It is easier to count the total directly: you have five possibilities for the first digit (any digit except $0$), and six each for the remaining three digits, so the total number is $5\times 6^3$. This agrees with your count ($6^4-6^3 = 6^3(6-1) = 6^3\times 5$), but I think it's easier to just count them directly.
You can use the same method for counting the possibilities of the first two digits: five possibilities for the first digit, six for the second, for a total of $5\times 6$; this is the same as your count, $6^2-6^1 = 6(6-1)$.
The rest is almost correct; again, you can be a bit briefer with the count of the last two digits: the first digit can be any of the six possibilities. If the first digit is 0, 2, or 4, then the second digit must be 0 or 4 to get a multiple of $4$; so for each of the three possibilities you have two $2$-digit combinations, giving $3\times 2=6$ possibilities. If the first digit is $1$, $3$, or $5$, then the second digit must be $2$, so you have only three more. The total here is the sum of the two, so we have $6+3=9$ possibilities.
What you forgot is that $00$ also works.
Since the choices for the first two digits are independent of the choices for the last two, so you multiply the totals. The total will then be
$$5\times 6 \times 9 = 270.$$
The fact that the total number (1080) is larger than the count of those that are divisible by 4 should not be a surprise: there are lots of numbers among the 1080 4-digit numbers in which every digit is one of 0, 1, 2, 3, 4, and 5 that are not divisible by $4$. So I'm not sure what your last line is meant to represent.
Let's confirm your solution.
There are $10^9$ strings of length $9$ that can be formed by using the ten decimal digits with repetition. From these, we must exclude those strings in which at least one odd digit is missing.
Let $A_i$ be the set of outcomes in which the digit $i$ is excluded, where $i \in \{1, 3, 5, 7, 9\}$.
Then, by the Inclusion Principle, the number of strings in which at least one odd digit is missing is
$$\sum_{i} |A_i| - \sum_{i < j} |A_i \cap A_j| + \sum_{i < j < k} |A_i \cap A_j \cap A_k| - \sum_{i < j < k < l} |A_i \cap A_j \cap A_k \cap A_l| + \sum_{i < j < k < l < m} |A_i \cap A_j \cap A_k \cap A_l \cap A_m|$$
$|A_1|$: Since $1$ is excluded each of the nine positions in the string can be filled in nine ways. Hence, $|A_1| = 9^9$. By symmetry,
$$|A_1| = |A_3| = |A_5| = |A_7| = |A_9|$$
$|A_1 \cap A_3|$: Since both $1$ and $3$ are excluded, each of the nine positions in the string can be filled in eight ways. Hence, $|A_1 \cap A_3| = 8^9$. By symmetry,
$$|A_1 \cap A_3| = |A_1 \cap A_5| = |A_1 \cap A_7| = |A_1 \cap A_9| = |A_3 \cap A_5| = |A_3 \cap A_7| = |A_3 \cap A_9| = |A_5 \cap A_7| = |A_5 \cap A_9| = |A_7 \cap A_9|$$
$|A_1 \cap A_3 \cap A_5|$: Since $1$, $3$, and $5$ are all excluded, each of the nine positions in the string can be filled in seven ways. Hence, $|A_1 \cap A_3 \cap A_5| = 7^9$. By symmetry,
$$|A_1 \cap A_3 \cap A_5| = |A_1 \cap A_3 \cap A_7| = |A_1 \cap A_3 \cap A_9| = |A_1 \cap A_5 \cap A_7| = |A_1 \cap A_5 \cap A_9| = |A_1 \cap A_7 \cap A_9| = |A_3 \cap A_5 \cap A_7| = |A_3 \cap A_5 \cap A_9| = |A_3 \cap A_7 \cap A_9| = |A_5 \cap A_7 \cap A_9|$$
$|A_1 \cap A_3 \cap A_5 \cap A_7|$: Since $1$, $3$, $5$, $7$, and $9$ are all excluded, each of the nine positions in the string can be filled in six ways. Hence,
$|A_1 \cap A_3 \cap A_5 \cap A_7| = 6^9$. By symmetry,
$$|A_1 \cap A_3 \cap A_5 \cap A_7| = |A_1 \cap A_3 \cap A_5 \cap A_9| = |A_1 \cap A_3 \cap A_7 \cap A_9| = |A_1 \cap A_5 \cap A_7 \cap A_9| = |A_3 \cap A_5 \cap A_7 \cap A_9|$$
$|A_1 \cap A_3 \cap A_5 \cap A_7 \cap A_9|$: Since $1$, $3$, $5$, $7$, and $9$ are all excluded, each of the nine positions in the string can be filled in five ways. Hence, $|A_1 \cap A_3 \cap A_5 \cap A_7 \cap A_9| = 5^9$.
Thus, by the Inclusion-Exclusion Principle, the number of strings of length $9$ in which at least one odd digit is missing is
$$5 \cdot 9^9 - 10 \cdot 8^9 + 10 \cdot 7^9 - 5 \cdot 6^9 + 5^9$$
Therefore, the number of strings of length $9$ in which no odd digits are missing is
\begin{align*}
10^9 - (5 \cdot 9^9 - 10 \cdot 8^9 + 10 \cdot 7^9 - 5 \cdot 6^9 + 5^9) & = 10^9 - 5 \cdot 9^9 + 10 \cdot 8^9 - 10 \cdot 7^9 + 5 \cdot 6^9 - 5^9\\
& = \binom{5}{0}10^9 - \binom{5}{1}9^9 + \binom{5}{2}8^9 - \binom{5}{3}7^9 + \binom{5}{4}6^9 - \binom{5}{5}5^9\\
& = \sum_{k = 0}^{5} (-1)^{k} \binom{5}{k}(10 - k)^9
\end{align*}
where $\binom{5}{k}$ is the number of ways of excluding $k$ of the $5$ odd digits and $(10 - k)^9$ is the number of ways of filling the nine positions of the string with the remaining $10 - k$ decimal digits.
Best Answer
There are a total of $5$ digits
The $3$ digit block $4,5,6$ can be placed in $3$ positions. ($1,2,3$ and $2,3,4$ and $3,4,5$)
$3×10×10$
So there are a total of $300$ numbers in which the digits $4,5,6$ appear in this order consecutively