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