splitting counting problem 
How many even four-digit numbers can be formed from the digits 0, 1, 2, 5, 6, and 9 if each digit can be used only once?
We consider the unit position by two parts, 0 or not 0.

If the units position is 0 :
 5 choices for the thousands positions,
 4 choices for the hundreds positions,
 3 choices for the tens positions.
a total of   choices 60
If the units position is not 0 :
 4 choices for the thousands positions,
 4 choices for the hundreds positions,
 3 choices for the tens positions.
a total of   choices. 96
The total number of even four-digit numbers is 60 + 96 = 156


i dont understand why it was splitted into two parts.  i understand the first part (since if u fixed the units position to be 0, then theres 1 choice for it, and then the rest is simply n-1). For the second part, why was the thousand positions 4, not 5? isn't it he will just choose between (2 and 6) and then just use n-1 all the way. Also why was 4 repeated in the thousands and hundreds position? Thanks
 A: The reason for spliting final $0$ from final non-$0$ is because of the problem of avoiding a leading $0$ combined with the final digit being even to make the whole number even.  
You seem to have missed a line in the calculation of $96$:

If the units position is not $0$ :
  
  
*
  
*$2$ choices for the units positions ($2$ or $6$)
  
*$4$ choices for the thousands positions ($1$, $5$, $9$ and whatever is left from $2$ or $6$)
  
*$4$ choices for the hundreds positions (the original six less the two already chosen)
  
*$3$ choices for the tens positions (the original six less the three already chosen)

Then $2 \times 4 \times 4 \times 3 = 96$ and you add the $60$ where the final digit is $0$ to get $156$.
A: In the standard way of writing numbers, there are no four-digit numbers that begin with $0$.  Thus $0$ and the other two even digits we were given are not equivalent, $0$ is special.
That specialness can be dealt with in other ways.  For example, let us first count how many strings of four distinct digits there are such that the final digit is even.
The rightmost digit can be chosen in $3$ ways.  For every way of choosing the rightmost digit, there are $5$ ways to choose the next digit to the left, for a total of $3\cdot 5$ choices.  For every such choice, there are $4$ ways to choose the next digit to the left, for a total so far of $3\cdot 5\cdot 4$. And for every choice so far, there are $3$ ways to choose the next digit to the left, for a total of $3\cdot \cdot 5\cdot 4\cdot 3$.  Multiply. We get $180$.
Our count of $180$, however, includes some bad strings, namely those that have leftmost digit $0$ and even rightmost digit. How many such bad strings are there? The rightmost digit can be chosen in $2$ ways (remember, the $0$ is already in front).  For every such choice, the next digit to the left can be chosen in $4$ ways. And for every one of the choices made so far, the next digit to the left can be chosen in $3$ ways, for a total of $2\cdot 4\cdot 3$. Multiply. We get $24$.
Thus the total number of even four-digit numbers is $180-24$.
Another way: We can split into cases in a way different from the one you were given. We split into two cases: (i) Leftmost digit is odd, and (ii) Leftmost digit is even. 
In case (i), the leftmost digit can be chosen in $3$ ways. For every such choice, the rightmost digit can be chosen in $3$ ways. And for every one of these $3\cdot 3$ choices, there are $4\cdot 3$ ways to fill in the rest, for a total of $108$.
In case (ii), the leftmost digit can be chosen in $2$ ways. For every such choice, the rightmost digit can be chosen in $2$ ways. And for each of these $2\cdot 2$ choices, there are $4\cdot 3$ ways to fill in the rest, for a total of $48$.
Thus the total number of even four-digit numbers is $108+48$.
Remark: We try to explain "why split" further, by giving a wrong argument that tries to capture the specialness of $0$, but fails. The first (rightmost) digit can be chosen in $5$ ways. The last can be chosen in $3$ ways, and the rest can be chosen in $4\cdot 3$ ways, for a total of $180$.  What's wrong with this? Certainly it is true that there are $5$ choices for first digit. It is also true that there are $3$ choices for the last digit.  But it is not true that for every one of the $5$ choices of first digit, there are $3$ choices of last digit. If we choose an odd first digit, then yes, there are $3$ choices for the last.  But if we choose an even first digit, there are $2$ choices for the last. So the number of ways to choose first and last is not $5\cdot 3$. In fact it is $3\cdot 3+2\cdot 2$.  
