How many even 3 digit integers greater than 700 
How many even 3 digit integers greater than 700 with distinct non zero
  digits are there ?

My answer is: 


*

*the only hundred digit that are possible are 7, 8 and 9 (3)

*the only ten digit that are possible are 1, 2, 3, 4, 5, 6, 7, 8, 9
(9)

*the only unit digit that are possible are 2, 4, 6, 8 (4)


what I did is: $$ 3 \times 9 \times 4 = 108 $$
Where is my problem? Answer is given 77. 
 A: Your problem is not taking into account the word "distinct".
Edit: I'm adding some in response to further questioning in the comments.
Because of the "distinct" condition (combined with the even condition), you have to consider cases more carefully. The digit choices are not independent, so you can't just multiply.  But if you break things up into mutually exclusive cases where the digit choices are independent, you can multiply for each, then add them all together.
For example, counting numbers with first digit odd, second digit odd yields $2\times 4\times 4$.
There are essentially $4$ cases to work with in this way. Count each then add them up.
A: It's a bit like combinations. You have possible first digits $7,8,9$. For $7$ there are $8$ possible second digits $1,2,3,4,5,6,8,9$ and for each of the latter either $3$ or $4$ third digits. I guess, you know hot to proceed from here.
A: There are 299 numbers between 700 and 999.
Out of those you need to exclude, digits having 0 in the units place and digits having 0 in the tens place. So that's 30+27=57.
There are 149 odd numbers between the range. So your premise reduces to 93. These are all the zero less even numbers.
Out of the remaining, remove the numbers having multiple of 11 in tens and unit place,and Which are even like 722,744,822 and so on. These are 4 per 100 range so that's 12.  So we reach to 81. 
Finally, remove remaining numbers with non distinct digits,which did not belong to above criteria. These are only four 988,998,898 and 848. So the answer is 77 even,non zero distinct digit numbers.
