Selecting $5$ out of $40$ floppy disks such that at least/at most $1$ is defective 
There are 40 floppy disks out of which 5 are defective. In how many ways we can select 5 disks containing at least and at most 1 defective disk.

 A: Combinatorial question.
Think of the question as two pools of floppy discs (ignore the "youth of today" comment): one that is all fine (35 of them) and the other that is all defective (5).
HARD WAY
Case 1: Exactly one is defective:
Five DIFFERENT ways to pick one from defective. ${5}\choose{1}$
And, 52360 ways of picking 4 out of 35 non defective. ${35}\choose{4}$
Case 2: Exactly two are defective:
10 different ways to pick exactly 2 from defective pile. ${5}\choose{2}$
And, 6545 ways of picking 3 out of 35 non defective. ${35}\choose{3}$
... all the way to case 5 where there is only one selection.
EASY WAY.
Complement trick. There are ${40}\choose{5}$ = 658008 cases.
But only 324632 = ${35}\choose{5}$ are non-defective selection.
So, do the subtraction to get the same answer as above: ${40}\choose{5}$ -  ${35}\choose{5}$. The final answer. 
A: The probabability for drawing exactly $k$ defective discs is given by the hypergeometric distribution. To get the number of possible drawings, simply multiply this probability with the number of ways to draw 5 out of 40 discs, i.e. ${40}\choose{5}$ . Depending on whether you actually want exactly one (or several) dics, add up.
A: Let me try and put it in a way that might hopefully be clear and plain.
So, you have a pile of 40 discs, $35 G + 5 D$, mixed up.
You randomly select the first and have $35/40$ chances that it is good.
Suppose the first is good, then select a second, and you have $34/39$ chances that it is good also.
Continuing in that way, the probabilty that you select a "GGGGD" sequence is $p = \frac{{35}}{{40}}\frac{{34}}{{39}}\frac{{33}}{{38}}\frac{{32}}{{37}}\frac{5}{{36}}$.
The probability to select a "GGGDG" sequence would instead be $p = \frac{{35}}{{40}}\frac{{34}}{{39}}\frac{{33}}{{38}}\frac{{5}}{{37}}\frac{32}{{36}}$.
You see that the total numerator and denominator does not change, therefore the probability to get any sequence with $4G+1D$ will be:
$$
\begin{gathered}
  p = 5\frac{{35}}
{{40}}\frac{{34}}
{{39}}\frac{{33}}
{{38}}\frac{{32}}
{{37}}\frac{5}
{{36}} = 5\,\frac{{35 \cdot 34 \cdot 33 \cdot 32}}
{{40 \cdot 39 \cdot 38 \cdot 37 \cdot 36}}\,\frac{5}
{1} =  \hfill \\
   = 5\,\frac{{\frac{{35 \cdot 34 \cdot 33 \cdot 32}}
{{5!}}}}
{{\frac{{40 \cdot 39 \cdot 38 \cdot 37 \cdot 36}}
{{5!}}}}\,\frac{5}
{1} = \,\frac{{\frac{{35 \cdot 34 \cdot 33 \cdot 32}}
{{4!}}}}
{{\frac{{40 \cdot 39 \cdot 38 \cdot 37 \cdot 36}}
{{5!}}}}\,\frac{5}
{1} =  \hfill \\
   = \left( \begin{gathered}
  35 \\ 
  4 \\ 
\end{gathered}  \right)\left( \begin{gathered}
  5 \\ 
  1 \\ 
\end{gathered}  \right)\mathop /\nolimits_{} \left( \begin{gathered}
  40 \\ 
  5 \\ 
\end{gathered}  \right) \hfill \\ 
\end{gathered} 
$$
Here the final binomial is the number of ways to choose $5$ discs out of $40$.
So the numerator is the number of ways to choose $4G+1D$  (irrespective of the order of extraction) out of the
pile of $40$ mixed discs.
That comes to be the same as if the total pile was separated in two piles, $35G$ and $5D$, and we choose $4$ from the first
and $1$ from the second. Which is what, possibly, raises doubts to you , and not to you only.
In fact, in the general case the situation is a bit more involved, and interesting.
Suppose you have two piles, one with $n$ good discs over a total of $N$, the other with $m$ good over $M$.
If now you make a random choice between the piles first, according to their weight on the total, 
i.e. with probability $\frac{N}{{N + M}}$ and $\frac{M}{{N + M}}$, and then you make a random choice
within the selected pile, the probability to select, e.g. a good disc, will be
$$
\frac{N}
{{N + M}}\frac{n}
{N} + \frac{M}
{{N + M}}\frac{m}
{M} = \frac{{n + m}}
{{N + M}}
$$
that is: nothing changes with respect to mixing up all the content into a single pile.
This turns somehow intuitive if you imagine the discs line-up, separated into two trunks and
put let's say under a uniform hail, and select them according the order they are hit (but not ..damaged).
Applying this consideration to the two piles $[35;35]$ and $[0;5]$ should help and clear the doubts.
