# Poker hands combination

In a poker game where we have a standard deck ($$52$$ cards) and each player is dealt $$5$$ cards, how many hands of each of the following types are possible ?

(Type 1, "Low face") one face card and $$4$$ other cards less than $$6$$.

(Type 2, "Wedding") Exactly two face cards and exactly one diamond.

Here is my approach for Type 1

we have $$3$$ different face cards for each color, namely (King,Queen and jack) and since we have $$4$$ different colors, we have $$12$$ different face cards in the deck.

Now we are choosing one of those $$12$$ cards so we have $$12\choose 1$$ but also for the cards that are less than $$6$$ we have $$5,4,3,2$$

I don't include the Ace here because the Ace is bigger than $$6$$ right ? Now again we have those 4 cards in each color, so we have $$16$$ cards less than $$6$$ in total and we choose $$4$$ cards of them so we have $$16\choose4$$ and so the answer should be $${12\choose1} \times {16\choose 4}$$

is that correct ?

For Type2,

we have exactly two face cards and so we have $$12\choose2$$ and we have exactly one diamond to choose, However, we have to subtract all the face cards diamond, so we have $$13-3 = 10$$ diamonds card to choose and so we have $$10\choose1$$, now we have $$2$$ other cards, but non of them must be a face nor a diamond so we have $$52-12-10 = 30$$ remaining cards to choose from, which is $$30\choose2$$

And so my answer would be $${12\choose2} \times {10\choose1} \times {30\choose2}$$

I just want to make sure I am on the right path here.

• both asnwers look good
– WW1
Sep 18 '15 at 20:14
• for type 2: does a face card that's diamond qualify?
– user271754
Sep 18 '15 at 20:30
• @DannyC. hmmm, so what Can I do about that ? Sep 18 '15 at 23:42
• Try set up two cases: case one = one of the face card is diamond; case two = the diamond is not a face card.
– user271754
Sep 19 '15 at 1:54
• So for case 1 we would have ${9 \choose 1} \times {30 \choose 3}$ and for case 2 we would have ${12 \choose 2} \times {10 \choose 1} \times {30 \choose 3}$ and then we add those two cases ? @DannyC. Sep 19 '15 at 5:01

Taking ace as a high card,

Type 1 ans is ok.

For Type 2,

either 1 non-diamond face card, a diamond face card , and 3 non-diamond non-face

or

2 non-diamond face cards, 1 diamond non-face card, and 2 non-diamond non-face

$$= {9\choose1}{3\choose 1}{30\choose3} + {9\choose2}{10\choose1}{30\choose2}$$

• For case 1, you already have a non-diamond face, why are u still choosing one face card of the remanining 3 ? $3 \choose 1$ ? case 2 looks right for me Sep 19 '15 at 15:55
• Because you get a required diamond through a face card. Sep 19 '15 at 16:54