Combinatorics and elementary probability 
$1$) Using the numbers $0,1,2,...,9$  as digits, how many $4$-digit numbers exist in which all digit are different with two digits as even numbers and two as odd numbers?

(Note: $0$ is an even number, and that $4$-digit number do not start with $0$)
 A: First pick 2 odd numbers and 2 even numbers. There are $\binom{5}{2}^2$ ways to do this. Now order them. There are $4!$ ways. So there are $4!\binom{5}{2}^2$ total numbers, but we have included 4-digit numbers that start with 0. Now let's count the number that start with $0$. There are $4$ choices for the other even digit, $\binom{5}{2}$ choices for the odd digits, and $3!$ orders. So for our final answer we have: $$4!\binom{5}{2}^2-3!(4)\binom{5}{2}$$
A: Another possibility would be this : consider the $4$-digit numbers without $0$ in them. There is $4$ possible even numbers and $5$ possible odd numbers, so $\binom 52 \binom 42 = 60$ possibilities. There are $4! = 24$ ways to place them, hence you have $60 \times 24 = 1440$ such numbers.
Now consider those with $0$ in them. The zero must be one of the $3$ first digits. Now for the number possibilities, there are $4$ choices for the even number and $\binom 52 = 10$ choices for the odd numbers, hence $40$ choices for the numbers. Afterwards, there are $3$ possibilities for the position of $0$ and $3!=6$ choices for the position of the three other digits, hence $18$ possibilities in total. Therefore there is $40 \times 18 = 720$ such digits. 
Summing up gives $1440 + 720 = 2160 = 4! \binom 52^2 - 3!(4) \binom 52$ possibilities. 
Hope that helps,
