Struggling with these Combination Questions

Question

So far combination has been the hardest chapter we've covered in my Data Management class. I somewhat struggled with permutation but at least I knew how to set the questions up. However, when it comes to combinations I'm completely lost. I'm going through my review sheet and I'm stuck on a couple of questions.

1.Given a deck of 52 cards, in how many ways can you select 4 cards such that

A) There is at least 1 Ace and at least 2 Kings

I couldn't get the answer for this so I looked at the answer key and I don't understand why the bolded part was included [Ace card-$\binom{4}{1}$] [King card $\binom{4}{2}$] [Other-$\binom{44}{1}$]

I know that we need to have a total of four cards, hence why we had to include the third one, but what I'm struggling with is how would I know that?

Edit: Reading the comments below I found out the answer for this question is incorrect. I will double check with my teacher and re-edit my post once I get the correct question.

B) All the cards are of different ranks

The answer key says [Ranks-$\binom{13}{4}$] [Suits-$\binom{4}{1}$ $\binom{4}{1}$ $\binom{4}{1}$ $\binom{4}{1}$]. I get where $\binom{13}{4}$ comes from, but what I don't get is why were suits included? The question asks us to select ranks, not to select ranks and suits.

2.In how many ways can you distribute cards from a deck of 52 cards so that

A) Alan receives 5 cards, Barbara receives 6 cards, Carl receives 7 cards, and Diana receives 8 cards.

Answer: $\binom{52}{5}$ $\binom{47}{6}$ $\binom{41}{7}$ $\binom{34}{8}$

B) One of the players receives 5 cards, another player receives 6 cards, a 3rd player receives 7 cards, a 4th player receives 8 cards

Answer: $\binom{52}{5}\binom{47}{6}\binom{41}{7}\binom{34}{8} \times 4!$

Why was the answer for part B multiplied by 4!? What's the difference between these two questions?

Answer 1

1.Given a deck of 52 cards, in how many ways can you select 4 cards such that
A) There is at least 1 Ace and at least 2 Kings

"at least N" means you can have N ... OR MORE: 1 or 2 A's, 2 or 3 K's.
You can have: AKKK, AKKA, AKKx. Order is not important (x = any other rank).

AKKK: 4 ways to choose left out king, 4 ways to choose A = 16
AKKA: C(4,2) for the two A's, C(4,2) for the two K's = 6, and 6 * 6 = 36
AKKx: 4 ways to choose A, C(4,2) for the two K's, 44 for the X = 4 * 6 * 44
Grand total = 16 + 36 + 24*44 = 1108.

You need the 44 because in the last case, the last card can be any of the 44 non-A, non-K cards.

The given answer is for EXACTLY 1 A and 2 K's - not AT LEAST.

B) All the cards are of different ranks
The answer key says [Ranks-C(13,4) ] [Suits-C(4,1)*C(4,1)*C(4,1)*C(4,1)]. I get where C(13,4) comes from, but what I don't get is why we had to include the four suits. The answer asks us to select four cards, but we have 8 cards now (four from the ranks and 1 card from each suit).

Let's translate the expression into English.
Ranks: Choose 4 ranks for the cards: C(13, 4) ways to do that
Suits: Now, choose the suit for EACH OF THE FOUR CARDS: 4 ways for each one, multiplied together.

Answer 2

Hints for $2$:

1) In (a), we know the exact players who got a distinct amount of cards. However, in (b), each of the players could get $5, 6, 7 \text{ or } 8$ cards, and since there are four people, how do we account for all combinations?

2) How many distinct (non-repeating) ways can we arrange the numbers $\{{5, 6, 7, 8}\}$?