilities for the hundred's: zero or non-zero, if it's zero then there are $(4)$ numbers, otherwise $(4)(3)$.
So putting all together: $6*5*4+3*(5*4+5*(4+4*3))=420$
A: If units digit is zero there are P(6,3)=120 ways.
If units digit is one of the other three, there are P(6,3)=120 ways.
BUT P(5,2)=20 of these start with a zero.
Ergo there are P(6,3)-P(5,2)= 100 for the others.
There are 3 (2,4,6,) of these, giving 300, add this to 120 for a total of 420
A: The answers provided by André Nicolas and N. F. Taussig are correct. However, I found a simpler way of counting the even numbers. 
First count the odd numbers. 
There are (3) ways of choosing the unit's place. 
There are (5) ways of choosing the thousand's place(0 cannot be used and one digit is already chosen in unit's place). 
There are (5) ways to choose the hundred's place.(2 already chosen). 
Similarly, there are (4) ways to choose the ten's place. 
Therefore, there are (5)(5)(4)(3) odd numbers. ie, 300 odd numbers. Total even numbers = 720-300 = 420. 
