Probability: Card distribution problem

Question

1. A deck of 52 cards is equally dealt to 4 players.

Find the number of ways to distribute the cards so that each player has exactly one card from each rank.

[Note: A deck of 52 cards consists of 13 ranks, A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, each has four suits

2. A deck of 52 cards is equally dealt to 4 players.

Find the number of ways to distribute the cards so that the cards for each player are of the same colour.

[Note: in a deck of 52 cards, 26 of them are red and 26 of them are black]

Edit: I am so sorry, I am new to this environment and it is so ignorant of me to do that without sharing my thought.

For first question, what I can think now:

(4C1 × 3C1 × 2C1 × 1C1)^13 = (4!)^13 What I think is from each rank, first person take one, second person take one, and so on till last person. Then I power it to 13 since there are 13 ranks.

For second question, what I can think now:

(26C13 × 4C1 × 13C13 × 3C1 × 26C13 × 2C1 × 13C13 × 1C1) What I think is the deck divided to two groups, red and black. Then we distribute the red to the first person and second person, then black to tge rest of the group.

Answer 1

1. Distribution of the cards among four hands such that each receives one card from each rank.

(4C1 × 3C1 × 2C1 × 1C1)^13 = (4!)^13 What I think is from each rank, first person take one, second person take one, and so on till last person. Then I power it to 13 since there are 13 ranks.

Yes. The four suits from each of the thirteen ranks may be distributed in $4!$ ways, so the total count is $4!^{13}$ ways.

2. Distribute the cards among the four hands such that each is comprised of one from the two colours.

26C13 × 4C1 × 13C13 × 3C1 × 26C13 × 2C1 × 13C13 × 1C1

Almost. There are $^{4}\mathcal C_2$ ways to select two hands to receive black cards, $^{26}\mathcal C_{13}$ ways to distribute those twenty-six cards among those two hands, and likewise $^{26}\mathcal C_{13}$ ways to distribute the red cards among the remaining two hands.$${^{4}\mathcal C_{2}}{^{26}\mathcal C_{13}}{^{26}\mathcal C_{13}} =\dfrac{4!~26!^2}{2!^2~13!^4}$$

Modifying your method: Select a hand to receive the Ace of Spades and twelve other black cards, select a second hand to receive the remaining thirteen black cards, repeat with the Ace of Hearts and the remaining twenty-five red cards.

$${^{4}\mathcal C_{1}}{^{25}\mathcal C_{12}}{^{3}\mathcal C_{1}}{^{13}\mathcal C_{13}}~{^{2}\mathcal C_{1}}{^{25}\mathcal C_{12}}{^{1}\mathcal C_{1}}{^{13}\mathcal C_{13}}$$

Note: ${^{4}\mathcal C_{1}}{^{3}\mathcal C_{1}}{^{2}\mathcal C_{1}}= 4\times{^{4}\mathcal C_{2}}$ and $2\times{^{25}\mathcal C_{12}} = {^{26}\mathcal C_{13}}$