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 ?