Password length is 5 or 6 characters. Characters allowed are $a-z$, $A-Z$, $0-9$.
The password must contain at least one lowercase letter, at least one uppercase letter and at least one digit.
Use the inclusion, exclusion principal. $|\{lud\}| - |\{lu\}|-|\{ld\}|-|\{ud\}| + |\{l\}|+|\{u\}|+|\{d\}|$
$$C= (26\!+\!26\!+\!10)^5(26\!+\!26\!+\!10\!+\!1)\!-\!(26\!+\!26)^5(26\!+\!26\!+\!1)\!- 2(26\!+\!10)^5(26\!+\!10\!+\!1)\!+\!2(26)^5(26\!+\!1)\!+\!(10)^5(10\!+\!1)$$
$$C= (62)^5(63) - (52)^5(53) - 2(36)^5(37) + 2(26)^5(27)+(10)^5(11)$$
$$C= 57,\!716,\!368,\!416 - 20,\!150,\!813,\!696 - 4,\!474,\!497,\!024 + 641,\!594,\!304+1,\!100,\!000$$
$$C= 57,\!716,\!368,\!416 -24,\!625,\!310,\!720 + 642,\!694,\!304$$
$$C= 33,\!733,\!752,\!000$$
but it doesn't match this:
$(26∗26∗10∗62∗62)+(26∗26∗10∗62∗62∗62)=1,637,082,720$
That only counts passwords starting with one upper case, one lower case, and one digit, --in that order-- followed by 2 or 3 more symbols from any group.
It does not count, for instance, "$\mathrm{HeLL0}$"
There are $63$ ($26 + 26 + 10 +1$) allowed characters. Only one is forbidden on the first place (hyphen).
So there are 62 passwords of length 1.
There are $62 \times 63$ passwords of length 2.
There are $62 \times 63 \times 63$ passwords of length 3, etc.
Now sum up to and including length 8 (and use a formula for a finite sum of this type, if you know it).
To expand on the last bit: note that for $a \neq 1$: $1 + a + a^2 + \ldots + a^k = \frac{a^{k+1}-1}{a-1}$ (proof: multiply both sides by $a-1$). In the sum, get the 62 out, and use this result to see we can write is directly as $63^8 - 1$.
The latter suggests a direct proof of that fact. Consider all sequences of all 63 allowed characters of length 8. Remove all consecutive hyphens from the start onwards. This gives all allowed passwords exactly once. The only problem is the all hyphen word, which would result in the empty password (which I assume is not allowed). Hence $63^8 - 1$.
Best Answer
No, Inclusion-Exclusion is not that simple. See this article for an introduction to Inclusion-Exclusion. Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula.
For any set $~E~$ with a finite number of elements, let $~|E|~$ denote the number of elements in the set $~E.$
Let $~S~$ denote all of the passwords possible, if the requirements of digits, special characters, upper case letters, and lower case letters are ignored.
Let $~S_1~$ denote the subset of $~S~$ that violates the constraint that at least one digit must be present.
Let $~S_2~$ denote the subset of $~S~$ that violates the constraint that at least one special character must be present.
Let $~S_3~$ denote the subset of $~S~$ that violates the constraint that at least one lower case letter must be present.
Let $~S_4~$ denote the subset of $~S~$ that violates the constraint that at least one upper case letter must be present.
Then, the desired computation is
$$|S| - |S_1 \cup S_2 \cup S_3 \cup S_4|. \tag1 $$
Let $~T_0~$ denote $~|S|.$
For $~r \in \{1,2,3,4\},$
let $~T_r~$ denote $\displaystyle \sum_{1 \leq i_1 < i_2 < \cdots < i_r \leq 4} | ~S_{i_1} \cap S_{i_2} \cap \cdots \cap S_{i_r} ~|.$
That is, $~T_r~$ denotes the sum of $~\displaystyle \binom{4}{r}~$ terms.
Then, in accordance with Inclusion-Exclusion theory, the computation in (1) above equals
$$\sum_{r=0}^4 (-1)^r T_r. \tag2 $$
So, the problem reduces to computing $~T_0, ~T_1, ~T_2, ~T_3, ~T_4.$
$\underline{\text{Computation of} ~T_0}$
There are $~77~$ choices for each of $~8~$ letter positions.
Therefore,
$$T_0 = (77)^8.$$
$\underline{\text{Computation of} ~T_1}$
To compute $~|S_1|~$ note that there are $~67~$ choices for each of the $~8~$ character positions. This is because $~S_1~$ represents the subset of $~S~$ where all digits are excluded. This means that instead of there being $~77~$ choices for each character position, there are only $~(77 - 10) = 67~$ choices for each character position.
Therefore, $~|S_1| = (67)^8.$
Very similarly:
$~|S_2| = (62)^8.$
$~|S_3| = (51)^8.$
$~|S_4| = (51)^8.$
Therefore,
$$T(1) = |S_1| + |S_2| + |S_3| + |S_4| \\ = (67)^8 + (62)^8 + (51)^8 + (51)^8.$$
$\underline{\text{Computation of} ~T_2}$
Similar to the analysis in the previous section,
to compute $|S_1 \cap S_2|,~$ you must reason that with no digits or special characters, there are $~52~$ choices for each of the 8 character positions.
Therefore, $~|S_1 \cap S_2| = (52)^8.$
Similarly:
$~|S_1 \cap S_3| = (41)^8.$
$~|S_1 \cap S_4| = (41)^8.$
$~|S_2 \cap S_3| = (36)^8.$
$~|S_2 \cap S_4| = (36)^8.$
$~|S_3 \cap S_4| = (25)^8.$
Therefore,
$$T_2 = (52)^8 + (41)^8 + (41)^8 + (36)^8 + (36)^8 + (25)^8.$$
$\underline{\text{Computation of} ~T_3}$
$|S_1 \cap S_2 \cap S_3| = (26)^8,~$
since this represents the set where only capital letters are used in each of the $~8~$ character positions.
Similarly:
$|S_1 \cap S_2 \cap S_4| = (26)^8.$
$|S_1 \cap S_3 \cap S_4| = (15)^8.$
$|S_2 \cap S_3 \cap S_4| = (10)^8.$
Therefore,
$$T_3 = (26)^8 + (26)^8 + (15)^8 + (10)^8.$$
$\underline{\text{Computation of} ~T_4}$
$S_1 \cap S_2 \cap S_3 \cap S_4 ~$ represents the set of all 8 character sequences with no digits, special characters, lower case letters or upper case letters.
Therefore, $~S_1 \cap S_2 \cap S_3 \cap S_4 = \emptyset.$
Therefore,
$$T_4 = 0.$$
$\underline{\text{Final Computation}}$
$$T_0 = (77)^8.$$
$$T(1) = (67)^8 + (62)^8 + (51)^8 + (51)^8.$$
$$T_2 = (52)^8 + (41)^8 + (41)^8 + (36)^8 + (36)^8 + (25)^8.$$
$$T_3 = (26)^8 + (26)^8 + (15)^8 + (10)^8.$$
$$T_4 = 0.$$
The expression in (1) above equals
$$T_0 - T_1 + T_2 - T_3 + T_4.$$