How many six-digit positive integers contain exactly two even digits?
Remember that an integer $n$ is said to be even if there exists an integer $m$ such that $n = 2m$. An integer that is not even is said to be odd. Hence, there are five even digits: $$0, 2, 4, 6, 8$$ and five odd digits: $$1, 3, 5, 7, 9$$
Your strategy of breaking the problem into two cases is correct. Let's correct your calculations for those cases.
Case 1: The leading digit is even.
Since the leading cannot be zero, there are four ways to fill the leading digit. There are five ways to choose the position of the other even digit and five ways to fill that position with an even digit. There are five ways to fill each of the four remaining positions with an odd digit.
$$4 \binom{5}{1} \cdot 5 \cdot 5^4 = 4 \binom{5}{1} 5^5$$
Case 2: The leading digit is odd.
Two of the remaining five positions must be filled with even digits. There are $\binom{5}{2}$ ways to select the positions of the even digits and five ways to fill each of those positions with an even digit. There are five ways to fill each of the four other positions with an odd digit.
$$\binom{5}{2} \cdot 5^2 \cdot 5^4 = \binom{5}{2} 5^6$$
Total: Since the above cases are mutually exclusive and exhaustive, the number of six-digit positive integers that contain exactly two even digits is
$$4 \binom{5}{1} 5^5 + \binom{5}{2} 5^6$$
In how many five-digit positive integers are there digits that appear more than once?
Hint: Subtract the number of five-digit positive integers in which all the digits are distinct from the total number of five-digit positive integers.
The total number of five-digit positive integers is $$9 \cdot 10 \cdot 10 \cdot 10 \cdot 10 = 90000$$ since the leading digit can be chosen in nine ways (as $0$ is prohibited) and each of the remaining digits can be selected in $10$ ways. Alternatively, the largest five-digit number is $99999$ and the largest number with fewer than five digits is $9999$, so the number of five-digit positive integers is $$99999 - 9999 = 90000$$ The number of five-digit positive integers with distinct digits is $$9 \cdot 9 \cdot 8 \cdot 7 \cdot 6 = 27216$$ since we have nine choices for the leading digit (as $0$ is prohibited), nine choices for the thousands digit (since we cannot use the leading digit), eight choices for the hundreds digit (since we cannot use the tens thousands or thousands digits), seven choices for the tens digit, and six choices for the units digit. Therefore, the number of five-digit positive integers that do contain a digit that appears more than once is $$9 \cdot 10 \cdot 10 \cdot 10 \cdot 10 - 9 \cdot 9 \cdot 8 \cdot 7 \cdot 6 = 90000 - 27216 = 62784$$
Best Answer
Let us define our notation. Let $\,A_n\,$ count the positive numbers with $\,n\,$ decimal digits such that no two adjacent digits are even. Among the numbers counted by $\,A_n\,$ let $\,B_n\,$ count those numbers which begin with an even digit and $\,C_n\,$ count those numbers which begin with an odd digit. Thus $\,A_n = B_n + C_n\,$ by definition. Now $\,B_n = 4 C_{n-1}\,$ because if a number begins with an even digit then the next digit must be odd. Now $\,C_n = 5 A_{n-1} + 5 C_{n-2}\,$ because if the first digit is odd and the second digit is not $0$ then we are in the $\,A_{n-1}\,$ case and there are $5$ odd digits while if the second digit is $0$ (which is even) then the next digit must be odd and we are in the $\,C_{n-2}\,$ case. In the equation for $\,C_n\,$ substitute the other two equations to get $$ C_n = 5(B_{n-1} + C_{n-1} + C_{n-2}) = $$ $$ 5(4C_{n-2} + C_{n-1} + C_{n-2}) = 5C_{n-1} + 25C_{n-2}. $$ Now $\,B_n\,$ satisfies the same recursion. Since $\,A_n = B_n + C_n\,$ so does $\,A_n.$ The initial values are $\, B_0 = 0,\; C_0 = 1,\; C_1 = 5.$