52-Deck 3 Cards Drawn Possible Combinations Question

Question

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

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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

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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!

Answer 1

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.

Answer 2

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 ${4\choose 1}{3\choose 1}{2\choose 1}=4!$

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

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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?