How many 5 letter different words can be formed using the letters AAASSSBB I saw that these sort of questions have been asked before, but not exactly this kind of problem: creating 5 DIFFERENT words, not just all possible combinations.
The words don't need to have any meanings.
I'd appreciate an explanation in addition to the solution.
 A: $$
\begin{array}{c|c|c}
\text{Letters}  & \text{Calculation} & \text{# Possibilities}\\
\hline\\
AAASS &  \displaystyle\binom{5}{3} & 10 \\\\
AAASB &  \displaystyle\binom{5}{3} \cdot 2 & 20\\\\
AAABB &  \displaystyle\binom{5}{3} & 10\\\\
AASSS &  \displaystyle\binom{5}{2} & 10\\\\
AASSB &  \displaystyle\binom{5}{2} \cdot \binom{3}{2} & 30\\\\
AASBB &  \displaystyle\binom{5}{2} \cdot 3 & 30\\\\
ASSSB &  5\cdot \displaystyle\binom{4}{3} & 20\\\\
ASSBB &  5\cdot \displaystyle\binom{4}{2} & 30\\\\
SSSBB &  \displaystyle\binom{5}{3} & 10\\\\
\end{array}
$$

This gives a total of $$\boxed{170}$$
A: One way is to use generating functions. Find the coefficient of $x^5$ in: 
$$f(x) = 5! \cdot (1 + x + x^2/2! + x^3/3!)^2 \cdot (1+x + x^2/2!) $$
In general, the number of length-$n$ words that can be achieved from an alphabet of $k$ distinct letters containing $b_i$ counts of letter $i$ (over all $1 \leq i \leq k$), is found by taking the coefficient of $x^n$ of $f(x)$ where:
$$f(x) = n!\prod_{i=1}^{k}\left(\sum_{j=0}^{b_i}\frac{x^j}{j!}\right)$$ 
One nice thing about this approach (in my opinion) is that it is generalizable and easily adjustable, and doesn't require individual case-counting.
A: There are words that use only two different letters, and there are words that use all three.
$2$ letters:
It must be three of one and two of another.
$3$ $A$'s, $2$ $S$'s
$3$ $A$'s, $2$ $B$'s
$3$ $S$'s, $2$ $A$'s
$3$ $S$'s, $2$ $B$'s
In any case, the number of words is $\binom53=\frac{5!}{3!2!}=\frac{5\cdot4}{2\cdot1}=10$
With four choices of letter-pairs, that gives us $40$ words.

$3$ letters:
The possible formats here are $3$ of one letter, and $1$ each of the other two, or $1$ of one letter, and $2$ each of the other two.
$3,1,1$:
The letter occurring $3$ times could be $A$ or $S$, so there are two lists of letters to work with. In either case, you can form $20$ words. That's $\binom53=10$ choices for where the three copies of one letter go, and $2$ ways to fill the remaining blanks.
That's another $40$ words.
$2,2,1$:
Now any of the three letters can be the one that occurs once. Given a letter-list, We have $5$ ways to choose where the lone letter goes, and $\binom42=6$ ways to place the remaining ones. We have $3\cdot 5\cdot 6=90$

Finally, $40+40+90=170$
A: If we were just forming length-$5$ words that consisted of the three letters $A,S,B$, there would be $3^5$ options - each letter position has $3$ independent choices.
We need to restrict that total to reflect the restrictions on count of each letter. In this case you cannot have a word that breaks constraint on $2$ letters at the same time. 
How many of those unconstrained words have $4$ or $5$ copies of $A$? Clearly the answer is $1+2\binom 5 4 = 11$:
$AAAAA$
$BAAAA$
$SAAAA$
$ABAAA$
etc.
and similarly for $S$. Then for $B$-constraint-breaking words, there are those $11$ plus $\binom 5 3(2+2) = 40$. So we end with $3^5-73=170$ valid options.
A: $${5!\over 3!2!0!}+ {5!\over 3!1!1!}+{5!\over 3!0!2!}+{5!\over 2!3!0!}+{5!\over 2!2!1!}+{5!\over 2!1!2!}+{5!\over 1!3!1!}+{5!\over 1!2!2!}+{5!\over 0!3!2!} =$$
$$=4{5!\over 3!2!}+ 2{5!\over 3!}+3{5!\over 2!2!} = 170$$
