Card game, combinations and permutations (suits, aces, none, some spades, minimum spades)

Question

I do really need help with this question, and it might be a newby question but unfortunately I am unsure of the answer.

In this game each player gets 5 cards from a standard deck of 52 cards. How many sets of 5 cards are there where:

$\textbf{a) All 5 are the same suit?}$

$\textbf{b) 4 aces are part of the 5 cards?}$

$\textbf{c) none of the 5 cards have the same number?}$

$\textbf{d) Exactly 2 of the 5 cards are spades?}$

$\textbf{e) At least two of the 5 cards are spades?}$

I don't know if I am approaching this right although I did this $C(52, 3) \cdot C(52 - 5, 3) \cdot C(47 - 5, 3) \cdot C(42 - 5, 3) \cdot C(37 - 5, 3) \cdot C(32 - 5, 3) \cdot C(27 - 5, 3)*C(22 - 5, 3) \cdot C(17 - 5, 3) \cdot C(12 - 5, 3) \cdot C(7 - 5, 3)$

I did this because they are sets of cards, but I am probably misunderstanding the question?

b) $C(52+5-1,52-1)$

c) $C(52, 5)$

d) $C(5, 2)$

e) $P(5, 2)$

I don't know if the word sets is trowing me off or something, I am really confused and really would like to find the answer because I have an exam in 4 days on this and things were explained in a huge rush.

Answer 1

All are the same suit.

We need $5$ cards of the same the suit. How do we get that? First we choose a suit, there are $4$ to chose from. Then from the $13$ cards in the suit, we choose $5$ cards. So the answer by the multiplication principle is,

$$4 {13 \choose 5}$$

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4 aces are part of the five cards.

We need four aces then any other card. How do we do this. Out of the $4$ aces we choose all $4$. Then we chose any other card, there are $48$ other cards to choose from.

$$48 {4 \choose 4}$$

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Exactly two are spades.

We need two spades and three other cards. How do we get this? Choose $2$ spades from the $13$ in the deck. Then choose $3$ non-spades from the $52-13=39$ non-spades in the deck.

$${13 \choose 2} {39 \choose 3}$$

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At least two are spades.

We can get: two spades, three non-spades. Three spades, two non-spades. Four spades, one non-spade. Five spades, zero non-spades.

To get $k$ spades. Choose $k$ out of the $13$ in the deck. Then to get $5-k$ non-spades choose $5-k$ from the $39$ non-spades in the deck. We may get $k=2,3,4,5$

$$\sum_{k=2}^{5} {13 \choose k}{39 \choose 5-k}$$