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I have a HW problem I'm trying to pin down and I think I'm confusing myself...

Question: In a card game w/ a standard 52 card deck, a hand is a set of 3 cards. Count the # of hands that are...

a) Jack,Queen,King of 3 different suits

b) 3 Aces of different suits

c) Jack,Queen,King of 2 different suits

My thoughts...

a) $ {13 \choose 3}{4 \choose 3}$ 1st is for the J,Q,K and 2nd for the suits?

b)${4 \choose 3}$ 4 possible aces, choosing 3?

c) ${13 \choose 3}{4 \choose 2}$ 1st is for the J,Q,K and 2nd for the suits?

I've been staring at it for too long to make anymore sense of it. Anyone know the answer and maybe some guidance? Thanks!

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  • $\begingroup$ a) should be $\binom{4}{3}\cdot3!$ $\endgroup$
    – barak manos
    Commented Nov 12, 2014 at 7:37

2 Answers 2

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A few observations.

You did b) correctly.

The number ${13 \choose 3}$ would be appropriate for the number of ways you could pick three diamonds, drawing from just the diamonds. So, that doesn't belong in a) or c).

For a), you need three different suits. Pick the suit for the jack ($4$ choices), then pick the suit for the queen ($3$ choices), and finally pick the suit for the king ($2$ choices). $24$ possibilities.

For c), you need two different suits, so there will be one card of one suit, and two of another suit. So, first pick the card that is the single suit ($3$ choices) and then pick the suit for that card ($4$ choices). Then, pick the suit for the remaining two cards ($3$ choices). $36$ possibilities.

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  • $\begingroup$ Thanks a lot! As a follow up to my progress... if the question was J,Q,K all same suit. I would just do ${4 \choose 1}$ because you have 4 ways to choose the suit on the 1st card and after that each other card only has 1 possibility right? I know this is a simple example you could just count 4, but is the logic right? $\endgroup$
    – John
    Commented Nov 12, 2014 at 16:38
  • $\begingroup$ Exactly correct. $\endgroup$
    – John
    Commented Nov 12, 2014 at 18:44
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For (a) you don't have to count ways to select the faces, as there is just 1 way: Pick Jack, Queen, King.

You do have to count ways to select the suits for each face so they are all different. So select 1 of 4 suits for Jack, 1 of 3 for the Queen, and 1 of 2 for the King.

That's just ${4\choose 1}{3\choose 1}{2\choose 1}=4!$

For (b), that answer is good. $\checkmark$ There are $4$ ways to select the suit the three aces don't have.

For (c), pick the faces: Jack, Queen, King. Now select which two faces need the same suit, select that suit for them, and select a different suit for the third card.

Can you do this now?

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