Question

10) Five cards are dealt from a standard deck of 52 cards. How many hands are possible that contain three 3's and two Kings (a full house)?

Answer 1

Answer:

24

Explanation:

In a standard deck, there are 4 suits. Therefore, there are:

• 4 cards numbered 3

,

• 4 Kings.

The number of ways we can select 3 out of the 4 cards labeled 3 is:

$^4C_3$

The number of ways we can select 2 Kings out of the 4 Kings is:

$^4C_2$

Therefore, the number of possible hands are:

$\begin{gathered} \begin{aligned} & ^4C_3\times^4C_2=\frac{4 !}{(3 !)(1 !)}\frac{4 !}{(2 !)(2 !)}=\frac{4 \times3 !}{(3 !)(1 !)}\frac{4 \times3 \times2 !}{(2 !)(2 !)} \\ & =4\times\frac{4\times3}{2}\end{aligned} \\ =24 \end{gathered}$

There are 24 possible hands.