This is simillar to another question that was asked on here but with 4 digits instead, I have already seen that question and used the same method, but for some reason I am still getting this question wrong.
Here is my thinking
XXXXX 5 spots
$9*10^4$ total numbers
17XXX
X17XX
XX17X
XXX17
$(10^3+9*10^2+9^2*10+9^3)$ = total number of numbers with 17 in them.
But since we counted some numbers twice
$(10^3+9*10^2+9^2*10+9^3 - 29)$
$9*10^4 - (10^3+9*10^2+9^2*10+9^3 - 29) = 86590$ using calculator
It says my answer is wrong, but I can't figure out what I have failed to take into account.
As TonyK pointed out in the comments, you overlooked the fact that while the leading digit cannot be equal to zero, the remaining digits can be equal to zero.
Let $A_i$, $1 \leq i \leq 4$, be the set of five-digit positive integers in which the sequence $17$ appears beginning in the $i$th position.
Since there are $9 \cdot 10^4$ positive integers with five digits, the number of five-digit positive integers in which the sequence $17$ does not appear is $$9 \cdot 10^4 - |A_1 \cup A_2 \cup A_3 \cup A_4|$$ By the Inclusion-Exclusion Principle, \begin{align*} |A_1 \cup A_2 \cup A_3 \cup A_4| & = \sum_{i = 1}^{4} |A_i| - \sum_{1 \leq i < j \leq 4} |A_i \cap A_j|\\ & \qquad + \sum_{1 \leq i < j < k \leq 4} |A_i \cap A_j \cap A_k| - |A_1 \cap A_2 \cap A_3 \cap A_4\\ & = |A_1| + |A_2| + |A_3| + |A_4|\\ & \quad - |A_1 \cap A_2| - |A_1 \cap A_3| - |A_1 \cap A_4| - |A_2 \cap A_3| - |A_2 \cap A_4| - |A_3 \cap A_4|\\ & \qquad + |A_1 \cap A_2 \cap A_3| + |A_1 \cap A_2 \cap A_4| + |A_1 \cap A_3 \cap A_4| + |A_2 \cap A_3 \cap A_4|\\ & \quad\qquad - |A_1 \cap A_2 \cap A_3 \cap A_4| \end{align*}
$|A_1|$: Since the sequence $17$ appears in the first two positions, there are $10$ choices for each of the remaining digits. Hence, $|A_1| = 10^3$.
$|A_2|$: Since the sequence $17$ appears in the second and third positions, there are nine choices for the leading digit and $10$ choices for each of the remaining digits. Hence, $|A_2| = 9 \cdot 10^2$.
By symmetry, $|A_2| = |A_3| = |A_4|$.
$|A_1 \cap A_2|$: This means that the sequence $17$ appears in both the first two positions and the second and third positions, which is impossible since the second digit would have to be both $1$ and $7$. Hence, $|A_1 \cap A_2| = 0$.
By symmetry, $|A_1 \cap A_2| = |A_2 \cap A_3| = |A_3 \cap A_4|$.
$|A_1 \cap A_3|$: This means that the sequence $17$ appears in both the first and second positions and third and fourth positions. There are $10$ choices for the fifth position. Hence, $|A_1 \cap A_3| = 10$.
By symmetry, $|A_1 \cap A_3| = |A_1 \cap A_4|$.
$|A_2 \cap A_4|$: This means that the sequence $17$ appears in both the second and third positions and in the fourth and fifth positions. There are $9$ choices for the leading digit. Hence, $|A_2 \cap A_4| = 9$.
Since there cannot be more than two appearances of the sequence $17$ in a five-digit positive integer, each of the remaining terms is equal to zero.
Hence, $$|A_1 \cup A_2 \cup A_3 \cup A_4| = 10^3 + 3 \cdot 9 \cdot 10^2 - 2 \cdot 10 - 9$$ Therefore, the number of five-digit positive integers that do not contain the sequence $17$ is $$9 \cdot 10^4 - 10^3 - 3 \cdot 9 \cdot 10^2 + 2 \cdot 10 + 9 = 86,329$$