How many even numbers of four distinct digits greater than 5000 are possible How many even numbers of four distinct digits greater than 5000 are possible? Please help me
The only thousand digit that are possible 5,6,7,8,9. 
The only hundred digit that are possible are 1, 2, 3, 4, 5, 6, 7, 8, 9 
The only ten digit that are possible are 1, 2, 3, 4, 5, 6, 7, 8, 9 
The only unit digit that are possible are 2, 4, 6, 8 
5x9x9x4=1620
 A: There might be a prettier solution to this problem, but I would use some variation of a decision tree:

The layers being the digits chosen left to right starting with the thousands, and the beige and blue circles being the odd and even options respectively. Each branch has the product taken and the different (exclusive) alternatives summed to 1288.

edit
fleablood's answer gives a far better order of choosing the digits: first, then last, then others. A diagram for this might be:

A: THe first digit can be $5,6,7,8,9$.  Those are four posibilities.
The second digit can be any of the ten $0,1,2..., 9$ (for some inexplicable reason you didn't include $0$) but the second digit must be different from the first.  So there are $9$ options.
The third digit must be different from the first two so there are $8$ options.
The fourth must be different for the first three so so there are $7$ options.
The last digit must be even so it is $0,2,4,6,8$ and it must be different that the first four and ... we have no idea how many of the first four are even or not.  So we are found in the Alps.  Dang.
Start over.  
Do two cases.  Either the first number is even $6,8$ (2 options) or it is $5,7,9$ (3 options).
That last digit must be even so if the first is even then the second must be different so there are $4$ options because it must be different.  If the first is not even there are $5$ options.
The second digit is different that the first or last so there are $8$ options.
The third digit must be different than the other three so there are $7$ options.
So if the first digit is even there are $2*4*8*7$.  And if the first digit is not even there are $3*5*8*7$.
So there are $2*4*8*7 + 3*5*8*7 = (2*4 + 3*5)*8*7 = (8+15)*56 = 23*56= 1288$ such numbers.
A: The question says distinct digits, so if one digit appears in the hundreds place, it should not appear again in the tens place as example.
Take the case for 5 in the thousands place. We have five choices for the ones place $\{ 0,2,4,6,8 \}$. If we choose any digit for the ones place, that digit will be prohibited in the tens and hundreds places. 
This means that for the hundreds place two digits are prohibited (5 for the thousands place and a digit for the ones place), so we have 8 choices for the hundreds place. 
Similarly, now for the tens place three digits are prohibited (5 for the thousands place, a digit for the ones place, and another for the hundreds place), so we have 7 choices for the tens place. 
So, we have 1 * 5 * 8 * 7 choices for even numbers of four distinct digits greater than 5000 (that starts with 5). 
You can consider the cases where the thousands place takes 6, 7, 8, or 9.
