Question

Find the number of possible 5-card hands that contain the cards specified from a standard 52-card deck.

Answer 1

A. A "red" card has either the heart or diamond suit. There are 26 of these. If we don't care about the order in which the cards are drawn (which is the assumption I'm using throughout this answer), then there are

$\dbinom{26}5=\dfrac{26!}{5!(26-5)!}=65,780$

such hands. To be clearer about the fact that we don't want "black" cards, we could also have written

$\dbinom{26}5\dbinom{26}0$

That is, we want 5 red cards but 0 black cards in our hand; there are 26 of each color to choose from. But the second term reduces to 1 and makes no difference anyway.

b. Now we want 4 spades (there are 13 in the deck) and 1 non-spade (39 non-spades in the deck) so we can make

$\dbinom{13}4\dbinom{39}1=27,885$

such hands.

c. Now we want 3 face cards (there are 3 per suit, with 4 suits altogether, giving a total of 12 face cards in the deck) and 2 non-face cards (40 in total), so we have

$\dbinom{12}3\dbinom{40}2=171,600$

possible hands.

d. Now we want 2 aces (4 total in the deck) and 3 non-aces (48 remaining), so we have

$\dbinom42\dbinom{48}3=103,776$

e. Now, we want at most 1 diamond. So either there are no diamonds or exactly 1 diamond in the hand. There are

$\dbinom{13}0\dbinom{39}5+\dbinom{13}1\dbinom{39}4=1,645,020$

such possible hands.

f. Finally, we want at least 1 king in the hand, which means we can have between 1 and 4 kings. We can add up the possible hands like in part (e),

$\dbinom41\dbinom{48}4+\dbinom42\dbinom{48}3+\dbinom43\dbinom{48}2+\dbinom44\dbinom{48}1$

but computing that can get tedious very quickly.

Instead, let's consider the number of hands *not* containing any kings at all. There are

$\dbinom40\dbinom{48}5=1,712,304$

such possible hands.

With the event of getting at least 1 king in mind, we can see that every possible 5 card hand either has at least 1 king or it does not. These events are disjoint, so

$\text{total 5 card hands}=\text{total hands with at least 1 king}+\text{total hands without kings}$

So the number of at-least-1-king hands is

$\dbinom{52}5-\dbinom40\dbinom{48}5=886,656$