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.

  • $\begingroup$ both asnwers look good $\endgroup$
    – WW1
    Sep 18 '15 at 20:14
  • 2
    $\begingroup$ for type 2: does a face card that's diamond qualify? $\endgroup$
    – user271754
    Sep 18 '15 at 20:30
  • $\begingroup$ @DannyC. hmmm, so what Can I do about that ? $\endgroup$
    – alkabary
    Sep 18 '15 at 23:42
  • $\begingroup$ Try set up two cases: case one = one of the face card is diamond; case two = the diamond is not a face card. $\endgroup$
    – user271754
    Sep 19 '15 at 1:54
  • $\begingroup$ 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. $\endgroup$
    – alkabary
    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


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

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

  • $\begingroup$ 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 $\endgroup$
    – alkabary
    Sep 19 '15 at 15:55
  • $\begingroup$ Because you get a required diamond through a face card. $\endgroup$ Sep 19 '15 at 16:54

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