# Question 4b3a7

Sep 26, 2016

$4512$ ways if the order in which the cards are drawn does not matter.

$541440$ ways if the order in which the cards are drawn does matter.

#### Explanation:

For this problem, we will use the following:

• The multiplication principle, which states that if there are $m$ ways of doing one thing and $n$ ways of doing another, there are $m n$ ways of doing both.

• Choose notation, ((n),(k)) = (n!)/(k!(n-k)!), which calculates the number of ways of choosing a subset of size $k$ from a set of size $n$.

Depending on if we also are distinguishing between drawing the same cards in a different order, we may also use that the number of permutations of a set of $n$ objects is given by n!.

Proceeding, note that there are $\left(\begin{matrix}4 \\ 3\end{matrix}\right)$ ways of selecting three of the four aces to be in the hand, and then $\left(\begin{matrix}48 \\ 2\end{matrix}\right)$ ways of selecting two more cards from the forty eight remaining non-ace cards in the deck. Thus, by the multiplication principle, the total number of hands is given by

((4),(3))*((48),(2)) = (4!)/(3!1!) ((48!)/(2!46!))

$= 4 \cdot \frac{48 \cdot 47}{2}$

$= 4512$

If the order in which the cards are drawn does not matter, then we are done. If it does, then we note that for each of the above hands, there are 5! different orders (permutations) in which the cards can be drawn, meaning the total becomes

4512*5! = 541440#