Chapter 6.1, Problem 15E

Solution

To find: the number of possible 5-card hands that contains the cards specified.

  $778,320\text{ hands}$

Given information:

The cards should be: 1 ace and 4 cards that are not aces

These cards are taken from a standard 52-card deck.

Formula Used:

Combinations of n Objects Taken r at a Time

The number of combinations of r objects taken from a group of n distinct objects is denoted by $C{}_r$ and is given by the formula:

  $C{}_r=\frac{n!}{\left(n-r\right)!\cdot r!}$

Some known values are:

  $\begin{array}{l}1!=1\\ 0!=1\end{array}$

Explanation:

It is known that there is 1 ace in each of the four suits (spades, clubs, hearts and diamonds).

So that's 4 ace cards out of 52 cards.

The number of ways to choose 1 ace card from these 4 ace cards is: $C{}_1$

Now, the remaining 4 cards are to be chosen such that none of them is ace. So, 4 cards must be chosen from any of the remaining $\left(52-4=48\right)$ cards.

The number of ways to choose 4 cards that are not aces is: $C{}_4$

Thus, the number of possible 5-card hands that contains 1 ace and 4 cards that are not aces is given by:

  $C{}_1\times C{}_4$

Use the above formula for finding combinations value:

  $\begin{array}{c}C{}_1\times C{}_4=\left[\frac{4!}{\left(4-1\right)!\cdot 1!}\right]\left[\frac{48!}{\left(48-4\right)!\cdot 4!}\right]\\ =\left[\frac{4!}{3!\cdot 1!}\right]\left[\frac{48!}{44!\cdot 4!}\right]\\ =\left[\frac{4\cdot \cancel{3!}}{\cancel{3!}\cdot 1}\right]\left[\frac{48\cdot 47\cdot 46\cdot 45\cdot \cancel{44!}}{\cancel{44!}\cdot \left(4\cdot 3\cdot 2\cdot 1\right)}\right]\\ =4\cdot 194,580\\ =778,320\text{ hands}\end{array}$