what's the difference between these two different combinations? $$\frac{{}_{17}C_5 \cdot{}_3C_0}{{}_{20}C_5}$$
and
$$\frac{{}_{17}C_5}{{}_{20}C_5}$$
They give different numbers of course, but they seem to both calculate the same thing. Can you give like write two English sentences that shows what exactly they might be computing here, so I can see the difference and know when to use which? 
 A: They don't give different numbers, since ${}_3 C_0 = 1$.
Suppose you will randomly choose a subcommittee of five members from a committee 17 Republicans and 3 Democrats.  You ask what is the probability that all five are Republicans.
You can say it's the number of ways of choosing 5 Republicans out of 17, divided by the number of ways of choosing 5 members out of 20;
OR
You can say it's the number of ways of choosing 5 Republicans out of 17 AND 0 Democrats out of 3, divided by the number of ways of choosing 5 members out of 20.
The second way requires you to multiply by ${}_3 C_0 = 1$.
But the answer should be the same either way.  That's one reason why it matters that the number of ways to choose $0$ out of $3$ is $1$.  It's also a reason why it matters that multiplying by $1$ is the same as not multiplying by anything at all.  Thus
$$
\prod_{x\in\varnothing} x = 1.
$$
The product of no numbers is $1$, just as the sum of no numbers is $0$.
Here's another way to look at it:  You're going to choose 3 members of the committee above.   Here are some possible results:
$$
\begin{array}{cccc}
ddd & rdd & rrd & rrr \\
    & drd & rdr \\
    & ddr & drr
\end{array}
$$
So:
The number of ways to get 3 Democrats out of 3 is 1.
The number of ways to get 2 Democrats out of 3 is 3.
The number of ways to get 1 Democrat out of 3 is 3.
The number of ways to get 0 Democrats out of 3 is 1.
Thus ${}_3C_3=1,\qquad {}_3C_2=3,\qquad {}_3C_1=3,\qquad{}_3C_0=1$.
