# Five cards are dealt from a standard deck of 52 cards. How many different 5-card hands will contain exactly 3 kings and exactly 2 aces?

Feb 2, 2016

There are $10$ possible $5$-card hands with exactly $3$ kings and exactly $2$ aces.

#### Explanation:

As there are less aces than kings in our $5$-card hand, let's focus on those. I will write the first ace as $\textcolor{red}{\text{A}}$ and the second as $\textcolor{b l u e}{\text{A}}$, the kings will be $\text{K}$.

• Let's say that the first card is an ace.
In this case, there are $4$ different hands possible:

color(red)("A")color(blue)("A")"KKK", color(red)("A")"K"color(blue)("A")"KK", color(red)("A")"KK"color(blue)("A")"K" and color(red)("A")"KKK"color(blue)("A")""

• Now let's imagine that the second card is an ace. We don't need to consider the first card since this is already covered by the first case.
Thus, we only have $3$ different hands possible:

$\text{K"color(red)("A")color(blue)("A")"KK}$, $\text{K"color(red)("A")"K"color(blue)("A")"K}$ and "K"color(red)("A")"KK"color(blue)("A")

• Now let the third card be an ace.
This time we have only $2$ possible hands (the others we have already counted):

$\text{KK"color(red)("A")color(blue)("A")"K}$ and "KK"color(red)("A")"K"color(blue)("A")

• Last but not least, if the fourth card is an ace, we have just $1$ possible hand:

"KKK"color(red)("A")color(blue)("A")

Thus, in total, we have

$4 + 3 + 2 + 1 = 10$

possible $5$-card hands with $3$ kings and $2$ aces.

Feb 27, 2016

$24$ five-card hands contain exactly 3 kings and 2 aces.

#### Explanation:

There are 6 choices for the 2 Aces based on 4 suits in a standard deck: Clubs, Hearts, Diamonds, Spades.
$\left\{C + H , C + D , C + S , H + D , H + S , S + S\right\}$
In general this can be calculated as " "_4C_2 =(4!)/(2!(4-2)!) = 6

For each of these choices there are 4 choices for the 3 Kings (basically one choice for each suit not included).

This gives a combination of $6 \times 4 = 24$ possible hands.