Since this problem appears in a section on the Inclusion-Exclusion Principle, your first interpretation is almost certainly not what the author intended.
As N. Shales pointed out, one interpretation could be:
How many four-digit positive integers are there in which no digit is repeated or are odd numbers that contain repeated digits?
As you correctly calculated, the number of four-digit positive integers with no repeated digits is $9 \cdot 9 \cdot 8 \cdot 7 = 4536$. The number of odd four-digit integers is $9 \cdot 10 \cdot 10 \cdot 5 = 4500$ since there are five choices for the units digit (which must be odd), nine choices for the thousands digit (since zero is excluded), and ten choices each for the hundreds and tens digits. The number of odd numbers with no repeated digits is $8 \cdot 8 \cdot 7 \cdot 5 = 2240$ since we have five choices for the units digit, eight choices for the thousands digit (which can be neither zero nor equal to the units digit), eight choices for the hundreds digit (which cannot be equal to the thousands digit or the units digit), and seven choices for the tens digit (which cannot be equal to the thousands digit, hundreds digit, or units digit). By the Inclusion-Exclusion Principle, there are
$$9 \cdot 9 \cdot 8 \cdot 7 + 9 \cdot 10 \cdot 10 \cdot 5 - 8 \cdot 8 \cdot 7 \cdot 5 = 4536 + 4500 - 2240 = 6796$$
four-digit positive integers in which no digit is repeated or are odd numbers that contain repeated digits.
How many four-digit positive integers are there in which no digit is repeated or are numbers in which all the digits are odd and repeated digits may appear?
This is your second interpretation. You correctly calculated that there are
$$9 \cdot 8 \cdot 7 \cdot 6 + 5^4 - 5 \cdot 4 \cdot 3 \cdot 2 = 4536 + 625 - 120 = 5041$$
such numbers. We can confirm this by calculating the number of four-digit positive integers consisting of only odd digits that contain a repeated digit. This is a bit messy. We consider cases:
Exactly one digit is used in all four positions: There are $5$ such numbers since there are $5$ odd digits.
Exactly two digits are used, with one used three times and the other once: There are $5$ choices for the repeated digit, $4$ choices for the remaining digit, and $4$ choices for the placement of the digit that is used exactly once, so there are
$$5 \cdot 4 \cdot 4 = 80$$
such numbers.
Exactly two digits are used, with each used twice: There are $\binom{5}{2}$ ways of choosing the two odd digits and $\binom{4}{2}$ ways to select the positions occupied by the larger of these two digits. Hence, there are
$$\binom{5}{2}\binom{4}{2} = 60$$
such numbers.
Exactly three digits are used, with one number used twice: There are five ways of selecting the repeated digit, $\binom{4}{2}$ ways of selecting the positions occupied by these digits, four ways of filling the leftmost open position, and three ways of filling the remaining position. Hence, there are
$$5 \cdot \binom{4}{2} \cdot 4 \cdot 3 = 360$$
such numbers.
Thus, in total, there are
$$5 + 80 + 60 + 360 = 505$$
four-digit integers consisting of only odd digits that contain repeated digits. Adding this to the $4536$ four-digit integers that do not contain repeated digits yields the
$$4536 + 505 = 5041$$
numbers you found more easily by using the Inclusion-Exclusion Principle.
How many four-digit positive integers are there in which no digit is repeated or in which the only digits that are repeated are odd numbers?
This is your first interpretation. You calculated this incorrectly. Before I explain why, let's calculate the number of four-digit positive integers in which the only repeated digits are odd numbers.
Exactly one odd digit is used in all four positions: As explained above, there are $5$ such numbers.
Exactly two digits are used, with an odd digit used in three of the four positions: We have to consider cases, depending on whether the thousands place is occupied by the repeated digit.
If the thousands place is not occupied by the repeated digit, there are eight choices for the thousands place (since it cannot be zero or the repeated digit) and five choices for the repeated digit. Hence, there are $8 \cdot 5 = 40$ such numbers.
If the thousands place is occupied by the repeated digit, there are $\binom{3}{2}$ ways of selecting the other two positions of that digit, five ways of choosing the digit, and $9$ ways of filling the remaining position. Hence, there are
$$\binom{3}{2} \cdot 5 \cdot 9 = 135$$
such numbers.
In total, there are $40 + 135 = 175$ four-digit positive integers in which exactly two digits are used and one odd digit is used in three of the four positions.
Exactly two digits are used, with two odd digits each filling two of the four positions: There are $\binom{5}{2}$ ways of selecting two of the five odd digits and $\binom{4}{2}$ ways of choosing which two of the four positions will be filled by the larger of them. Hence, there are
$$\binom{5}{2}\binom{4}{2} = 60$$
such numbers.
Exactly three digits are used, with one odd digit used twice: We have to consider cases, depending on whether the repeated digit is used in the thousands place.
