Count the number of combinations of a pair if the deck is a pinochle deck.

Question

the problem goes like this,

A pinochle deck consists of 48 cards. There are six denominations (or values) of the cards: 9, 10, Jack (J), Queen (Q), King (K) and Ace (A). There are two of each denomination for each of the standard suits, ♣,♦,♥,♠. So there are two 9 of ♠s, two Ace of ♣s, etc. A “pair” of cards is two cards with the same denomination. Examples include 9♦ 9♣, Q♠ Q♣ and K♣K♣. A pair is “clean” if the two suits are different. Similarly, a “trip” (triple) is three cards of the same denomination, and a “clean trip” has all three of the cards with different suits. There are also “quads” (four cards), “quints” (five cards) and “hexs” (6 cards of the same denomination). You are dealt 6 cards. For each of the following, indicate how many ways that hand can be dealt. Leave your answers as a combinatorial expression, and explain how you got your answers. Like poker, a hand has a “pair” when it has exactly two cards of the same denomination, no other cards of that denomination, and no other cards forming a pair or a larger group. Unlike poker, straights, flushes, etc. do not count for anything.

I'm currently working on this question and will edit continuously with all my answers, i only want feedback on questions I've attempted so far, thank you.

1. Exactly one pair. Examples include{9♠,9♣,J♣,Q♦,K♥,A♥}and{9♠,9♠,10♦,J♠,K♥,A♣}.

2. Exactly one clean pair. The first example above is clean, the second isn’t.

3. Exactly one pair and one quad.

4. Exactly one clean pair and one clean quad.
5. Exactly one hex.
6. Exactly one clean hex.
7. Exactly one pair and one clean pair. The pairs can not be the same, that would be a quad.
8. Exactly one pair and one clean trip. The pair and trip can not be the same, that would be a quint.
9. Nothing. No pairs, no trips, etc.

Here are my Answers

1) I got C(6/1)*C(8/2)*C(5/4)*8^4

2) C(6/1) *C(4/1)C(3/1) C(5/4)*8^4

3) C(6/1) *C(8/2) * C(5/1) * C(8/4)

4)

5)

6)

7)

8)

9)

Answer 1

For 3, we:

Choose the value for the quad, then the pair $\binom{6}{1}\binom{5}{1}=30$.

Choosing two cards out of 8, there are $\binom{8}{2}=28$ possibilities, and 4 of them are un-clean pairs, so we have 24.

To have a clean quad, we just need to choose one from each suit four times, so $2^4=16.$

Multiplying these together, we have 11520.

For 4, exactly one hex, we have:

Choose one value: $\binom{6}{1}=6$.

Choose 6 cards out of the 8 of the value: $\binom{8}{6}=28$.

Multiplying the two, we have 168.

For 5, it is impossible to have 6 of the same value cards of different suits, because there are only 4 suits, so 0.

For 6 and 7, I'm not sure if you mean one non-clean pair and one clean pair, or one any pair and one clean pair, so I won't answer now.

For 8, we just choose one from each value, as otherwise, we would have at least a pair. Thus, we have $8^6=262144$.

Hope this helps.

Answer 2

I got for 2) (6C1) * (8C1) * (6C1) * (5C4) * (8C1)^4. For the (8C1) * (6C1), you begin with choosing 1 suit from a possible 8. By definition of clean, you want a different suit, so you can pick 1 more suit from a possible 6 that are left available. So you multiply (6C1). The rest of the expression I got the same.