Chapter 6.1, Problem 18E

Solution

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

  $2,023,203\text{ hands}$

Given information:

The cards should be: at least 1 spade

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:

The given case of at least 1 spade actually consists of 5 cases: exactly 1 spade or exactly 2 spades or exactly 3 spades or exactly 4 spades of exactly five spades.

Rather than calculating all these possibilities individually and adding them to find the required number of hands, it is better to calculate the number of possibilities of getting no spade and subtracting it from the total number of possibility of choosing 5 cards from the standard 52-card deck.

It is known that there are 13 cards of the four suits (spades, clubs, hearts and diamonds).

No spade:

It is known that there are 13 cards of the four suits (spades, clubs, hearts and diamonds).

In this case all the 5 cards are to be chosen such that none of them is spade. So, 5 cards must be chosen from any of the remaining $\left(52-13=39\right)$ cards.

The number of ways to choose 5 cards that are not spades is: $C{}_5$

Use the above formula for finding combinations value:

  $\begin{array}{c}C{}_5=\frac{39!}{\left(39-5\right)!\cdot 5!}\\ =\frac{39!}{34!\cdot 5!}\\ =\frac{39\cdot 38\cdot 37\cdot 36\cdot 35\cdot \cancel{34!}}{\cancel{34!}\cdot \left(5\cdot 4\cdot 3\cdot 2\cdot 1\right)}\\ =575,757\text{ hands}\end{array}$

Total number:

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

The number of ways to choose any 5 random from the 52-card deck is: $C{}_5$

Use the above formula for finding combinations value:

  $\begin{array}{c}C{}_5=\frac{52!}{\left(52-5\right)!\cdot 5!}\\ =\frac{52!}{47!\cdot 5!}\\ =\frac{52\cdot 51\cdot 50\cdot 49\cdot 48\cdot \cancel{47!}}{\cancel{47!}\cdot \left(5\cdot 4\cdot 3\cdot 2\cdot 1\right)}\\ =2,598,960\text{ hands}\end{array}$

So, total number of possible hands is:

  $\begin{array}{c}C{}_5-C{}_5=2598960-575757\\ =2,023,203\text{ hands}\end{array}$