If the thousands place is not occupied by the repeated digit, there are eight ways to fill the thousands place (since it cannot be zero or the repeated digit). There are five choices for the repeated digit, $\binom{3}{2}$ ways of selecting which two of the remaining three positions it will fill, and eight choices for the remaining digit (since it cannot be the repeated digit or the thousands digit). Hence, there are
$$8 \cdot 5 \cdot \binom{3}{2} \cdot 8 = 960$$
such numbers.
If the thousands place is occupied by the repeated digit, there are five ways to fill the thousands place (since it must be odd), three ways to select the other digit occupied by the repeated digit, nine ways to fill the leftmost open position, and eight ways to fill the remaining open position. Hence, there are
$$5 \cdot 3 \cdot 9 \cdot 8 = 1080$$
such numbers.
In total, there are $960 + 1080 = 2040$ four-digit positive integers using exactly three digits in which one odd digit is repeated.
Total: Adding these cases yields
$$5 + 175 + 60 + 2040 = 2280$$
four-digit positive integers containing repeated digits in which the only digits that are repeated are odd digits.
Your tried to use the Inclusion-Exclusion Principle to do this. Let's see what went wrong.
As you noted, there are $10^4$ four-digit strings. However, the number of such strings with no repeated digits is $10 \cdot 9 \cdot 8 \cdot 7 = 5040$. Hence, there are $$10^4 - 10 \cdot 9 \cot 8 \cdot 7 = 1000 - 5040 = 4960$$ such strings with repeated digits. You attempted to use symmetry to eliminate those that contain repeated even digits by dividing by $2$. Doing so yields
$2480$. As you realized, we are still counting strings with two even and two odd digits. There are actually
$$5 \cdot 5 \cdot \binom{4}{2} = 150$$
such strings since there are five choices of even digit, five choices of odd digit, and $\binom{4}{2}$ ways of selecting the positions of the even digit. Half of these were eliminated when we divided by $2$, leaving $75$ more to be eliminated. This leaves us with $$2480 - 75 = 2405$$ four-digit strings which contain only repeated odd digits. However, not all these strings are four-digit positive integers since the first digit may be zero. We must eliminate these.
Three-digit positive integers in which a single odd digit is used in all three positions: There are $5$ such numbers since there are five odd digits.
Three-digit odd positive integers in which exactly two digits are used and an odd digit is repeated: Notice that we have already eliminated numbers in which a zero appears since it would be a repeated even digit (as zero also occupies the thousands place). There are five ways of choosing the repeated digit, $\binom{3}{2}$ ways of selecting two of the three locations for that digit, and eight ways to fill the remaining digit. Hence, there are
$$5 \cdot \binom{3}{2} \cdot 8 = 120$$
such numbers.
Hence, there are
$$\frac{10^4 - 10 \cdot 9 \cdot 8 \cdot 7}{2} - \frac{5 \cdot 5 \cdot \binom{4}{2}}{2} - 5 - 5 \cdot \binom{3}{2} \cdot 8 = 2480 - 75 - 5 - 120 = 2280$$ four-digit positive integers in which the only repeated digits are odd, which agrees with the result we obtained above.
Thus, there are a total of
$$4536 + 2280 = 6816$$
four-digit positive integers in which no digit is repeated or the only digits that are repeated are odd numbers.
Best Answer
There are two options for doing this problem. You can either place the digits or you can select the digits and then arrange them.
In your attempt, you effectively placed the two digits, then arranged them.
Placing the digits: There are four ways to place the $2$ and three ways to place the $5$ in one of the three remaining positions. There are eight ways to fill the first open position with one of the remaining digits and seven ways to fill the final open position with one of the remaining digits. Hence, there are $$4 \cdot 3 \cdot 8 \cdot 7 = 672$$ $4$-digit pin codes with distinct digits that include both a $2$ and a $5$.
Selecting the digits: You must select both $2$ and $5$. You must also select two of the remaining eight digits, which can be done in $\binom{8}{2}$ ways. Once you have selected the four digits which are to be used, you can arrange the selected digits in $4!$ ways, giving $$\binom{2}{2}\binom{8}{2}4! = 672$$ $4$-digit pin codes with distinct digits that include both a $2$ and a $5$.
Why is your approach wrong?
In your approach, you imposed an order on the selection of the additional digits by selecting the first additional digit in eight ways and the second additional digit in seven ways. If you select $3$ as your first additional digit and then $4$ as your second additional digit, your method gives $4!$ ways to arrange the digits $2, 3, 4, 5$. However, if you were to select $4$ as the first additional digit and $3$ as the second additional digit, your method also gives $4!$ arrangements of the digits $2, 3, 4, 5$. You only want to count such arrangements once. By selecting the additional digits in order, you end up counting each admissible arrangement twice.