Combinatorics Password Selection The Problem
We want to know the amount of 8 digit passwords we can make using digits, uppercase, lowercase, and special characters. For this problem there are only 16 special characters. The rules are that the passwords must contain at least 1 uppercase, 1 digit, and 1 special character.

My Approach
After drawing a diagram Venn Diagram of 4 sets and assigning Uppercase to 'A', Digits to 'B', Special Characters to 'C', and Lowercase to 'D' I noticed that the total number we are interested in are the intersects of:  
|A $\cap$ B $\cap$ C $\cap$ D| + |A $\cap$ B $\cap$ C|.

Based on the inclusion/exclusion principle the total number of combinations for 4 sets are:
          $|A∪B∪C∪D|=$
$|A|+|B|+|C|+|D|$ all singletons
$-(|A∩B|+|A∩C|+|A∩D|+|B∩C|+|B∩D|+|C∩D|)$ all pairs
$+(|A∩B∩C|+|A∩B∩D|+|A∩C∩D|+|B∩C∩D|)$ all triples
$-|A∩B∩C∩D|$ all quadruples.  
$(26+10+16+26)^8$(singles)$-[42^8+36^8+52^8+26^8+42^8+36^8]$(pairs)$+[52^8+68^8+62^8+52^8]$(tripled pairs)$-[26^8+16^8+10^8+26^8]$(quadrupled pairs) 
=20734390444484096.

This is where I began getting confuzzled.
My intuition tells me to start subtracting amounts that we don't need from the total, which is everything but |A $\cap$ B $\cap$ C $\cap$ D| + |A $\cap$ B $\cap$ C| which feels the same as just substituting values $(26^8+16^8+10^8+26^8) + (52^8) =$ 53881777627904.

Is this the correct answer?
Is there a better way of going about this problem?
 A: I think it is a bit easier than what you did if you handle $D$ more freely.


*

*Take all combinations: $(A+B+C+D)^8$

*Exclude combinations that do not have (at least) one of $A, B, C$ = $ (B+C+D)^8 + (A+C+D)^8 + (A+B+D)^8$

*Include combinations that do not have (at least) two of $A, B, C$ = $ (C+D)^8 + (B+D)^8 + (A+D)^8$

*Exclude combinations that do not have three of $A, B, C$ = $ D^8$


The inclusion/exclusion principle lies in the fact that if you exclude and item because it does not have an $A$, but it also does not have a $B$, you have to include it again.
A: Starting from a similar point as you, but with slightly different logic from there, I got...
$$78^8-68^8-52^8-62^8+42^8+52^8+36^8-26^8 = 706905960284160 \approx 7.07E14$$
I started with the total number of possibilities.
$$78^8$$
 Then, I subtracted all possibilities without a digit, without an upper case, and without a special, respectively.
$$-68^8-52^8-62^8$$
Then, using inclusion-exclusion, I added back in all the possibilities having no digits or upper cases, no digits or specials, and no upper cases or specials.
$$+42^8+52^8+36^8$$
Finally, once again because of inclusion-exclusion, I subtracted the number of possibilities that have no digits no upper cases and no specials.
$$-26^8$$
A: Here's a somewhat simpler approach. We need to use 1 uppercase, 1 digit and 1 special character. There are $26\cdot 10\cdot 16$ ways to choose which 3 symbols we want to use. Then we have to choose a spot for them in the password. There are 8 possible spots for the uppercase, 7 for the digit and 6 for the special character. So far we have
$$26\cdot 10\cdot 16\cdot (8\cdot 7\cdot 6)$$
different ways of choosing these things. Now the other 5 symbols are completely free, they can be anything. There is a total of $26+26+10+16=78$ symbols, so we can choose these 5 symbols in $78^5$ ways. The answer is then
$$26\cdot 10\cdot 16\cdot (8\cdot 7\cdot 6)\cdot 78^5\simeq 4,03\cdot 10^{15}$$
