Using the numbers $0,1,2,\dots,9$ as digits, how many four-digit numbers exist for which the following three conditions hold simultaneously:
(1) all digits are different,
(2) two digits are even numbers, and
(3) two digits are odd numbers?
Recall that $0$ is an even number, and that a four-digit number by definition does not start by $0$.
Best Answer
Hint: The answer is $$4!\times\binom{5}{2}\times\binom{5}{2}-3!\times\binom{5}{2}\times\binom{4}{1}$$