Need to clarify the "At-least Concept" in Combination. I managed to solve this question but I had some inquiries regarding the solution.


If two cards are chosen at random from a standard deck of playing cards, how many different ways are there to draw the two cards if at least one card is a jack, queen or a king?


Here is how I solved it:
Jack/Queen/King = $4 \times 3 = 12$ cards
Other Cards = $52-12 = 40$ cards
Now we can only pick two cards so:


*

*It can be either from the Jack/King/Queen so $\binom{12}{1}=12$ 

*Both cards can be from Jack/King/Queen so $\binom{12}{2} = 66$ 

*One card can be from the remaining stack (Non - jack,king or queen) so
$\binom{40}{1}=40$
Now the only problem I have with this question is when getting the final value initially I was doing
$1$ from Remaining Cards $\times$ [  ($1$ from Jack/King/Queen) + 2(from Jack/King/Queen) ]= $\binom{40}{1} \times (  \binom{12}{1} + \binom{12}{2} ) $
But The actual answer comes if we do the following
[$1$ from Remaining Cards $\times$  $1$ from Jack/King/Queen ] + 2(from Jack/King/Queen) ]= $( \binom{40}{1} \times   \binom{12}{1}) + \binom{12}{2}  $
I would really appreciate it if someone could clarify why do we do it the second way and not the first way ? Which part is added/multiplied to which part ? Is there an easier way to know how its done. Am I missing some important concept here ?
Edit : 
While trying to understand this I also looked up the definition of disjoint events which means "Two events are disjoint if they can't both happen at the same time" so then again
here is what I did (Special here means Jack/king/Queen)
(1 from the 40 and 1 from special) or (1 from the 40 and 2 from special)
$(\binom{40}{1} \times \binom{12}{1}) + (\binom{40}{1} \times \binom{12}{2}) ) $
Which simplifies to $\binom{40}{1} \times (  \binom{12}{1} + \binom{12}{2} ) $
So is the representation of the problem using 
(1 from the 40 and 1 from special) or (1 from the 40 and 2 from special) wrong ?
 A: If you expand your first answer, you have
$$
\binom{12}{1} \cdot \binom{40}{1} + \binom{12}{2} \cdot \binom{40}{1}.
$$
Notice in the term on the right, you are choosing a total of three cards, which does not count what you want.

A rule of thumb is to turn "at least" problems into several instances of "exactly".
For the problem at hand, "at least one jack/queen/king" translates into "either exactly one jack/queen/king or exactly two jack/queen/king".
The number of ways to get exactly one jack/queen/king and one other card is
$$
\binom{12}{1} \cdot \binom{40}{1}.
$$
We multiply here because you choose a card from the jack/queen/king pile and a card from the "other" pile.
The number of ways to get exactly two from jack/queen/king is
$$
\binom{12}{2}.
$$
Since these cases are disjoint (this is important), we simply add the results from the two cases. Thus, the number of ways to get at least one Jack, Queen, or King is
$$
\binom{12}{1} \cdot \binom{40}{1} + \binom{12}{2}.
$$
A: The easiest way to solve this problem is to note that subtracting the number of ways of selecting any two cards neither of which is a Jack, Queen, or King, from the number of ways of selecting any two cards, leaves with the number of ways to pick two cards of which at least one of them is a Jack, Queen, or King.
So,
$$\mbox{(Ways to select any 2 cards from the deck)} - \mbox{(Ways to select 2 cards where neither one is J/Q/K)} = \mbox{Ways to select 2 cards where at least one is J/Q/K}$$
$$\binom{52}{2} - \binom{40}{2} = 1326 - 780 = 546$$
