I have a question about combination and why some approaches don't work Let's say Computer keyboard failures can be attributed to electrical defects or mechanical defects. A repair facility currently has 25 failed keyboards, 6 of which have electrical defects and 19 of which have mechanical defects.
The question asks you If a sample of 5 keyboards is randomly selected, what is the probability that at least 4 of these will have a mechanical defect? (Round your answer to four decimal places.)
I tried to solve this using the logic as such 
(19! / (4!15!)) * (21! / (1!20!)) which is 3876 * 21 but it isn't correct 
because it is larger than (25! / (5!20!))
I looked at the solution and it's the union of (19! / (4!15!)) * (6! / (1!5!)) + (19! / (5!14!))
I understand the logic where you basically add the probability of having only 4 mechanical defect and 5 mechanical defect but why does my method calculate a result that is completely incorrect?
 A: Your choices are not independent, so you cannot naively multiply to do the counting.
To get my head round things like this I often imagine labelling the keyboards. Imagine they are labelled $A,B,C,\dots,Y$ and that those with electrical defects are $\{A,B,C,D,E,F\}$ and that the others are the ones with mechanical defects.
Your logic seems to suppose that to pick a valid set of 5 keyboards (i.e. such that at least four have mechanical defects) you need to make two choices):
1. Choose 4 of the ones with mechanical defects.
2. Choose any keyboard from the remaining 21 keyboards.
I like to imagine a kind of table: Let's say that as you go down the columns, you change choice 1., and as you go along the rows, you change choice 2.. Each 5-set that you count is an entry in the table made from the union of a 4-set from choice 1. and a 1-set from choice 2.:
$$
\begin{matrix}
\{G,H,I,J\} \cup \{A\} & \{G,H,I,J\} \cup \{B\} & \{G,H,I,J\} \cup \{C\} & \cdots & \mathbf{\{G,H,I,J\} \cup \{K\}} & \cdots \\
\{G,H,I,K\} \cup \{A\} & \{G,H,I,K\} \cup \{B\} & \{G,H,I,K\} \cup \{C\} & \cdots & \mathbf{\{G,H,I,K\} \cup \{J\}} & \cdots\\
\{G,H,I,L\} \cup \{A\} & \{G,H,I,L\} \cup \{A\} & \{G,H,I,L\} \cup \{A\} & \cdots & \ddots & \vdots \\
\vdots & \vdots & \vdots
\end{matrix}
$$
It should be clear now that you can now end up counting $\{G,H,I,J\} \cup \{K\}$ and then, separately, taking $\{G,H,I,K\} \cup \{J\}$. 
I have counted the valid 5 set $\{G,H,I,J,K\}$ more than once.
Notice that there are 
$$
\text{No. of columns} =  \binom{21}{1} = \frac{21!}{1!20!}
$$
and
$$
\text{No. of rows} =  \binom{19}{4} = \frac{19!}{4!15!}
$$
So, if it were the case that every valid 5-set appeared exactly once in the matrix, then the total number of 5-sets would be as you computed.
