Problem Solving

If three cards are chosen at random from a standard deck of playing cards, how many different ways are there to draw the three cards if at least 2 cards are a jack, queen or king
[spoiler]
OA: 2925
IMO:2860[/spoiler]

Can you explain your logic..Plz...

i am getting an answer nowhere near what you are getting! Could you pls check if you posted the question correctly? The words, meaning, etc.

I agree with 2860, because;

12C2 * 40C1 + 12C3*40C0

12C2 * 40C1 = 66*40 = 2640 + 220 = 2860. Maybe, there is a trick??

By the way 2860 is the right answer. I looked it up on another website;

https://www.omegamath.net/Data/hwkd2.2.html

what is the logic behind adding 12C3*40C0 ?

It works like this.

We have total of 52 cards in a deck of playing cards which contains 4 jack, 4 king and 4 queen.

From the question we need to pick 3 cards such that atleast two cards are jack , queen or a king.

This can be split into 2 parts. First part where only two cards are jack,queen or king and second part where all three cards are jack, queen or a king.

First part can be solved as below:

Adding jack, queen and king cards we get 12 i.e 4+4+4

i.e 12C2 since condition says atleast 2 cards are jack, queen or king.

Now we have remaining 40 cards from which we need to select 1 so 40c1

So it is 12C2*+40C1=2640...1

Similarly Second part in which we need to select all three cards are jack, queen or king

so 12C3 *40C0=220......2

Both 1 and 2 adds to 2860..

Hope logic is clear. Please correct me if I'm wrong.

Thanks,
Shobha